Compute $sumlimits_{j = 0}^{m - 1} left(c_j + 1right)lnleft(c_j + 1right)$ where $c_j =...












9












$begingroup$


As part of solving:



begin{equation}
I_m = int_0^1 lnleft(1 + x^{2m}right):dx.
end{equation}
where $m in mathbb{N}$. I found an unresolved component that I'm unsure how to start:



begin{equation}
G_m = sum_{j = 0}^{m - 1} left(c_j + 1right)lnleft(c_j + 1right),
end{equation}



where $c_j = cosleft(frac{pi}{2m}left(1 + 2jright) right)$



I'm just looking for a starting point. Any tips would be greatly appreciated.



By the way, I was able to show (and this was part of the solution too) :



begin{equation}
sum_{j = 0}^{m - 1} c_j = 0
end{equation}



Edit: For those that may be interested. In collaboration with clathratus, we found that for $n > 1$



begin{equation}
int_{0}^{1} frac{1}{t^n + 1}:dt = frac{1}{n}left[frac{pi}{sinleft(frac{pi}{n} right)}- Bleft(1 - frac{1}{n}, frac{1}{n}, frac{1}{2}right)right]
end{equation}



Or for any positive upper bound $x$:
begin{align}
I_n(x) &= int_{0}^{x} frac{1}{t^n + 1}:dt = frac{1}{n}left[Gammaleft(1 - frac{1}{n} right)Gammaleft(frac{1}{n} right)- Bleft(1 - frac{1}{n}, frac{1}{n}, frac{1}{x^n + 1}right)right]
end{align}



Here though, I was curious to investigate when $n$ was an even integer. This is my work:



Here we will consider $r = 2m$ where $m in mathbb{N}$. In doing so, we observe that the roots of the function are $m$ pairs of complex roots $(z, c(z))$ where $c(z)$ is the conjugate of $z$. To verify this:



begin{align}
x^{2m} + 1 = 0 rightarrow x^{2m} = e^{pi i}
end{align}



By De Moivre's formula, we observe that:



begin{align}
x = expleft({frac{pi + 2pi j}{2m} i} right) mbox{ for } j = 0dots 2m - 1,
end{align}



which we can express as the set



begin{align}
S &= Bigg{ expleft({frac{pi + 2pi cdot 0}{2m} i} right) , :expleft({frac{pi + 2pi cdot 1}{2m} i} right),dots,:expleft({frac{pi + 2pi cdot (2m - 2)}{2m} i} right)\
&qquad:expleft({frac{pi + 2pi cdot (2m - 1)}{2m} i} right)Bigg},
end{align}



which can be expressed as the set of $2$-tuples



begin{align}
S &= left{ left( expleft({frac{pi + 2pi j}{2m} i} right) , :expleft({frac{pi + 2pi(2m - 1 - j )}{2m} i} right)right): bigg|: j = 0 dots m - 1right}\
& = left{ (z_j, cleft(z_jright):|: j = 0 dots m - 1 right}
end{align}



From here, we can factor $x^{2m} + 1$ into the form



begin{align}
x^{2m} + 1 &= prod_{r in S} left(x + r_jright)left(x + c(r_j)right) \
&= prod_{i = 0}^{m - 1} left(x^2 + left(r_j + c(r_j)right)x + r_j c(r_j)right) \
&= prod_{i = 0}^{m - 1} left(x^2 + 2Releft(r_jright)x + left|r_j right|^2right)
end{align}



For our case here $left|r_j right|^2 = 1$ and $Releft(r_jright) = cosleft(frac{pi + 2pi j}{2m} right)= cosleft(frac{pi}{2m}left(1 + 2jright)right) = c_j$



begin{align}
int_0^1 logleft( x^{2m} + 1right):dx &= int_0^1 logleft(prod_{r in S} left(x^2 + 2c_jx+ left|r_j right|^2right)right)\
&= sum_{j = 0}^{m - 1} int_0^1 logleft(x^2 + 2c_jx + 1 right)\
&= sum_{j = 0}^{m - 1} left[2sqrt{1 - c_j^2}arctanleft(frac{x + c_j}{sqrt{1 - c_j^2}}right) + left(x + c_jright)logleft(x^2 + 2c_jx + 1right) - 2x right]_0^1 \
&= sum_{j = 0}^{m - 1} left[ 2sqrt{1 - c_j^2}arctanleft(sqrt{frac{1 - c_j}{1 + c_j}} right) + log(2)c_j + left(log(2) - 2right) + left(c_j + 1right)logleft(c_j + 1right) right] \
&= 2sum_{j = 0}^{m - 1}sqrt{1 - c_j^2}arctanleft(sqrt{frac{1 - c_j}{1 + c_j}} right) + log(2)sum_{j = 0}^{m - 1} c_j + mleft(log(2) - 2right)\
&qquad+ sum_{j = 0}^{m - 1}left(c_j + 1right)logleft(c_j + 1right)
end{align}



Thus,



begin{align}
int_0^1 logleft( x^{2m} + 1right):dx &=sum_{j = 0}^{m - 1}c_jsinleft(frac{pi}{2m}left(1 + 2jright)right) + log(2)sum_{j = 0}^{m - 1} c_j + mleft(log(2) - 2right)\
&qquad+ sum_{j = 0}^{m - 1}left(c_j + 1right)logleft(c_j + 1right)
end{align}






integration sequences-and-series definite-integrals logarithms trigonometric-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    @Mason - Yes, thanks for the pickup, I will edit now
    $endgroup$
    – user150203
    Dec 27 '18 at 4:46










  • $begingroup$
    And $G_n$ should be $G_m$? Sorry. I don't have a lot to add to this question except catching pedantic typos... Here's the closed forms for $I_0,I_1,I_2$. Does $I_m$ have a nice closed form? And is it easy to see how $I_m$ and $G_m$ are connected?
    $endgroup$
    – Mason
    Dec 27 '18 at 4:58












  • $begingroup$
    @Mason - Yes, sorry, I'm not on my game at the moment. I will correct. Thanks for the pickup. Have been trying to find the relationship through brute force examination (at present) and haven't been able to find anything.
    $endgroup$
    – user150203
    Dec 27 '18 at 5:02






  • 1




    $begingroup$
    Thanks for the shout-out David ;) I'll try to answer the question
    $endgroup$
    – clathratus
    Dec 27 '18 at 9:08






  • 1




    $begingroup$
    @clathratus - Thank you for posting the question and your work on it. I've been fascinated with the question since your initial solution (the same that Claude) came to.
    $endgroup$
    – user150203
    Dec 27 '18 at 9:13
















9












$begingroup$


As part of solving:



begin{equation}
I_m = int_0^1 lnleft(1 + x^{2m}right):dx.
end{equation}
where $m in mathbb{N}$. I found an unresolved component that I'm unsure how to start:



begin{equation}
G_m = sum_{j = 0}^{m - 1} left(c_j + 1right)lnleft(c_j + 1right),
end{equation}



where $c_j = cosleft(frac{pi}{2m}left(1 + 2jright) right)$



I'm just looking for a starting point. Any tips would be greatly appreciated.



By the way, I was able to show (and this was part of the solution too) :



begin{equation}
sum_{j = 0}^{m - 1} c_j = 0
end{equation}



Edit: For those that may be interested. In collaboration with clathratus, we found that for $n > 1$



begin{equation}
int_{0}^{1} frac{1}{t^n + 1}:dt = frac{1}{n}left[frac{pi}{sinleft(frac{pi}{n} right)}- Bleft(1 - frac{1}{n}, frac{1}{n}, frac{1}{2}right)right]
end{equation}



Or for any positive upper bound $x$:
begin{align}
I_n(x) &= int_{0}^{x} frac{1}{t^n + 1}:dt = frac{1}{n}left[Gammaleft(1 - frac{1}{n} right)Gammaleft(frac{1}{n} right)- Bleft(1 - frac{1}{n}, frac{1}{n}, frac{1}{x^n + 1}right)right]
end{align}



Here though, I was curious to investigate when $n$ was an even integer. This is my work:



Here we will consider $r = 2m$ where $m in mathbb{N}$. In doing so, we observe that the roots of the function are $m$ pairs of complex roots $(z, c(z))$ where $c(z)$ is the conjugate of $z$. To verify this:



begin{align}
x^{2m} + 1 = 0 rightarrow x^{2m} = e^{pi i}
end{align}



By De Moivre's formula, we observe that:



begin{align}
x = expleft({frac{pi + 2pi j}{2m} i} right) mbox{ for } j = 0dots 2m - 1,
end{align}



which we can express as the set



begin{align}
S &= Bigg{ expleft({frac{pi + 2pi cdot 0}{2m} i} right) , :expleft({frac{pi + 2pi cdot 1}{2m} i} right),dots,:expleft({frac{pi + 2pi cdot (2m - 2)}{2m} i} right)\
&qquad:expleft({frac{pi + 2pi cdot (2m - 1)}{2m} i} right)Bigg},
end{align}



which can be expressed as the set of $2$-tuples



begin{align}
S &= left{ left( expleft({frac{pi + 2pi j}{2m} i} right) , :expleft({frac{pi + 2pi(2m - 1 - j )}{2m} i} right)right): bigg|: j = 0 dots m - 1right}\
& = left{ (z_j, cleft(z_jright):|: j = 0 dots m - 1 right}
end{align}



From here, we can factor $x^{2m} + 1$ into the form



begin{align}
x^{2m} + 1 &= prod_{r in S} left(x + r_jright)left(x + c(r_j)right) \
&= prod_{i = 0}^{m - 1} left(x^2 + left(r_j + c(r_j)right)x + r_j c(r_j)right) \
&= prod_{i = 0}^{m - 1} left(x^2 + 2Releft(r_jright)x + left|r_j right|^2right)
end{align}



For our case here $left|r_j right|^2 = 1$ and $Releft(r_jright) = cosleft(frac{pi + 2pi j}{2m} right)= cosleft(frac{pi}{2m}left(1 + 2jright)right) = c_j$



begin{align}
int_0^1 logleft( x^{2m} + 1right):dx &= int_0^1 logleft(prod_{r in S} left(x^2 + 2c_jx+ left|r_j right|^2right)right)\
&= sum_{j = 0}^{m - 1} int_0^1 logleft(x^2 + 2c_jx + 1 right)\
&= sum_{j = 0}^{m - 1} left[2sqrt{1 - c_j^2}arctanleft(frac{x + c_j}{sqrt{1 - c_j^2}}right) + left(x + c_jright)logleft(x^2 + 2c_jx + 1right) - 2x right]_0^1 \
&= sum_{j = 0}^{m - 1} left[ 2sqrt{1 - c_j^2}arctanleft(sqrt{frac{1 - c_j}{1 + c_j}} right) + log(2)c_j + left(log(2) - 2right) + left(c_j + 1right)logleft(c_j + 1right) right] \
&= 2sum_{j = 0}^{m - 1}sqrt{1 - c_j^2}arctanleft(sqrt{frac{1 - c_j}{1 + c_j}} right) + log(2)sum_{j = 0}^{m - 1} c_j + mleft(log(2) - 2right)\
&qquad+ sum_{j = 0}^{m - 1}left(c_j + 1right)logleft(c_j + 1right)
end{align}



Thus,



begin{align}
int_0^1 logleft( x^{2m} + 1right):dx &=sum_{j = 0}^{m - 1}c_jsinleft(frac{pi}{2m}left(1 + 2jright)right) + log(2)sum_{j = 0}^{m - 1} c_j + mleft(log(2) - 2right)\
&qquad+ sum_{j = 0}^{m - 1}left(c_j + 1right)logleft(c_j + 1right)
end{align}






integration sequences-and-series definite-integrals logarithms trigonometric-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    @Mason - Yes, thanks for the pickup, I will edit now
    $endgroup$
    – user150203
    Dec 27 '18 at 4:46










  • $begingroup$
    And $G_n$ should be $G_m$? Sorry. I don't have a lot to add to this question except catching pedantic typos... Here's the closed forms for $I_0,I_1,I_2$. Does $I_m$ have a nice closed form? And is it easy to see how $I_m$ and $G_m$ are connected?
    $endgroup$
    – Mason
    Dec 27 '18 at 4:58












  • $begingroup$
    @Mason - Yes, sorry, I'm not on my game at the moment. I will correct. Thanks for the pickup. Have been trying to find the relationship through brute force examination (at present) and haven't been able to find anything.
    $endgroup$
    – user150203
    Dec 27 '18 at 5:02






  • 1




    $begingroup$
    Thanks for the shout-out David ;) I'll try to answer the question
    $endgroup$
    – clathratus
    Dec 27 '18 at 9:08






  • 1




    $begingroup$
    @clathratus - Thank you for posting the question and your work on it. I've been fascinated with the question since your initial solution (the same that Claude) came to.
    $endgroup$
    – user150203
    Dec 27 '18 at 9:13














9












9








9


3



$begingroup$


As part of solving:



begin{equation}
I_m = int_0^1 lnleft(1 + x^{2m}right):dx.
end{equation}
where $m in mathbb{N}$. I found an unresolved component that I'm unsure how to start:



begin{equation}
G_m = sum_{j = 0}^{m - 1} left(c_j + 1right)lnleft(c_j + 1right),
end{equation}



where $c_j = cosleft(frac{pi}{2m}left(1 + 2jright) right)$



I'm just looking for a starting point. Any tips would be greatly appreciated.



By the way, I was able to show (and this was part of the solution too) :



begin{equation}
sum_{j = 0}^{m - 1} c_j = 0
end{equation}



Edit: For those that may be interested. In collaboration with clathratus, we found that for $n > 1$



begin{equation}
int_{0}^{1} frac{1}{t^n + 1}:dt = frac{1}{n}left[frac{pi}{sinleft(frac{pi}{n} right)}- Bleft(1 - frac{1}{n}, frac{1}{n}, frac{1}{2}right)right]
end{equation}



Or for any positive upper bound $x$:
begin{align}
I_n(x) &= int_{0}^{x} frac{1}{t^n + 1}:dt = frac{1}{n}left[Gammaleft(1 - frac{1}{n} right)Gammaleft(frac{1}{n} right)- Bleft(1 - frac{1}{n}, frac{1}{n}, frac{1}{x^n + 1}right)right]
end{align}



Here though, I was curious to investigate when $n$ was an even integer. This is my work:



Here we will consider $r = 2m$ where $m in mathbb{N}$. In doing so, we observe that the roots of the function are $m$ pairs of complex roots $(z, c(z))$ where $c(z)$ is the conjugate of $z$. To verify this:



begin{align}
x^{2m} + 1 = 0 rightarrow x^{2m} = e^{pi i}
end{align}



By De Moivre's formula, we observe that:



begin{align}
x = expleft({frac{pi + 2pi j}{2m} i} right) mbox{ for } j = 0dots 2m - 1,
end{align}



which we can express as the set



begin{align}
S &= Bigg{ expleft({frac{pi + 2pi cdot 0}{2m} i} right) , :expleft({frac{pi + 2pi cdot 1}{2m} i} right),dots,:expleft({frac{pi + 2pi cdot (2m - 2)}{2m} i} right)\
&qquad:expleft({frac{pi + 2pi cdot (2m - 1)}{2m} i} right)Bigg},
end{align}



which can be expressed as the set of $2$-tuples



begin{align}
S &= left{ left( expleft({frac{pi + 2pi j}{2m} i} right) , :expleft({frac{pi + 2pi(2m - 1 - j )}{2m} i} right)right): bigg|: j = 0 dots m - 1right}\
& = left{ (z_j, cleft(z_jright):|: j = 0 dots m - 1 right}
end{align}



From here, we can factor $x^{2m} + 1$ into the form



begin{align}
x^{2m} + 1 &= prod_{r in S} left(x + r_jright)left(x + c(r_j)right) \
&= prod_{i = 0}^{m - 1} left(x^2 + left(r_j + c(r_j)right)x + r_j c(r_j)right) \
&= prod_{i = 0}^{m - 1} left(x^2 + 2Releft(r_jright)x + left|r_j right|^2right)
end{align}



For our case here $left|r_j right|^2 = 1$ and $Releft(r_jright) = cosleft(frac{pi + 2pi j}{2m} right)= cosleft(frac{pi}{2m}left(1 + 2jright)right) = c_j$



begin{align}
int_0^1 logleft( x^{2m} + 1right):dx &= int_0^1 logleft(prod_{r in S} left(x^2 + 2c_jx+ left|r_j right|^2right)right)\
&= sum_{j = 0}^{m - 1} int_0^1 logleft(x^2 + 2c_jx + 1 right)\
&= sum_{j = 0}^{m - 1} left[2sqrt{1 - c_j^2}arctanleft(frac{x + c_j}{sqrt{1 - c_j^2}}right) + left(x + c_jright)logleft(x^2 + 2c_jx + 1right) - 2x right]_0^1 \
&= sum_{j = 0}^{m - 1} left[ 2sqrt{1 - c_j^2}arctanleft(sqrt{frac{1 - c_j}{1 + c_j}} right) + log(2)c_j + left(log(2) - 2right) + left(c_j + 1right)logleft(c_j + 1right) right] \
&= 2sum_{j = 0}^{m - 1}sqrt{1 - c_j^2}arctanleft(sqrt{frac{1 - c_j}{1 + c_j}} right) + log(2)sum_{j = 0}^{m - 1} c_j + mleft(log(2) - 2right)\
&qquad+ sum_{j = 0}^{m - 1}left(c_j + 1right)logleft(c_j + 1right)
end{align}



Thus,



begin{align}
int_0^1 logleft( x^{2m} + 1right):dx &=sum_{j = 0}^{m - 1}c_jsinleft(frac{pi}{2m}left(1 + 2jright)right) + log(2)sum_{j = 0}^{m - 1} c_j + mleft(log(2) - 2right)\
&qquad+ sum_{j = 0}^{m - 1}left(c_j + 1right)logleft(c_j + 1right)
end{align}






integration sequences-and-series definite-integrals logarithms trigonometric-series







share|cite|improve this question











$endgroup$




As part of solving:



begin{equation}
I_m = int_0^1 lnleft(1 + x^{2m}right):dx.
end{equation}
where $m in mathbb{N}$. I found an unresolved component that I'm unsure how to start:



begin{equation}
G_m = sum_{j = 0}^{m - 1} left(c_j + 1right)lnleft(c_j + 1right),
end{equation}



where $c_j = cosleft(frac{pi}{2m}left(1 + 2jright) right)$



I'm just looking for a starting point. Any tips would be greatly appreciated.



By the way, I was able to show (and this was part of the solution too) :



begin{equation}
sum_{j = 0}^{m - 1} c_j = 0
end{equation}



Edit: For those that may be interested. In collaboration with clathratus, we found that for $n > 1$



begin{equation}
int_{0}^{1} frac{1}{t^n + 1}:dt = frac{1}{n}left[frac{pi}{sinleft(frac{pi}{n} right)}- Bleft(1 - frac{1}{n}, frac{1}{n}, frac{1}{2}right)right]
end{equation}



Or for any positive upper bound $x$:
begin{align}
I_n(x) &= int_{0}^{x} frac{1}{t^n + 1}:dt = frac{1}{n}left[Gammaleft(1 - frac{1}{n} right)Gammaleft(frac{1}{n} right)- Bleft(1 - frac{1}{n}, frac{1}{n}, frac{1}{x^n + 1}right)right]
end{align}



Here though, I was curious to investigate when $n$ was an even integer. This is my work:



Here we will consider $r = 2m$ where $m in mathbb{N}$. In doing so, we observe that the roots of the function are $m$ pairs of complex roots $(z, c(z))$ where $c(z)$ is the conjugate of $z$. To verify this:



begin{align}
x^{2m} + 1 = 0 rightarrow x^{2m} = e^{pi i}
end{align}



By De Moivre's formula, we observe that:



begin{align}
x = expleft({frac{pi + 2pi j}{2m} i} right) mbox{ for } j = 0dots 2m - 1,
end{align}



which we can express as the set



begin{align}
S &= Bigg{ expleft({frac{pi + 2pi cdot 0}{2m} i} right) , :expleft({frac{pi + 2pi cdot 1}{2m} i} right),dots,:expleft({frac{pi + 2pi cdot (2m - 2)}{2m} i} right)\
&qquad:expleft({frac{pi + 2pi cdot (2m - 1)}{2m} i} right)Bigg},
end{align}



which can be expressed as the set of $2$-tuples



begin{align}
S &= left{ left( expleft({frac{pi + 2pi j}{2m} i} right) , :expleft({frac{pi + 2pi(2m - 1 - j )}{2m} i} right)right): bigg|: j = 0 dots m - 1right}\
& = left{ (z_j, cleft(z_jright):|: j = 0 dots m - 1 right}
end{align}



From here, we can factor $x^{2m} + 1$ into the form



begin{align}
x^{2m} + 1 &= prod_{r in S} left(x + r_jright)left(x + c(r_j)right) \
&= prod_{i = 0}^{m - 1} left(x^2 + left(r_j + c(r_j)right)x + r_j c(r_j)right) \
&= prod_{i = 0}^{m - 1} left(x^2 + 2Releft(r_jright)x + left|r_j right|^2right)
end{align}



For our case here $left|r_j right|^2 = 1$ and $Releft(r_jright) = cosleft(frac{pi + 2pi j}{2m} right)= cosleft(frac{pi}{2m}left(1 + 2jright)right) = c_j$



begin{align}
int_0^1 logleft( x^{2m} + 1right):dx &= int_0^1 logleft(prod_{r in S} left(x^2 + 2c_jx+ left|r_j right|^2right)right)\
&= sum_{j = 0}^{m - 1} int_0^1 logleft(x^2 + 2c_jx + 1 right)\
&= sum_{j = 0}^{m - 1} left[2sqrt{1 - c_j^2}arctanleft(frac{x + c_j}{sqrt{1 - c_j^2}}right) + left(x + c_jright)logleft(x^2 + 2c_jx + 1right) - 2x right]_0^1 \
&= sum_{j = 0}^{m - 1} left[ 2sqrt{1 - c_j^2}arctanleft(sqrt{frac{1 - c_j}{1 + c_j}} right) + log(2)c_j + left(log(2) - 2right) + left(c_j + 1right)logleft(c_j + 1right) right] \
&= 2sum_{j = 0}^{m - 1}sqrt{1 - c_j^2}arctanleft(sqrt{frac{1 - c_j}{1 + c_j}} right) + log(2)sum_{j = 0}^{m - 1} c_j + mleft(log(2) - 2right)\
&qquad+ sum_{j = 0}^{m - 1}left(c_j + 1right)logleft(c_j + 1right)
end{align}



Thus,



begin{align}
int_0^1 logleft( x^{2m} + 1right):dx &=sum_{j = 0}^{m - 1}c_jsinleft(frac{pi}{2m}left(1 + 2jright)right) + log(2)sum_{j = 0}^{m - 1} c_j + mleft(log(2) - 2right)\
&qquad+ sum_{j = 0}^{m - 1}left(c_j + 1right)logleft(c_j + 1right)
end{align}






integration sequences-and-series definite-integrals logarithms trigonometric-series




integration sequences-and-series definite-integrals logarithms trigonometric-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 29 '18 at 11:02









Did

249k23228466




249k23228466










asked Dec 27 '18 at 4:44







user150203



















  • $begingroup$
    @Mason - Yes, thanks for the pickup, I will edit now
    $endgroup$
    – user150203
    Dec 27 '18 at 4:46










  • $begingroup$
    And $G_n$ should be $G_m$? Sorry. I don't have a lot to add to this question except catching pedantic typos... Here's the closed forms for $I_0,I_1,I_2$. Does $I_m$ have a nice closed form? And is it easy to see how $I_m$ and $G_m$ are connected?
    $endgroup$
    – Mason
    Dec 27 '18 at 4:58












  • $begingroup$
    @Mason - Yes, sorry, I'm not on my game at the moment. I will correct. Thanks for the pickup. Have been trying to find the relationship through brute force examination (at present) and haven't been able to find anything.
    $endgroup$
    – user150203
    Dec 27 '18 at 5:02






  • 1




    $begingroup$
    Thanks for the shout-out David ;) I'll try to answer the question
    $endgroup$
    – clathratus
    Dec 27 '18 at 9:08






  • 1




    $begingroup$
    @clathratus - Thank you for posting the question and your work on it. I've been fascinated with the question since your initial solution (the same that Claude) came to.
    $endgroup$
    – user150203
    Dec 27 '18 at 9:13


















  • $begingroup$
    @Mason - Yes, thanks for the pickup, I will edit now
    $endgroup$
    – user150203
    Dec 27 '18 at 4:46










  • $begingroup$
    And $G_n$ should be $G_m$? Sorry. I don't have a lot to add to this question except catching pedantic typos... Here's the closed forms for $I_0,I_1,I_2$. Does $I_m$ have a nice closed form? And is it easy to see how $I_m$ and $G_m$ are connected?
    $endgroup$
    – Mason
    Dec 27 '18 at 4:58












  • $begingroup$
    @Mason - Yes, sorry, I'm not on my game at the moment. I will correct. Thanks for the pickup. Have been trying to find the relationship through brute force examination (at present) and haven't been able to find anything.
    $endgroup$
    – user150203
    Dec 27 '18 at 5:02






  • 1




    $begingroup$
    Thanks for the shout-out David ;) I'll try to answer the question
    $endgroup$
    – clathratus
    Dec 27 '18 at 9:08






  • 1




    $begingroup$
    @clathratus - Thank you for posting the question and your work on it. I've been fascinated with the question since your initial solution (the same that Claude) came to.
    $endgroup$
    – user150203
    Dec 27 '18 at 9:13
















$begingroup$
@Mason - Yes, thanks for the pickup, I will edit now
$endgroup$
– user150203
Dec 27 '18 at 4:46




$begingroup$
@Mason - Yes, thanks for the pickup, I will edit now
$endgroup$
– user150203
Dec 27 '18 at 4:46












$begingroup$
And $G_n$ should be $G_m$? Sorry. I don't have a lot to add to this question except catching pedantic typos... Here's the closed forms for $I_0,I_1,I_2$. Does $I_m$ have a nice closed form? And is it easy to see how $I_m$ and $G_m$ are connected?
$endgroup$
– Mason
Dec 27 '18 at 4:58






$begingroup$
And $G_n$ should be $G_m$? Sorry. I don't have a lot to add to this question except catching pedantic typos... Here's the closed forms for $I_0,I_1,I_2$. Does $I_m$ have a nice closed form? And is it easy to see how $I_m$ and $G_m$ are connected?
$endgroup$
– Mason
Dec 27 '18 at 4:58














$begingroup$
@Mason - Yes, sorry, I'm not on my game at the moment. I will correct. Thanks for the pickup. Have been trying to find the relationship through brute force examination (at present) and haven't been able to find anything.
$endgroup$
– user150203
Dec 27 '18 at 5:02




$begingroup$
@Mason - Yes, sorry, I'm not on my game at the moment. I will correct. Thanks for the pickup. Have been trying to find the relationship through brute force examination (at present) and haven't been able to find anything.
$endgroup$
– user150203
Dec 27 '18 at 5:02




1




1




$begingroup$
Thanks for the shout-out David ;) I'll try to answer the question
$endgroup$
– clathratus
Dec 27 '18 at 9:08




$begingroup$
Thanks for the shout-out David ;) I'll try to answer the question
$endgroup$
– clathratus
Dec 27 '18 at 9:08




1




1




$begingroup$
@clathratus - Thank you for posting the question and your work on it. I've been fascinated with the question since your initial solution (the same that Claude) came to.
$endgroup$
– user150203
Dec 27 '18 at 9:13




$begingroup$
@clathratus - Thank you for posting the question and your work on it. I've been fascinated with the question since your initial solution (the same that Claude) came to.
$endgroup$
– user150203
Dec 27 '18 at 9:13










4 Answers
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6












$begingroup$

This does not answer the question as asked in the post.



Consider
$$J_m=int log(1+x^{2m}),dx$$ One integration by parts gives
$$J_m=x log left(1+x^{2 m}right)-2mint frac{ x^{2 m}+1-1}{x^{2 m}+1},dx=x log left(1+x^{2 m}right)-2mx+2mint frac{dx}{x^{2 m}+1}$$ and
$$int frac{dx}{x^{2 m}+1}=x , _2F_1left(1,frac{1}{2 m};1+frac{1}{2 m};-x^{2 m}right)$$ where appears the Gaussian or ordinary hypergeometric function.



So
$$K_m=int_0^a log(1+x^{2m}),dx=a log left(1+a^{2 m}right)-2ma+2ma , _2F_1left(1,frac{1}{2 m};1+frac{1}{2 m};-a^{2 m}right)$$ and, if $a=1$,
$$I_m=int_0^1 log(1+x^{2m}),dx= log left(2right)-2m+2m , _2F_1left(1,frac{1}{2 m};1+frac{1}{2 m};-1right)$$ which can write
$$I_m=log (2)-Phi left(-1,1,1+frac{1}{2 m}right)$$
where appears the Lerch transcendent function.



Now, (this is something I never looked at), a few (ugly) expressions for $f_m=Phi left(-1,1,1+frac{1}{2 m}right)$ before any simplification
$$f(1)=frac{pi }{2}-2$$
$$f(2)=frac{1}{4} left(pi tan left(frac{pi }{8}right)+pi cot left(frac{pi
}{8}right)-4 sqrt{2} log left(sin left(frac{pi }{8}right)right)+4
sqrt{2} log left(cos left(frac{pi }{8}right)right)right)-4$$
$$f(3)=frac{2 left(pi -sqrt{3} log left(sqrt{3}-1right)+sqrt{3} log
left(1+sqrt{3}right)right)}{left(sqrt{3}-1right)
left(1+sqrt{3}right)}-6$$
$$f(4)=frac{1}{4} left(pi tan left(frac{pi }{16}right)+pi cot left(frac{pi
}{16}right)-8 sin left(frac{pi }{8}right) log left(sin left(frac{3
pi }{16}right)right)+8 cos left(frac{pi }{8}right) log left(cos
left(frac{pi }{16}right)right)-8 cos left(frac{pi }{8}right) log
left(sin left(frac{pi }{16}right)right)+8 sin left(frac{pi
}{8}right) log left(cos left(frac{3 pi }{16}right)right)right)-8$$






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  • $begingroup$
    Yes, my collaborator and I came to the same result (very nice closed form indeed). If I may ask - what are your thoughts on the special case I've discussed?
    $endgroup$
    – user150203
    Dec 27 '18 at 8:49










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    @DavidG. Your post is very interesting. I need to spend more time with it !
    $endgroup$
    – Claude Leibovici
    Dec 27 '18 at 8:54










  • $begingroup$
    Please do! I look forward to any comments you may have. Thanks again for your post.
    $endgroup$
    – user150203
    Dec 27 '18 at 9:48










  • $begingroup$
    Mathematica calculates the same result
    $endgroup$
    – stocha
    Mar 22 at 19:11



















4





+250







$begingroup$

$color{brown}{textbf{Preliminary notes.}}$




Calculation of the required sum looks more complex task than of the
issue integral.



Approach with calculations of the sum via the inttegral allows to get close form both of the integral and the sum.




Denote
$$I_n=intlimits_0^1log(1+x^n)mathrm dx.tag1$$
Easy to see that the issue integral is $I_{2m}.$



$color{brown}{textbf{Closed form of the integrals.}}$



The first step is the integration by parts:
$$I_n = xlog(1+x^n)biggr|_0^1-nint_0^1xdfrac{x^{n-1}}{1+x^n}mathrm dx,$$
$$I_n = log2 - n + nintlimits_0^1dfrac{mathrm dx}{1+x^n}.tag2$$
(see also Claude Leibovici).



Then, the substitution
$$x=e^{-t}tag3$$
gives
$$J_n=intlimits_0^1dfrac{mathrm dx}{1+x^n}=intlimits_0^inftydfrac{e^{-t},mathrm dt}{1+e^{-nt}} = sumlimits_{k=0}^infty(-1)^kintlimits_0^infty e^{-(kn+1)t},mathrm dt=sumlimits_{k=0}^inftydfrac{(-1)^k}{kn+1} ,$$
$$J_n = intlimits_0^1dfrac{mathrm dx}{1+x^n}= dfrac1{2n}left(psi_0left(dfrac{n+1}{2n}right) - psi_0left(dfrac{1}{2n}right)right),tag4$$
where $psi_0(x)$ is the polygamma function.



Using $(2),(4),$ easy to get the expression for the OP integral in the form of
$$boxed{I_{2m} = log2 - 2m +dfrac1{2}left(psi_0left(dfrac{2m+1}{4m}right) - psi_0left(dfrac{1}{4m}right)right).}tag5$$



$color{brown}{textbf{Testing of the solution.}}$



Solution $(4)$ can be expressed via the elementary functions for a lot of the given values of n.



Immediate calculations allow to check obtained expressions in the simple cases
begin{align}
&J_1 = intlimits_0^1dfrac{mathrm dx}{1+x} = log 2,\[4pt]
&J_2 = intlimits_0^1dfrac{mathrm dx}{1+x^2} = arctan 1 = dfracpi4,\[4pt]
&J_3 = intlimits_0^1dfrac{mathrm dx}{1+x^3}
= dfrac16intlimits_0^1left(dfrac{2}{1+x} - dfrac{2x-1}{1-x+x^2} + dfrac{3}{1-x+x^2}right),mathrm dx\
&= left(dfrac13log(1+x) - dfrac16 log(1-x+x^2) + dfrac1{sqrt3}arctandfrac{2x-1}{sqrt3}right)bigg|_0^1
= dfracpi{3sqrt3} + frac13 log2.\[4pt]
end{align}
More hard cases are
begin{align}
&J_4 = intlimits_0^1dfrac{mathrm dx}{1+x^4}
= dfrac{sqrt2}8left(logdfrac{x^2 + sqrt2 x + 1}{x^2 - sqrt2 x + 1}
- 2arctan(1-sqrt2x) + 2arctan(sqrt2x+1)right)bigg|_0^1\[4pt]
&= dfrac{sqrt2}8left(logdfrac{(x^2 + sqrt2 x + 1)^2}{1+x^4} + 2arctandfrac{xsqrt2}{1-x^2}right)bigg|_0^1
= dfrac{sqrt2}8(pi+log(3+2sqrt2)),\[4pt]
&J_4=dfrac18Bigg(dfrac12sqrt{dfrac{2-sqrt2}{2+sqrt2}}pi+dfrac12sqrt{dfrac{2+sqrt2}{2-sqrt2}}pi
-2sqrt2left(-log2+dfrac12log(2-sqrt2)right)\[4pt]
&+2sqrt2left(-log2+dfrac12log(2+sqrt2)right)Bigg)\[4pt]
&=dfrac18Bigg(dfracpi2left(tandfracpi8+cotdfracpi8right)
+sqrt2logdfrac{2+sqrt2}{2-sqrt2}Bigg)
= dfrac{sqrt2}8(pi+log(3+2sqrt2)),
end{align}
and $n>4.$



Numeric calculations of the issue integral $I_{2m}$ and its closed form $(5)$ also confirm the correctness of the closed form.



$color{brown}{textbf{The alternative approach.}}$



The alternative approach is considered in OP. Let us repeat it with some differences.



Polynomial factorization can be presented in the form of



begin{align}
&1+x^{2m} = prodlimits_{j=0}^{2m-1}{large left(x-e^{frac{2j+1}{2m}pi i}right)}
= prodlimits_{j=0}^{m-1}left(x^2-2xcosfrac{2j+1}{2m}pi+1right),
end{align}
so
begin{align}
&I_{2m} = intlimits_0^1 ln(1+x^{2m}),mathrm dx
= sumlimits_{j=0}^{m-1} T_j,quadtext{where}\[4pt]
&T_j = intlimits_0^1 lnleft(x^2-2xcosfrac{2j+1}{2m}pi+1right),mathrm dx,\[4pt]
&T_j = xlnleft(x^2-2xcosfrac{2j+1}{2m}pi+1right)bigg|_0^1
- 2intlimits_0^1 dfrac{xleft(x-cosfrac{2j+1}{2m}piright)}{x^2-2xcosfrac{2j+1}{2m}pi+1},mathrm dx\[4pt]
& = lnleft(2-2cosfrac{2j+1}{2m}piright)
- 2-2intlimits_0^1 dfrac{xcosfrac{2j+1}{2m}pi-1}{x^2-2xcosfrac{2j+1}{2m}pi+1},mathrm dx\[4pt]
& = lnleft(2-2cosfrac{2j+1}{2m}piright) - 2
- 2cosfrac{2j+1}{2m}pi intlimits_0^1 dfrac{x-cosfrac{2j+1}{2m}pi}{x^2-2xcosfrac{2j+1}{2m}pi+1},mathrm dx\[4pt]
& + 2intlimits_0^1 dfrac{sin^2frac{2j+1}{2m}pi}{left(x-cosfrac{2j+1}{2m}piright)^2+sin^2frac{2j+1}{2m}pi},mathrm dx\[4pt]
& = lnleft(2-2cosfrac{2j+1}{2m}piright) - 2
- cosfrac{2j+1}{2m}pi lnleft(x^2-2xcosfrac{2j+1}{2m}pi+1right)bigg|_0^1\[4pt]
& + 2sinfrac{2j+1}{2m}pi arctandfrac{x-cosfrac{2j+1}{2m}pi}{sinfrac{2j+1}{2m}pi}bigg|_0^1\[4pt]
& = lnleft(2-2cosfrac{2j+1}{2m}piright) - 2
- cosfrac{2j+1}{2m}pi lnleft(2-2cosfrac{2j+1}{2m}piright)\[4pt]
& + 2sinfrac{2j+1}{2m}pi
left(arctandfrac{1-cosfrac{2j+1}{2m}pi}{sinfrac{2j+1}{2m}pi}
+ arctandfrac{cosfrac{2j+1}{2m}pi}{sinfrac{2j+1}{2m}pi}right)\[4pt]
& = lnleft(2-2cosfrac{2j+1}{2m}piright) - 2
- cosfrac{2j+1}{2m}pi lnleft(2-2cosfrac{2j+1}{2m}piright)\[4pt]
& + 2sinfrac{2j+1}{2m}pi
arctandfrac{sinfrac{2j+1}{2m}pi}
{sin^2frac{2j+1}{2m}pi - left(1 - cosfrac{2j+1}{2m}piright)cosfrac{2j+1}{2m}pi}\[4pt]
& = left(1-cosfrac{2j+1}{2m}piright)lnleft(2-2cosfrac{2j+1}{2m}piright) - 2 + 2sinfrac{2j+1}{2m}piarctancotfrac{2j+1}{4m}pi,\[4pt]
&T_j = left(1-cosfrac{2j+1}{2m}piright)lnleft(2-2cosfrac{2j+1}{2m}piright) - 2 + dfrac{2(m-j)-1}{2m}pisinfrac{2j+1}{2m}pi.
end{align}
Taking in account that
begin{align}
&sumlimits_{j=0}^{m-1}cosfrac{2j+1}{2m}pi
= Re sumlimits_{j=0}^{m-1}e^{frac{2j+1}{2m}pi i}
= Re {large dfrac{1-e^{pi i}}{1-e^{fracpi{m}i}}e^{fracpi{2m}i}}
=Redfrac {i}{sinfracpi{2m}} = 0,\[4pt]
&sumlimits_{j=0}^{m-1}sinfrac{2j+1}{2m}pi
= Im sumlimits_{j=0}^{m-1}e^{frac{2j+1}{2m}pi i}
=Imdfrac{i}{sinfracpi{2m}} = dfrac1{sinfracpi{2m}},\[4pt]
&sumlimits_{j=0}^{m-1}frac{2j+1}{2m}pisinfrac{2j+1}{2m}pi
= dfracpi{2sinfracpi{2m}}
end{align}
(see also Wolfram Alpha calculations),



one can get
$$boxed{I_{2m} = m(ln2-2) + dfracpi2cscfracpi{2m}
+ sumlimits_{j=0}^{m-1}left(1pmcosfrac{2j+1}{2m}piright)
logleft(1pmcosfrac{2j+1}{2m}piright).}tag6$$



$color{brown}{textbf{Closed form for the sum.}}$



From $(5)-(6)$ should
$$color{green}{boxed{sumlimits_{j=0}^{m-1}left(1pm c_jright)
logleft(1pm c_jright)
= dfrac1{2}left(psi_0left(dfrac{2m+1}{4m}right) - psi_0left(dfrac{1}{4m}right)right) - (m-1)ln2 - dfracpi2cscfracpi{2m}
,}}tag7$$
where
$$c_j = cosfrac{2j+1}{2m}pi.$$



Numeric calculations confirm the expression $(6)$ for the both variants of the sign "$pm$".



Also, numeric calculations of the issue sum and its closed form $(7)$ confirm the correctness of the closed form for the both variants of the sign "$pm$".






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  • 1




    $begingroup$
    Very nice presentation, already before the alternate approach (+1).
    $endgroup$
    – Markus Scheuer
    Mar 28 at 21:46










  • $begingroup$
    @MarkusScheuer Thanks, you are welcome!
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    – Yuri Negometyanov
    Mar 28 at 22:16



















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I did it!



I actually have no idea whether or not this works, but this is how I did it.



$ninBbb N$



Define the sequence ${r_k^{(n)}}_{k=1}^{k=n}$ such that
$$x^n+1=prod_{k=1}^{n}big(x-r^{(n)}_{k}big)$$
We then know that $$r_k^{(n)}=expbigg[frac{ipi}{n}(2k-1)bigg]$$
Then we define
$$S_n={r_k^{(n)}:kin[1,n]capBbb N}$$
So we have that
$$frac1{x^n+1}=prod_{rin S_n}frac1{x-r}=prod_{k=1}^nfrac1{x-r_k^{(n)}}$$
Then we assume that we can write
$$prod_{rin S_n}frac1{x-r}=sum_{rin S_n}frac{b(r)}{x-r}$$
Multiplying both sides by $prod_{ain S_n}(x-a)$,
$$1=sum_{rin S_n}b(r)prod_{ain S_n\ aneq r}(x-a)$$
So for any $omegain S_n$,
$$1=b(omega)prod_{ain S_n\ aneq omega}(omega-a)$$
$$b(omega)=prod_{ain S_n\ aneq omega}frac1{omega-a}$$
$$b(r_k^{(n)})=prod_{p=1\ pneq k}^nfrac1{r_k^{(n)}-r_p^{(n)}}$$
So we know that
$$I_n=int_0^1frac{mathrm{d}x}{1+x^n}=sum_{k=1}^{n}b(r_k^{(n)})int_0^1frac{mathrm{d}x}{x-r_k^{(n)}}$$
$$I_n=sum_{k=1}^{n}b(r_k^{(n)})logbigg|frac{r_k^{(n)}-1}{r_k^{(n)}}bigg|$$
$$I_n=sum_{k=1}^{n}logbigg|frac{r_k^{(n)}-1}{r_k^{(n)}}bigg|prod_{p=1\ pneq k}^nfrac1{r_k^{(n)}-r_p^{(n)}}$$
So we have
$$int_0^1log(1+x^n)mathrm{d}x=log2-n+nsum_{k=1}^{n}logbigg|frac{r_k^{(n)}-1}{r_k^{(n)}}bigg|prod_{p=1\ pneq k}^nfrac1{r_k^{(n)}-r_p^{(n)}}$$
along with a plethora of other identities...






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  • 1




    $begingroup$
    Cheers @clathratus - I will have to have a look over in detail, but on the surface it looks good. Thanks heaps!
    $endgroup$
    – user150203
    Dec 31 '18 at 6:16










  • $begingroup$
    I say this not as a critique of your post, but of my own... do I have defend the set $r_j$ as having size $n$ ?? i.e. that the $c_j$ are distinct...
    $endgroup$
    – user150203
    Dec 31 '18 at 9:47










  • $begingroup$
    Well any polynomial with degree $n$ is going to have $n$ distinct roots, so it would make sense that $|S_n|=n$
    $endgroup$
    – clathratus
    Dec 31 '18 at 20:28



















1












$begingroup$

Here's another, quicker, method (I also don't know if this one works)



Using the same $r_k^{(n)}$ as last time, we apply the $logprod_{i}a_i=sum_ilog a_i$ property to see that
$$log(1+x^n)=logprod_{k=1}^{n}(x-r_k^{(n)})=sum_{k=1}^{n}log(x-r_k^{(n)})$$
So
$$I_n=int_0^1log(1+x^n)mathrm dx=sum_{k=1}^{n}int_0^1log(x-r_k^{(n)})mathrm dx$$
This last integral boils down to
$$begin{align}
int_0^1log(x-a)mathrm dx=&alogfrac{a}{1+a}+log(1-a)-1\
=&logfrac{a^a(1-a)}{e(1+a)^a}
end{align}$$
So
$$I_n=sum_{rin S_n}logfrac{r^r(1-r)}{e(1+r)^r}$$
And you know how I love product representations, so we again use $logprod_{i}a_i=sum_ilog a_i$ to see that
$$
I_n=logprod_{rin S_n}frac{r^r(1-r)}{e(1+r)^r}\
prod_{rin S_n}frac{r^r(1-r)}{(1+r)^r}=exp(n+I_n)
$$
Which I just think is really neat.






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    $begingroup$

    This does not answer the question as asked in the post.



    Consider
    $$J_m=int log(1+x^{2m}),dx$$ One integration by parts gives
    $$J_m=x log left(1+x^{2 m}right)-2mint frac{ x^{2 m}+1-1}{x^{2 m}+1},dx=x log left(1+x^{2 m}right)-2mx+2mint frac{dx}{x^{2 m}+1}$$ and
    $$int frac{dx}{x^{2 m}+1}=x , _2F_1left(1,frac{1}{2 m};1+frac{1}{2 m};-x^{2 m}right)$$ where appears the Gaussian or ordinary hypergeometric function.



    So
    $$K_m=int_0^a log(1+x^{2m}),dx=a log left(1+a^{2 m}right)-2ma+2ma , _2F_1left(1,frac{1}{2 m};1+frac{1}{2 m};-a^{2 m}right)$$ and, if $a=1$,
    $$I_m=int_0^1 log(1+x^{2m}),dx= log left(2right)-2m+2m , _2F_1left(1,frac{1}{2 m};1+frac{1}{2 m};-1right)$$ which can write
    $$I_m=log (2)-Phi left(-1,1,1+frac{1}{2 m}right)$$
    where appears the Lerch transcendent function.



    Now, (this is something I never looked at), a few (ugly) expressions for $f_m=Phi left(-1,1,1+frac{1}{2 m}right)$ before any simplification
    $$f(1)=frac{pi }{2}-2$$
    $$f(2)=frac{1}{4} left(pi tan left(frac{pi }{8}right)+pi cot left(frac{pi
    }{8}right)-4 sqrt{2} log left(sin left(frac{pi }{8}right)right)+4
    sqrt{2} log left(cos left(frac{pi }{8}right)right)right)-4$$
    $$f(3)=frac{2 left(pi -sqrt{3} log left(sqrt{3}-1right)+sqrt{3} log
    left(1+sqrt{3}right)right)}{left(sqrt{3}-1right)
    left(1+sqrt{3}right)}-6$$
    $$f(4)=frac{1}{4} left(pi tan left(frac{pi }{16}right)+pi cot left(frac{pi
    }{16}right)-8 sin left(frac{pi }{8}right) log left(sin left(frac{3
    pi }{16}right)right)+8 cos left(frac{pi }{8}right) log left(cos
    left(frac{pi }{16}right)right)-8 cos left(frac{pi }{8}right) log
    left(sin left(frac{pi }{16}right)right)+8 sin left(frac{pi
    }{8}right) log left(cos left(frac{3 pi }{16}right)right)right)-8$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Yes, my collaborator and I came to the same result (very nice closed form indeed). If I may ask - what are your thoughts on the special case I've discussed?
      $endgroup$
      – user150203
      Dec 27 '18 at 8:49










    • $begingroup$
      @DavidG. Your post is very interesting. I need to spend more time with it !
      $endgroup$
      – Claude Leibovici
      Dec 27 '18 at 8:54










    • $begingroup$
      Please do! I look forward to any comments you may have. Thanks again for your post.
      $endgroup$
      – user150203
      Dec 27 '18 at 9:48










    • $begingroup$
      Mathematica calculates the same result
      $endgroup$
      – stocha
      Mar 22 at 19:11
















    6












    $begingroup$

    This does not answer the question as asked in the post.



    Consider
    $$J_m=int log(1+x^{2m}),dx$$ One integration by parts gives
    $$J_m=x log left(1+x^{2 m}right)-2mint frac{ x^{2 m}+1-1}{x^{2 m}+1},dx=x log left(1+x^{2 m}right)-2mx+2mint frac{dx}{x^{2 m}+1}$$ and
    $$int frac{dx}{x^{2 m}+1}=x , _2F_1left(1,frac{1}{2 m};1+frac{1}{2 m};-x^{2 m}right)$$ where appears the Gaussian or ordinary hypergeometric function.



    So
    $$K_m=int_0^a log(1+x^{2m}),dx=a log left(1+a^{2 m}right)-2ma+2ma , _2F_1left(1,frac{1}{2 m};1+frac{1}{2 m};-a^{2 m}right)$$ and, if $a=1$,
    $$I_m=int_0^1 log(1+x^{2m}),dx= log left(2right)-2m+2m , _2F_1left(1,frac{1}{2 m};1+frac{1}{2 m};-1right)$$ which can write
    $$I_m=log (2)-Phi left(-1,1,1+frac{1}{2 m}right)$$
    where appears the Lerch transcendent function.



    Now, (this is something I never looked at), a few (ugly) expressions for $f_m=Phi left(-1,1,1+frac{1}{2 m}right)$ before any simplification
    $$f(1)=frac{pi }{2}-2$$
    $$f(2)=frac{1}{4} left(pi tan left(frac{pi }{8}right)+pi cot left(frac{pi
    }{8}right)-4 sqrt{2} log left(sin left(frac{pi }{8}right)right)+4
    sqrt{2} log left(cos left(frac{pi }{8}right)right)right)-4$$
    $$f(3)=frac{2 left(pi -sqrt{3} log left(sqrt{3}-1right)+sqrt{3} log
    left(1+sqrt{3}right)right)}{left(sqrt{3}-1right)
    left(1+sqrt{3}right)}-6$$
    $$f(4)=frac{1}{4} left(pi tan left(frac{pi }{16}right)+pi cot left(frac{pi
    }{16}right)-8 sin left(frac{pi }{8}right) log left(sin left(frac{3
    pi }{16}right)right)+8 cos left(frac{pi }{8}right) log left(cos
    left(frac{pi }{16}right)right)-8 cos left(frac{pi }{8}right) log
    left(sin left(frac{pi }{16}right)right)+8 sin left(frac{pi
    }{8}right) log left(cos left(frac{3 pi }{16}right)right)right)-8$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Yes, my collaborator and I came to the same result (very nice closed form indeed). If I may ask - what are your thoughts on the special case I've discussed?
      $endgroup$
      – user150203
      Dec 27 '18 at 8:49










    • $begingroup$
      @DavidG. Your post is very interesting. I need to spend more time with it !
      $endgroup$
      – Claude Leibovici
      Dec 27 '18 at 8:54










    • $begingroup$
      Please do! I look forward to any comments you may have. Thanks again for your post.
      $endgroup$
      – user150203
      Dec 27 '18 at 9:48










    • $begingroup$
      Mathematica calculates the same result
      $endgroup$
      – stocha
      Mar 22 at 19:11














    6












    6








    6





    $begingroup$

    This does not answer the question as asked in the post.



    Consider
    $$J_m=int log(1+x^{2m}),dx$$ One integration by parts gives
    $$J_m=x log left(1+x^{2 m}right)-2mint frac{ x^{2 m}+1-1}{x^{2 m}+1},dx=x log left(1+x^{2 m}right)-2mx+2mint frac{dx}{x^{2 m}+1}$$ and
    $$int frac{dx}{x^{2 m}+1}=x , _2F_1left(1,frac{1}{2 m};1+frac{1}{2 m};-x^{2 m}right)$$ where appears the Gaussian or ordinary hypergeometric function.



    So
    $$K_m=int_0^a log(1+x^{2m}),dx=a log left(1+a^{2 m}right)-2ma+2ma , _2F_1left(1,frac{1}{2 m};1+frac{1}{2 m};-a^{2 m}right)$$ and, if $a=1$,
    $$I_m=int_0^1 log(1+x^{2m}),dx= log left(2right)-2m+2m , _2F_1left(1,frac{1}{2 m};1+frac{1}{2 m};-1right)$$ which can write
    $$I_m=log (2)-Phi left(-1,1,1+frac{1}{2 m}right)$$
    where appears the Lerch transcendent function.



    Now, (this is something I never looked at), a few (ugly) expressions for $f_m=Phi left(-1,1,1+frac{1}{2 m}right)$ before any simplification
    $$f(1)=frac{pi }{2}-2$$
    $$f(2)=frac{1}{4} left(pi tan left(frac{pi }{8}right)+pi cot left(frac{pi
    }{8}right)-4 sqrt{2} log left(sin left(frac{pi }{8}right)right)+4
    sqrt{2} log left(cos left(frac{pi }{8}right)right)right)-4$$
    $$f(3)=frac{2 left(pi -sqrt{3} log left(sqrt{3}-1right)+sqrt{3} log
    left(1+sqrt{3}right)right)}{left(sqrt{3}-1right)
    left(1+sqrt{3}right)}-6$$
    $$f(4)=frac{1}{4} left(pi tan left(frac{pi }{16}right)+pi cot left(frac{pi
    }{16}right)-8 sin left(frac{pi }{8}right) log left(sin left(frac{3
    pi }{16}right)right)+8 cos left(frac{pi }{8}right) log left(cos
    left(frac{pi }{16}right)right)-8 cos left(frac{pi }{8}right) log
    left(sin left(frac{pi }{16}right)right)+8 sin left(frac{pi
    }{8}right) log left(cos left(frac{3 pi }{16}right)right)right)-8$$






    share|cite|improve this answer









    $endgroup$



    This does not answer the question as asked in the post.



    Consider
    $$J_m=int log(1+x^{2m}),dx$$ One integration by parts gives
    $$J_m=x log left(1+x^{2 m}right)-2mint frac{ x^{2 m}+1-1}{x^{2 m}+1},dx=x log left(1+x^{2 m}right)-2mx+2mint frac{dx}{x^{2 m}+1}$$ and
    $$int frac{dx}{x^{2 m}+1}=x , _2F_1left(1,frac{1}{2 m};1+frac{1}{2 m};-x^{2 m}right)$$ where appears the Gaussian or ordinary hypergeometric function.



    So
    $$K_m=int_0^a log(1+x^{2m}),dx=a log left(1+a^{2 m}right)-2ma+2ma , _2F_1left(1,frac{1}{2 m};1+frac{1}{2 m};-a^{2 m}right)$$ and, if $a=1$,
    $$I_m=int_0^1 log(1+x^{2m}),dx= log left(2right)-2m+2m , _2F_1left(1,frac{1}{2 m};1+frac{1}{2 m};-1right)$$ which can write
    $$I_m=log (2)-Phi left(-1,1,1+frac{1}{2 m}right)$$
    where appears the Lerch transcendent function.



    Now, (this is something I never looked at), a few (ugly) expressions for $f_m=Phi left(-1,1,1+frac{1}{2 m}right)$ before any simplification
    $$f(1)=frac{pi }{2}-2$$
    $$f(2)=frac{1}{4} left(pi tan left(frac{pi }{8}right)+pi cot left(frac{pi
    }{8}right)-4 sqrt{2} log left(sin left(frac{pi }{8}right)right)+4
    sqrt{2} log left(cos left(frac{pi }{8}right)right)right)-4$$
    $$f(3)=frac{2 left(pi -sqrt{3} log left(sqrt{3}-1right)+sqrt{3} log
    left(1+sqrt{3}right)right)}{left(sqrt{3}-1right)
    left(1+sqrt{3}right)}-6$$
    $$f(4)=frac{1}{4} left(pi tan left(frac{pi }{16}right)+pi cot left(frac{pi
    }{16}right)-8 sin left(frac{pi }{8}right) log left(sin left(frac{3
    pi }{16}right)right)+8 cos left(frac{pi }{8}right) log left(cos
    left(frac{pi }{16}right)right)-8 cos left(frac{pi }{8}right) log
    left(sin left(frac{pi }{16}right)right)+8 sin left(frac{pi
    }{8}right) log left(cos left(frac{3 pi }{16}right)right)right)-8$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 27 '18 at 8:47









    Claude LeiboviciClaude Leibovici

    126k1158135




    126k1158135












    • $begingroup$
      Yes, my collaborator and I came to the same result (very nice closed form indeed). If I may ask - what are your thoughts on the special case I've discussed?
      $endgroup$
      – user150203
      Dec 27 '18 at 8:49










    • $begingroup$
      @DavidG. Your post is very interesting. I need to spend more time with it !
      $endgroup$
      – Claude Leibovici
      Dec 27 '18 at 8:54










    • $begingroup$
      Please do! I look forward to any comments you may have. Thanks again for your post.
      $endgroup$
      – user150203
      Dec 27 '18 at 9:48










    • $begingroup$
      Mathematica calculates the same result
      $endgroup$
      – stocha
      Mar 22 at 19:11


















    • $begingroup$
      Yes, my collaborator and I came to the same result (very nice closed form indeed). If I may ask - what are your thoughts on the special case I've discussed?
      $endgroup$
      – user150203
      Dec 27 '18 at 8:49










    • $begingroup$
      @DavidG. Your post is very interesting. I need to spend more time with it !
      $endgroup$
      – Claude Leibovici
      Dec 27 '18 at 8:54










    • $begingroup$
      Please do! I look forward to any comments you may have. Thanks again for your post.
      $endgroup$
      – user150203
      Dec 27 '18 at 9:48










    • $begingroup$
      Mathematica calculates the same result
      $endgroup$
      – stocha
      Mar 22 at 19:11
















    $begingroup$
    Yes, my collaborator and I came to the same result (very nice closed form indeed). If I may ask - what are your thoughts on the special case I've discussed?
    $endgroup$
    – user150203
    Dec 27 '18 at 8:49




    $begingroup$
    Yes, my collaborator and I came to the same result (very nice closed form indeed). If I may ask - what are your thoughts on the special case I've discussed?
    $endgroup$
    – user150203
    Dec 27 '18 at 8:49












    $begingroup$
    @DavidG. Your post is very interesting. I need to spend more time with it !
    $endgroup$
    – Claude Leibovici
    Dec 27 '18 at 8:54




    $begingroup$
    @DavidG. Your post is very interesting. I need to spend more time with it !
    $endgroup$
    – Claude Leibovici
    Dec 27 '18 at 8:54












    $begingroup$
    Please do! I look forward to any comments you may have. Thanks again for your post.
    $endgroup$
    – user150203
    Dec 27 '18 at 9:48




    $begingroup$
    Please do! I look forward to any comments you may have. Thanks again for your post.
    $endgroup$
    – user150203
    Dec 27 '18 at 9:48












    $begingroup$
    Mathematica calculates the same result
    $endgroup$
    – stocha
    Mar 22 at 19:11




    $begingroup$
    Mathematica calculates the same result
    $endgroup$
    – stocha
    Mar 22 at 19:11











    4





    +250







    $begingroup$

    $color{brown}{textbf{Preliminary notes.}}$




    Calculation of the required sum looks more complex task than of the
    issue integral.



    Approach with calculations of the sum via the inttegral allows to get close form both of the integral and the sum.




    Denote
    $$I_n=intlimits_0^1log(1+x^n)mathrm dx.tag1$$
    Easy to see that the issue integral is $I_{2m}.$



    $color{brown}{textbf{Closed form of the integrals.}}$



    The first step is the integration by parts:
    $$I_n = xlog(1+x^n)biggr|_0^1-nint_0^1xdfrac{x^{n-1}}{1+x^n}mathrm dx,$$
    $$I_n = log2 - n + nintlimits_0^1dfrac{mathrm dx}{1+x^n}.tag2$$
    (see also Claude Leibovici).



    Then, the substitution
    $$x=e^{-t}tag3$$
    gives
    $$J_n=intlimits_0^1dfrac{mathrm dx}{1+x^n}=intlimits_0^inftydfrac{e^{-t},mathrm dt}{1+e^{-nt}} = sumlimits_{k=0}^infty(-1)^kintlimits_0^infty e^{-(kn+1)t},mathrm dt=sumlimits_{k=0}^inftydfrac{(-1)^k}{kn+1} ,$$
    $$J_n = intlimits_0^1dfrac{mathrm dx}{1+x^n}= dfrac1{2n}left(psi_0left(dfrac{n+1}{2n}right) - psi_0left(dfrac{1}{2n}right)right),tag4$$
    where $psi_0(x)$ is the polygamma function.



    Using $(2),(4),$ easy to get the expression for the OP integral in the form of
    $$boxed{I_{2m} = log2 - 2m +dfrac1{2}left(psi_0left(dfrac{2m+1}{4m}right) - psi_0left(dfrac{1}{4m}right)right).}tag5$$



    $color{brown}{textbf{Testing of the solution.}}$



    Solution $(4)$ can be expressed via the elementary functions for a lot of the given values of n.



    Immediate calculations allow to check obtained expressions in the simple cases
    begin{align}
    &J_1 = intlimits_0^1dfrac{mathrm dx}{1+x} = log 2,\[4pt]
    &J_2 = intlimits_0^1dfrac{mathrm dx}{1+x^2} = arctan 1 = dfracpi4,\[4pt]
    &J_3 = intlimits_0^1dfrac{mathrm dx}{1+x^3}
    = dfrac16intlimits_0^1left(dfrac{2}{1+x} - dfrac{2x-1}{1-x+x^2} + dfrac{3}{1-x+x^2}right),mathrm dx\
    &= left(dfrac13log(1+x) - dfrac16 log(1-x+x^2) + dfrac1{sqrt3}arctandfrac{2x-1}{sqrt3}right)bigg|_0^1
    = dfracpi{3sqrt3} + frac13 log2.\[4pt]
    end{align}
    More hard cases are
    begin{align}
    &J_4 = intlimits_0^1dfrac{mathrm dx}{1+x^4}
    = dfrac{sqrt2}8left(logdfrac{x^2 + sqrt2 x + 1}{x^2 - sqrt2 x + 1}
    - 2arctan(1-sqrt2x) + 2arctan(sqrt2x+1)right)bigg|_0^1\[4pt]
    &= dfrac{sqrt2}8left(logdfrac{(x^2 + sqrt2 x + 1)^2}{1+x^4} + 2arctandfrac{xsqrt2}{1-x^2}right)bigg|_0^1
    = dfrac{sqrt2}8(pi+log(3+2sqrt2)),\[4pt]
    &J_4=dfrac18Bigg(dfrac12sqrt{dfrac{2-sqrt2}{2+sqrt2}}pi+dfrac12sqrt{dfrac{2+sqrt2}{2-sqrt2}}pi
    -2sqrt2left(-log2+dfrac12log(2-sqrt2)right)\[4pt]
    &+2sqrt2left(-log2+dfrac12log(2+sqrt2)right)Bigg)\[4pt]
    &=dfrac18Bigg(dfracpi2left(tandfracpi8+cotdfracpi8right)
    +sqrt2logdfrac{2+sqrt2}{2-sqrt2}Bigg)
    = dfrac{sqrt2}8(pi+log(3+2sqrt2)),
    end{align}
    and $n>4.$



    Numeric calculations of the issue integral $I_{2m}$ and its closed form $(5)$ also confirm the correctness of the closed form.



    $color{brown}{textbf{The alternative approach.}}$



    The alternative approach is considered in OP. Let us repeat it with some differences.



    Polynomial factorization can be presented in the form of



    begin{align}
    &1+x^{2m} = prodlimits_{j=0}^{2m-1}{large left(x-e^{frac{2j+1}{2m}pi i}right)}
    = prodlimits_{j=0}^{m-1}left(x^2-2xcosfrac{2j+1}{2m}pi+1right),
    end{align}
    so
    begin{align}
    &I_{2m} = intlimits_0^1 ln(1+x^{2m}),mathrm dx
    = sumlimits_{j=0}^{m-1} T_j,quadtext{where}\[4pt]
    &T_j = intlimits_0^1 lnleft(x^2-2xcosfrac{2j+1}{2m}pi+1right),mathrm dx,\[4pt]
    &T_j = xlnleft(x^2-2xcosfrac{2j+1}{2m}pi+1right)bigg|_0^1
    - 2intlimits_0^1 dfrac{xleft(x-cosfrac{2j+1}{2m}piright)}{x^2-2xcosfrac{2j+1}{2m}pi+1},mathrm dx\[4pt]
    & = lnleft(2-2cosfrac{2j+1}{2m}piright)
    - 2-2intlimits_0^1 dfrac{xcosfrac{2j+1}{2m}pi-1}{x^2-2xcosfrac{2j+1}{2m}pi+1},mathrm dx\[4pt]
    & = lnleft(2-2cosfrac{2j+1}{2m}piright) - 2
    - 2cosfrac{2j+1}{2m}pi intlimits_0^1 dfrac{x-cosfrac{2j+1}{2m}pi}{x^2-2xcosfrac{2j+1}{2m}pi+1},mathrm dx\[4pt]
    & + 2intlimits_0^1 dfrac{sin^2frac{2j+1}{2m}pi}{left(x-cosfrac{2j+1}{2m}piright)^2+sin^2frac{2j+1}{2m}pi},mathrm dx\[4pt]
    & = lnleft(2-2cosfrac{2j+1}{2m}piright) - 2
    - cosfrac{2j+1}{2m}pi lnleft(x^2-2xcosfrac{2j+1}{2m}pi+1right)bigg|_0^1\[4pt]
    & + 2sinfrac{2j+1}{2m}pi arctandfrac{x-cosfrac{2j+1}{2m}pi}{sinfrac{2j+1}{2m}pi}bigg|_0^1\[4pt]
    & = lnleft(2-2cosfrac{2j+1}{2m}piright) - 2
    - cosfrac{2j+1}{2m}pi lnleft(2-2cosfrac{2j+1}{2m}piright)\[4pt]
    & + 2sinfrac{2j+1}{2m}pi
    left(arctandfrac{1-cosfrac{2j+1}{2m}pi}{sinfrac{2j+1}{2m}pi}
    + arctandfrac{cosfrac{2j+1}{2m}pi}{sinfrac{2j+1}{2m}pi}right)\[4pt]
    & = lnleft(2-2cosfrac{2j+1}{2m}piright) - 2
    - cosfrac{2j+1}{2m}pi lnleft(2-2cosfrac{2j+1}{2m}piright)\[4pt]
    & + 2sinfrac{2j+1}{2m}pi
    arctandfrac{sinfrac{2j+1}{2m}pi}
    {sin^2frac{2j+1}{2m}pi - left(1 - cosfrac{2j+1}{2m}piright)cosfrac{2j+1}{2m}pi}\[4pt]
    & = left(1-cosfrac{2j+1}{2m}piright)lnleft(2-2cosfrac{2j+1}{2m}piright) - 2 + 2sinfrac{2j+1}{2m}piarctancotfrac{2j+1}{4m}pi,\[4pt]
    &T_j = left(1-cosfrac{2j+1}{2m}piright)lnleft(2-2cosfrac{2j+1}{2m}piright) - 2 + dfrac{2(m-j)-1}{2m}pisinfrac{2j+1}{2m}pi.
    end{align}
    Taking in account that
    begin{align}
    &sumlimits_{j=0}^{m-1}cosfrac{2j+1}{2m}pi
    = Re sumlimits_{j=0}^{m-1}e^{frac{2j+1}{2m}pi i}
    = Re {large dfrac{1-e^{pi i}}{1-e^{fracpi{m}i}}e^{fracpi{2m}i}}
    =Redfrac {i}{sinfracpi{2m}} = 0,\[4pt]
    &sumlimits_{j=0}^{m-1}sinfrac{2j+1}{2m}pi
    = Im sumlimits_{j=0}^{m-1}e^{frac{2j+1}{2m}pi i}
    =Imdfrac{i}{sinfracpi{2m}} = dfrac1{sinfracpi{2m}},\[4pt]
    &sumlimits_{j=0}^{m-1}frac{2j+1}{2m}pisinfrac{2j+1}{2m}pi
    = dfracpi{2sinfracpi{2m}}
    end{align}
    (see also Wolfram Alpha calculations),



    one can get
    $$boxed{I_{2m} = m(ln2-2) + dfracpi2cscfracpi{2m}
    + sumlimits_{j=0}^{m-1}left(1pmcosfrac{2j+1}{2m}piright)
    logleft(1pmcosfrac{2j+1}{2m}piright).}tag6$$



    $color{brown}{textbf{Closed form for the sum.}}$



    From $(5)-(6)$ should
    $$color{green}{boxed{sumlimits_{j=0}^{m-1}left(1pm c_jright)
    logleft(1pm c_jright)
    = dfrac1{2}left(psi_0left(dfrac{2m+1}{4m}right) - psi_0left(dfrac{1}{4m}right)right) - (m-1)ln2 - dfracpi2cscfracpi{2m}
    ,}}tag7$$
    where
    $$c_j = cosfrac{2j+1}{2m}pi.$$



    Numeric calculations confirm the expression $(6)$ for the both variants of the sign "$pm$".



    Also, numeric calculations of the issue sum and its closed form $(7)$ confirm the correctness of the closed form for the both variants of the sign "$pm$".






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Very nice presentation, already before the alternate approach (+1).
      $endgroup$
      – Markus Scheuer
      Mar 28 at 21:46










    • $begingroup$
      @MarkusScheuer Thanks, you are welcome!
      $endgroup$
      – Yuri Negometyanov
      Mar 28 at 22:16
















    4





    +250







    $begingroup$

    $color{brown}{textbf{Preliminary notes.}}$




    Calculation of the required sum looks more complex task than of the
    issue integral.



    Approach with calculations of the sum via the inttegral allows to get close form both of the integral and the sum.




    Denote
    $$I_n=intlimits_0^1log(1+x^n)mathrm dx.tag1$$
    Easy to see that the issue integral is $I_{2m}.$



    $color{brown}{textbf{Closed form of the integrals.}}$



    The first step is the integration by parts:
    $$I_n = xlog(1+x^n)biggr|_0^1-nint_0^1xdfrac{x^{n-1}}{1+x^n}mathrm dx,$$
    $$I_n = log2 - n + nintlimits_0^1dfrac{mathrm dx}{1+x^n}.tag2$$
    (see also Claude Leibovici).



    Then, the substitution
    $$x=e^{-t}tag3$$
    gives
    $$J_n=intlimits_0^1dfrac{mathrm dx}{1+x^n}=intlimits_0^inftydfrac{e^{-t},mathrm dt}{1+e^{-nt}} = sumlimits_{k=0}^infty(-1)^kintlimits_0^infty e^{-(kn+1)t},mathrm dt=sumlimits_{k=0}^inftydfrac{(-1)^k}{kn+1} ,$$
    $$J_n = intlimits_0^1dfrac{mathrm dx}{1+x^n}= dfrac1{2n}left(psi_0left(dfrac{n+1}{2n}right) - psi_0left(dfrac{1}{2n}right)right),tag4$$
    where $psi_0(x)$ is the polygamma function.



    Using $(2),(4),$ easy to get the expression for the OP integral in the form of
    $$boxed{I_{2m} = log2 - 2m +dfrac1{2}left(psi_0left(dfrac{2m+1}{4m}right) - psi_0left(dfrac{1}{4m}right)right).}tag5$$



    $color{brown}{textbf{Testing of the solution.}}$



    Solution $(4)$ can be expressed via the elementary functions for a lot of the given values of n.



    Immediate calculations allow to check obtained expressions in the simple cases
    begin{align}
    &J_1 = intlimits_0^1dfrac{mathrm dx}{1+x} = log 2,\[4pt]
    &J_2 = intlimits_0^1dfrac{mathrm dx}{1+x^2} = arctan 1 = dfracpi4,\[4pt]
    &J_3 = intlimits_0^1dfrac{mathrm dx}{1+x^3}
    = dfrac16intlimits_0^1left(dfrac{2}{1+x} - dfrac{2x-1}{1-x+x^2} + dfrac{3}{1-x+x^2}right),mathrm dx\
    &= left(dfrac13log(1+x) - dfrac16 log(1-x+x^2) + dfrac1{sqrt3}arctandfrac{2x-1}{sqrt3}right)bigg|_0^1
    = dfracpi{3sqrt3} + frac13 log2.\[4pt]
    end{align}
    More hard cases are
    begin{align}
    &J_4 = intlimits_0^1dfrac{mathrm dx}{1+x^4}
    = dfrac{sqrt2}8left(logdfrac{x^2 + sqrt2 x + 1}{x^2 - sqrt2 x + 1}
    - 2arctan(1-sqrt2x) + 2arctan(sqrt2x+1)right)bigg|_0^1\[4pt]
    &= dfrac{sqrt2}8left(logdfrac{(x^2 + sqrt2 x + 1)^2}{1+x^4} + 2arctandfrac{xsqrt2}{1-x^2}right)bigg|_0^1
    = dfrac{sqrt2}8(pi+log(3+2sqrt2)),\[4pt]
    &J_4=dfrac18Bigg(dfrac12sqrt{dfrac{2-sqrt2}{2+sqrt2}}pi+dfrac12sqrt{dfrac{2+sqrt2}{2-sqrt2}}pi
    -2sqrt2left(-log2+dfrac12log(2-sqrt2)right)\[4pt]
    &+2sqrt2left(-log2+dfrac12log(2+sqrt2)right)Bigg)\[4pt]
    &=dfrac18Bigg(dfracpi2left(tandfracpi8+cotdfracpi8right)
    +sqrt2logdfrac{2+sqrt2}{2-sqrt2}Bigg)
    = dfrac{sqrt2}8(pi+log(3+2sqrt2)),
    end{align}
    and $n>4.$



    Numeric calculations of the issue integral $I_{2m}$ and its closed form $(5)$ also confirm the correctness of the closed form.



    $color{brown}{textbf{The alternative approach.}}$



    The alternative approach is considered in OP. Let us repeat it with some differences.



    Polynomial factorization can be presented in the form of



    begin{align}
    &1+x^{2m} = prodlimits_{j=0}^{2m-1}{large left(x-e^{frac{2j+1}{2m}pi i}right)}
    = prodlimits_{j=0}^{m-1}left(x^2-2xcosfrac{2j+1}{2m}pi+1right),
    end{align}
    so
    begin{align}
    &I_{2m} = intlimits_0^1 ln(1+x^{2m}),mathrm dx
    = sumlimits_{j=0}^{m-1} T_j,quadtext{where}\[4pt]
    &T_j = intlimits_0^1 lnleft(x^2-2xcosfrac{2j+1}{2m}pi+1right),mathrm dx,\[4pt]
    &T_j = xlnleft(x^2-2xcosfrac{2j+1}{2m}pi+1right)bigg|_0^1
    - 2intlimits_0^1 dfrac{xleft(x-cosfrac{2j+1}{2m}piright)}{x^2-2xcosfrac{2j+1}{2m}pi+1},mathrm dx\[4pt]
    & = lnleft(2-2cosfrac{2j+1}{2m}piright)
    - 2-2intlimits_0^1 dfrac{xcosfrac{2j+1}{2m}pi-1}{x^2-2xcosfrac{2j+1}{2m}pi+1},mathrm dx\[4pt]
    & = lnleft(2-2cosfrac{2j+1}{2m}piright) - 2
    - 2cosfrac{2j+1}{2m}pi intlimits_0^1 dfrac{x-cosfrac{2j+1}{2m}pi}{x^2-2xcosfrac{2j+1}{2m}pi+1},mathrm dx\[4pt]
    & + 2intlimits_0^1 dfrac{sin^2frac{2j+1}{2m}pi}{left(x-cosfrac{2j+1}{2m}piright)^2+sin^2frac{2j+1}{2m}pi},mathrm dx\[4pt]
    & = lnleft(2-2cosfrac{2j+1}{2m}piright) - 2
    - cosfrac{2j+1}{2m}pi lnleft(x^2-2xcosfrac{2j+1}{2m}pi+1right)bigg|_0^1\[4pt]
    & + 2sinfrac{2j+1}{2m}pi arctandfrac{x-cosfrac{2j+1}{2m}pi}{sinfrac{2j+1}{2m}pi}bigg|_0^1\[4pt]
    & = lnleft(2-2cosfrac{2j+1}{2m}piright) - 2
    - cosfrac{2j+1}{2m}pi lnleft(2-2cosfrac{2j+1}{2m}piright)\[4pt]
    & + 2sinfrac{2j+1}{2m}pi
    left(arctandfrac{1-cosfrac{2j+1}{2m}pi}{sinfrac{2j+1}{2m}pi}
    + arctandfrac{cosfrac{2j+1}{2m}pi}{sinfrac{2j+1}{2m}pi}right)\[4pt]
    & = lnleft(2-2cosfrac{2j+1}{2m}piright) - 2
    - cosfrac{2j+1}{2m}pi lnleft(2-2cosfrac{2j+1}{2m}piright)\[4pt]
    & + 2sinfrac{2j+1}{2m}pi
    arctandfrac{sinfrac{2j+1}{2m}pi}
    {sin^2frac{2j+1}{2m}pi - left(1 - cosfrac{2j+1}{2m}piright)cosfrac{2j+1}{2m}pi}\[4pt]
    & = left(1-cosfrac{2j+1}{2m}piright)lnleft(2-2cosfrac{2j+1}{2m}piright) - 2 + 2sinfrac{2j+1}{2m}piarctancotfrac{2j+1}{4m}pi,\[4pt]
    &T_j = left(1-cosfrac{2j+1}{2m}piright)lnleft(2-2cosfrac{2j+1}{2m}piright) - 2 + dfrac{2(m-j)-1}{2m}pisinfrac{2j+1}{2m}pi.
    end{align}
    Taking in account that
    begin{align}
    &sumlimits_{j=0}^{m-1}cosfrac{2j+1}{2m}pi
    = Re sumlimits_{j=0}^{m-1}e^{frac{2j+1}{2m}pi i}
    = Re {large dfrac{1-e^{pi i}}{1-e^{fracpi{m}i}}e^{fracpi{2m}i}}
    =Redfrac {i}{sinfracpi{2m}} = 0,\[4pt]
    &sumlimits_{j=0}^{m-1}sinfrac{2j+1}{2m}pi
    = Im sumlimits_{j=0}^{m-1}e^{frac{2j+1}{2m}pi i}
    =Imdfrac{i}{sinfracpi{2m}} = dfrac1{sinfracpi{2m}},\[4pt]
    &sumlimits_{j=0}^{m-1}frac{2j+1}{2m}pisinfrac{2j+1}{2m}pi
    = dfracpi{2sinfracpi{2m}}
    end{align}
    (see also Wolfram Alpha calculations),



    one can get
    $$boxed{I_{2m} = m(ln2-2) + dfracpi2cscfracpi{2m}
    + sumlimits_{j=0}^{m-1}left(1pmcosfrac{2j+1}{2m}piright)
    logleft(1pmcosfrac{2j+1}{2m}piright).}tag6$$



    $color{brown}{textbf{Closed form for the sum.}}$



    From $(5)-(6)$ should
    $$color{green}{boxed{sumlimits_{j=0}^{m-1}left(1pm c_jright)
    logleft(1pm c_jright)
    = dfrac1{2}left(psi_0left(dfrac{2m+1}{4m}right) - psi_0left(dfrac{1}{4m}right)right) - (m-1)ln2 - dfracpi2cscfracpi{2m}
    ,}}tag7$$
    where
    $$c_j = cosfrac{2j+1}{2m}pi.$$



    Numeric calculations confirm the expression $(6)$ for the both variants of the sign "$pm$".



    Also, numeric calculations of the issue sum and its closed form $(7)$ confirm the correctness of the closed form for the both variants of the sign "$pm$".






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Very nice presentation, already before the alternate approach (+1).
      $endgroup$
      – Markus Scheuer
      Mar 28 at 21:46










    • $begingroup$
      @MarkusScheuer Thanks, you are welcome!
      $endgroup$
      – Yuri Negometyanov
      Mar 28 at 22:16














    4





    +250







    4





    +250



    4




    +250



    $begingroup$

    $color{brown}{textbf{Preliminary notes.}}$




    Calculation of the required sum looks more complex task than of the
    issue integral.



    Approach with calculations of the sum via the inttegral allows to get close form both of the integral and the sum.




    Denote
    $$I_n=intlimits_0^1log(1+x^n)mathrm dx.tag1$$
    Easy to see that the issue integral is $I_{2m}.$



    $color{brown}{textbf{Closed form of the integrals.}}$



    The first step is the integration by parts:
    $$I_n = xlog(1+x^n)biggr|_0^1-nint_0^1xdfrac{x^{n-1}}{1+x^n}mathrm dx,$$
    $$I_n = log2 - n + nintlimits_0^1dfrac{mathrm dx}{1+x^n}.tag2$$
    (see also Claude Leibovici).



    Then, the substitution
    $$x=e^{-t}tag3$$
    gives
    $$J_n=intlimits_0^1dfrac{mathrm dx}{1+x^n}=intlimits_0^inftydfrac{e^{-t},mathrm dt}{1+e^{-nt}} = sumlimits_{k=0}^infty(-1)^kintlimits_0^infty e^{-(kn+1)t},mathrm dt=sumlimits_{k=0}^inftydfrac{(-1)^k}{kn+1} ,$$
    $$J_n = intlimits_0^1dfrac{mathrm dx}{1+x^n}= dfrac1{2n}left(psi_0left(dfrac{n+1}{2n}right) - psi_0left(dfrac{1}{2n}right)right),tag4$$
    where $psi_0(x)$ is the polygamma function.



    Using $(2),(4),$ easy to get the expression for the OP integral in the form of
    $$boxed{I_{2m} = log2 - 2m +dfrac1{2}left(psi_0left(dfrac{2m+1}{4m}right) - psi_0left(dfrac{1}{4m}right)right).}tag5$$



    $color{brown}{textbf{Testing of the solution.}}$



    Solution $(4)$ can be expressed via the elementary functions for a lot of the given values of n.



    Immediate calculations allow to check obtained expressions in the simple cases
    begin{align}
    &J_1 = intlimits_0^1dfrac{mathrm dx}{1+x} = log 2,\[4pt]
    &J_2 = intlimits_0^1dfrac{mathrm dx}{1+x^2} = arctan 1 = dfracpi4,\[4pt]
    &J_3 = intlimits_0^1dfrac{mathrm dx}{1+x^3}
    = dfrac16intlimits_0^1left(dfrac{2}{1+x} - dfrac{2x-1}{1-x+x^2} + dfrac{3}{1-x+x^2}right),mathrm dx\
    &= left(dfrac13log(1+x) - dfrac16 log(1-x+x^2) + dfrac1{sqrt3}arctandfrac{2x-1}{sqrt3}right)bigg|_0^1
    = dfracpi{3sqrt3} + frac13 log2.\[4pt]
    end{align}
    More hard cases are
    begin{align}
    &J_4 = intlimits_0^1dfrac{mathrm dx}{1+x^4}
    = dfrac{sqrt2}8left(logdfrac{x^2 + sqrt2 x + 1}{x^2 - sqrt2 x + 1}
    - 2arctan(1-sqrt2x) + 2arctan(sqrt2x+1)right)bigg|_0^1\[4pt]
    &= dfrac{sqrt2}8left(logdfrac{(x^2 + sqrt2 x + 1)^2}{1+x^4} + 2arctandfrac{xsqrt2}{1-x^2}right)bigg|_0^1
    = dfrac{sqrt2}8(pi+log(3+2sqrt2)),\[4pt]
    &J_4=dfrac18Bigg(dfrac12sqrt{dfrac{2-sqrt2}{2+sqrt2}}pi+dfrac12sqrt{dfrac{2+sqrt2}{2-sqrt2}}pi
    -2sqrt2left(-log2+dfrac12log(2-sqrt2)right)\[4pt]
    &+2sqrt2left(-log2+dfrac12log(2+sqrt2)right)Bigg)\[4pt]
    &=dfrac18Bigg(dfracpi2left(tandfracpi8+cotdfracpi8right)
    +sqrt2logdfrac{2+sqrt2}{2-sqrt2}Bigg)
    = dfrac{sqrt2}8(pi+log(3+2sqrt2)),
    end{align}
    and $n>4.$



    Numeric calculations of the issue integral $I_{2m}$ and its closed form $(5)$ also confirm the correctness of the closed form.



    $color{brown}{textbf{The alternative approach.}}$



    The alternative approach is considered in OP. Let us repeat it with some differences.



    Polynomial factorization can be presented in the form of



    begin{align}
    &1+x^{2m} = prodlimits_{j=0}^{2m-1}{large left(x-e^{frac{2j+1}{2m}pi i}right)}
    = prodlimits_{j=0}^{m-1}left(x^2-2xcosfrac{2j+1}{2m}pi+1right),
    end{align}
    so
    begin{align}
    &I_{2m} = intlimits_0^1 ln(1+x^{2m}),mathrm dx
    = sumlimits_{j=0}^{m-1} T_j,quadtext{where}\[4pt]
    &T_j = intlimits_0^1 lnleft(x^2-2xcosfrac{2j+1}{2m}pi+1right),mathrm dx,\[4pt]
    &T_j = xlnleft(x^2-2xcosfrac{2j+1}{2m}pi+1right)bigg|_0^1
    - 2intlimits_0^1 dfrac{xleft(x-cosfrac{2j+1}{2m}piright)}{x^2-2xcosfrac{2j+1}{2m}pi+1},mathrm dx\[4pt]
    & = lnleft(2-2cosfrac{2j+1}{2m}piright)
    - 2-2intlimits_0^1 dfrac{xcosfrac{2j+1}{2m}pi-1}{x^2-2xcosfrac{2j+1}{2m}pi+1},mathrm dx\[4pt]
    & = lnleft(2-2cosfrac{2j+1}{2m}piright) - 2
    - 2cosfrac{2j+1}{2m}pi intlimits_0^1 dfrac{x-cosfrac{2j+1}{2m}pi}{x^2-2xcosfrac{2j+1}{2m}pi+1},mathrm dx\[4pt]
    & + 2intlimits_0^1 dfrac{sin^2frac{2j+1}{2m}pi}{left(x-cosfrac{2j+1}{2m}piright)^2+sin^2frac{2j+1}{2m}pi},mathrm dx\[4pt]
    & = lnleft(2-2cosfrac{2j+1}{2m}piright) - 2
    - cosfrac{2j+1}{2m}pi lnleft(x^2-2xcosfrac{2j+1}{2m}pi+1right)bigg|_0^1\[4pt]
    & + 2sinfrac{2j+1}{2m}pi arctandfrac{x-cosfrac{2j+1}{2m}pi}{sinfrac{2j+1}{2m}pi}bigg|_0^1\[4pt]
    & = lnleft(2-2cosfrac{2j+1}{2m}piright) - 2
    - cosfrac{2j+1}{2m}pi lnleft(2-2cosfrac{2j+1}{2m}piright)\[4pt]
    & + 2sinfrac{2j+1}{2m}pi
    left(arctandfrac{1-cosfrac{2j+1}{2m}pi}{sinfrac{2j+1}{2m}pi}
    + arctandfrac{cosfrac{2j+1}{2m}pi}{sinfrac{2j+1}{2m}pi}right)\[4pt]
    & = lnleft(2-2cosfrac{2j+1}{2m}piright) - 2
    - cosfrac{2j+1}{2m}pi lnleft(2-2cosfrac{2j+1}{2m}piright)\[4pt]
    & + 2sinfrac{2j+1}{2m}pi
    arctandfrac{sinfrac{2j+1}{2m}pi}
    {sin^2frac{2j+1}{2m}pi - left(1 - cosfrac{2j+1}{2m}piright)cosfrac{2j+1}{2m}pi}\[4pt]
    & = left(1-cosfrac{2j+1}{2m}piright)lnleft(2-2cosfrac{2j+1}{2m}piright) - 2 + 2sinfrac{2j+1}{2m}piarctancotfrac{2j+1}{4m}pi,\[4pt]
    &T_j = left(1-cosfrac{2j+1}{2m}piright)lnleft(2-2cosfrac{2j+1}{2m}piright) - 2 + dfrac{2(m-j)-1}{2m}pisinfrac{2j+1}{2m}pi.
    end{align}
    Taking in account that
    begin{align}
    &sumlimits_{j=0}^{m-1}cosfrac{2j+1}{2m}pi
    = Re sumlimits_{j=0}^{m-1}e^{frac{2j+1}{2m}pi i}
    = Re {large dfrac{1-e^{pi i}}{1-e^{fracpi{m}i}}e^{fracpi{2m}i}}
    =Redfrac {i}{sinfracpi{2m}} = 0,\[4pt]
    &sumlimits_{j=0}^{m-1}sinfrac{2j+1}{2m}pi
    = Im sumlimits_{j=0}^{m-1}e^{frac{2j+1}{2m}pi i}
    =Imdfrac{i}{sinfracpi{2m}} = dfrac1{sinfracpi{2m}},\[4pt]
    &sumlimits_{j=0}^{m-1}frac{2j+1}{2m}pisinfrac{2j+1}{2m}pi
    = dfracpi{2sinfracpi{2m}}
    end{align}
    (see also Wolfram Alpha calculations),



    one can get
    $$boxed{I_{2m} = m(ln2-2) + dfracpi2cscfracpi{2m}
    + sumlimits_{j=0}^{m-1}left(1pmcosfrac{2j+1}{2m}piright)
    logleft(1pmcosfrac{2j+1}{2m}piright).}tag6$$



    $color{brown}{textbf{Closed form for the sum.}}$



    From $(5)-(6)$ should
    $$color{green}{boxed{sumlimits_{j=0}^{m-1}left(1pm c_jright)
    logleft(1pm c_jright)
    = dfrac1{2}left(psi_0left(dfrac{2m+1}{4m}right) - psi_0left(dfrac{1}{4m}right)right) - (m-1)ln2 - dfracpi2cscfracpi{2m}
    ,}}tag7$$
    where
    $$c_j = cosfrac{2j+1}{2m}pi.$$



    Numeric calculations confirm the expression $(6)$ for the both variants of the sign "$pm$".



    Also, numeric calculations of the issue sum and its closed form $(7)$ confirm the correctness of the closed form for the both variants of the sign "$pm$".






    share|cite|improve this answer











    $endgroup$



    $color{brown}{textbf{Preliminary notes.}}$




    Calculation of the required sum looks more complex task than of the
    issue integral.



    Approach with calculations of the sum via the inttegral allows to get close form both of the integral and the sum.




    Denote
    $$I_n=intlimits_0^1log(1+x^n)mathrm dx.tag1$$
    Easy to see that the issue integral is $I_{2m}.$



    $color{brown}{textbf{Closed form of the integrals.}}$



    The first step is the integration by parts:
    $$I_n = xlog(1+x^n)biggr|_0^1-nint_0^1xdfrac{x^{n-1}}{1+x^n}mathrm dx,$$
    $$I_n = log2 - n + nintlimits_0^1dfrac{mathrm dx}{1+x^n}.tag2$$
    (see also Claude Leibovici).



    Then, the substitution
    $$x=e^{-t}tag3$$
    gives
    $$J_n=intlimits_0^1dfrac{mathrm dx}{1+x^n}=intlimits_0^inftydfrac{e^{-t},mathrm dt}{1+e^{-nt}} = sumlimits_{k=0}^infty(-1)^kintlimits_0^infty e^{-(kn+1)t},mathrm dt=sumlimits_{k=0}^inftydfrac{(-1)^k}{kn+1} ,$$
    $$J_n = intlimits_0^1dfrac{mathrm dx}{1+x^n}= dfrac1{2n}left(psi_0left(dfrac{n+1}{2n}right) - psi_0left(dfrac{1}{2n}right)right),tag4$$
    where $psi_0(x)$ is the polygamma function.



    Using $(2),(4),$ easy to get the expression for the OP integral in the form of
    $$boxed{I_{2m} = log2 - 2m +dfrac1{2}left(psi_0left(dfrac{2m+1}{4m}right) - psi_0left(dfrac{1}{4m}right)right).}tag5$$



    $color{brown}{textbf{Testing of the solution.}}$



    Solution $(4)$ can be expressed via the elementary functions for a lot of the given values of n.



    Immediate calculations allow to check obtained expressions in the simple cases
    begin{align}
    &J_1 = intlimits_0^1dfrac{mathrm dx}{1+x} = log 2,\[4pt]
    &J_2 = intlimits_0^1dfrac{mathrm dx}{1+x^2} = arctan 1 = dfracpi4,\[4pt]
    &J_3 = intlimits_0^1dfrac{mathrm dx}{1+x^3}
    = dfrac16intlimits_0^1left(dfrac{2}{1+x} - dfrac{2x-1}{1-x+x^2} + dfrac{3}{1-x+x^2}right),mathrm dx\
    &= left(dfrac13log(1+x) - dfrac16 log(1-x+x^2) + dfrac1{sqrt3}arctandfrac{2x-1}{sqrt3}right)bigg|_0^1
    = dfracpi{3sqrt3} + frac13 log2.\[4pt]
    end{align}
    More hard cases are
    begin{align}
    &J_4 = intlimits_0^1dfrac{mathrm dx}{1+x^4}
    = dfrac{sqrt2}8left(logdfrac{x^2 + sqrt2 x + 1}{x^2 - sqrt2 x + 1}
    - 2arctan(1-sqrt2x) + 2arctan(sqrt2x+1)right)bigg|_0^1\[4pt]
    &= dfrac{sqrt2}8left(logdfrac{(x^2 + sqrt2 x + 1)^2}{1+x^4} + 2arctandfrac{xsqrt2}{1-x^2}right)bigg|_0^1
    = dfrac{sqrt2}8(pi+log(3+2sqrt2)),\[4pt]
    &J_4=dfrac18Bigg(dfrac12sqrt{dfrac{2-sqrt2}{2+sqrt2}}pi+dfrac12sqrt{dfrac{2+sqrt2}{2-sqrt2}}pi
    -2sqrt2left(-log2+dfrac12log(2-sqrt2)right)\[4pt]
    &+2sqrt2left(-log2+dfrac12log(2+sqrt2)right)Bigg)\[4pt]
    &=dfrac18Bigg(dfracpi2left(tandfracpi8+cotdfracpi8right)
    +sqrt2logdfrac{2+sqrt2}{2-sqrt2}Bigg)
    = dfrac{sqrt2}8(pi+log(3+2sqrt2)),
    end{align}
    and $n>4.$



    Numeric calculations of the issue integral $I_{2m}$ and its closed form $(5)$ also confirm the correctness of the closed form.



    $color{brown}{textbf{The alternative approach.}}$



    The alternative approach is considered in OP. Let us repeat it with some differences.



    Polynomial factorization can be presented in the form of



    begin{align}
    &1+x^{2m} = prodlimits_{j=0}^{2m-1}{large left(x-e^{frac{2j+1}{2m}pi i}right)}
    = prodlimits_{j=0}^{m-1}left(x^2-2xcosfrac{2j+1}{2m}pi+1right),
    end{align}
    so
    begin{align}
    &I_{2m} = intlimits_0^1 ln(1+x^{2m}),mathrm dx
    = sumlimits_{j=0}^{m-1} T_j,quadtext{where}\[4pt]
    &T_j = intlimits_0^1 lnleft(x^2-2xcosfrac{2j+1}{2m}pi+1right),mathrm dx,\[4pt]
    &T_j = xlnleft(x^2-2xcosfrac{2j+1}{2m}pi+1right)bigg|_0^1
    - 2intlimits_0^1 dfrac{xleft(x-cosfrac{2j+1}{2m}piright)}{x^2-2xcosfrac{2j+1}{2m}pi+1},mathrm dx\[4pt]
    & = lnleft(2-2cosfrac{2j+1}{2m}piright)
    - 2-2intlimits_0^1 dfrac{xcosfrac{2j+1}{2m}pi-1}{x^2-2xcosfrac{2j+1}{2m}pi+1},mathrm dx\[4pt]
    & = lnleft(2-2cosfrac{2j+1}{2m}piright) - 2
    - 2cosfrac{2j+1}{2m}pi intlimits_0^1 dfrac{x-cosfrac{2j+1}{2m}pi}{x^2-2xcosfrac{2j+1}{2m}pi+1},mathrm dx\[4pt]
    & + 2intlimits_0^1 dfrac{sin^2frac{2j+1}{2m}pi}{left(x-cosfrac{2j+1}{2m}piright)^2+sin^2frac{2j+1}{2m}pi},mathrm dx\[4pt]
    & = lnleft(2-2cosfrac{2j+1}{2m}piright) - 2
    - cosfrac{2j+1}{2m}pi lnleft(x^2-2xcosfrac{2j+1}{2m}pi+1right)bigg|_0^1\[4pt]
    & + 2sinfrac{2j+1}{2m}pi arctandfrac{x-cosfrac{2j+1}{2m}pi}{sinfrac{2j+1}{2m}pi}bigg|_0^1\[4pt]
    & = lnleft(2-2cosfrac{2j+1}{2m}piright) - 2
    - cosfrac{2j+1}{2m}pi lnleft(2-2cosfrac{2j+1}{2m}piright)\[4pt]
    & + 2sinfrac{2j+1}{2m}pi
    left(arctandfrac{1-cosfrac{2j+1}{2m}pi}{sinfrac{2j+1}{2m}pi}
    + arctandfrac{cosfrac{2j+1}{2m}pi}{sinfrac{2j+1}{2m}pi}right)\[4pt]
    & = lnleft(2-2cosfrac{2j+1}{2m}piright) - 2
    - cosfrac{2j+1}{2m}pi lnleft(2-2cosfrac{2j+1}{2m}piright)\[4pt]
    & + 2sinfrac{2j+1}{2m}pi
    arctandfrac{sinfrac{2j+1}{2m}pi}
    {sin^2frac{2j+1}{2m}pi - left(1 - cosfrac{2j+1}{2m}piright)cosfrac{2j+1}{2m}pi}\[4pt]
    & = left(1-cosfrac{2j+1}{2m}piright)lnleft(2-2cosfrac{2j+1}{2m}piright) - 2 + 2sinfrac{2j+1}{2m}piarctancotfrac{2j+1}{4m}pi,\[4pt]
    &T_j = left(1-cosfrac{2j+1}{2m}piright)lnleft(2-2cosfrac{2j+1}{2m}piright) - 2 + dfrac{2(m-j)-1}{2m}pisinfrac{2j+1}{2m}pi.
    end{align}
    Taking in account that
    begin{align}
    &sumlimits_{j=0}^{m-1}cosfrac{2j+1}{2m}pi
    = Re sumlimits_{j=0}^{m-1}e^{frac{2j+1}{2m}pi i}
    = Re {large dfrac{1-e^{pi i}}{1-e^{fracpi{m}i}}e^{fracpi{2m}i}}
    =Redfrac {i}{sinfracpi{2m}} = 0,\[4pt]
    &sumlimits_{j=0}^{m-1}sinfrac{2j+1}{2m}pi
    = Im sumlimits_{j=0}^{m-1}e^{frac{2j+1}{2m}pi i}
    =Imdfrac{i}{sinfracpi{2m}} = dfrac1{sinfracpi{2m}},\[4pt]
    &sumlimits_{j=0}^{m-1}frac{2j+1}{2m}pisinfrac{2j+1}{2m}pi
    = dfracpi{2sinfracpi{2m}}
    end{align}
    (see also Wolfram Alpha calculations),



    one can get
    $$boxed{I_{2m} = m(ln2-2) + dfracpi2cscfracpi{2m}
    + sumlimits_{j=0}^{m-1}left(1pmcosfrac{2j+1}{2m}piright)
    logleft(1pmcosfrac{2j+1}{2m}piright).}tag6$$



    $color{brown}{textbf{Closed form for the sum.}}$



    From $(5)-(6)$ should
    $$color{green}{boxed{sumlimits_{j=0}^{m-1}left(1pm c_jright)
    logleft(1pm c_jright)
    = dfrac1{2}left(psi_0left(dfrac{2m+1}{4m}right) - psi_0left(dfrac{1}{4m}right)right) - (m-1)ln2 - dfracpi2cscfracpi{2m}
    ,}}tag7$$
    where
    $$c_j = cosfrac{2j+1}{2m}pi.$$



    Numeric calculations confirm the expression $(6)$ for the both variants of the sign "$pm$".



    Also, numeric calculations of the issue sum and its closed form $(7)$ confirm the correctness of the closed form for the both variants of the sign "$pm$".







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Mar 28 at 22:54

























    answered Mar 26 at 2:16









    Yuri NegometyanovYuri Negometyanov

    12.7k1729




    12.7k1729








    • 1




      $begingroup$
      Very nice presentation, already before the alternate approach (+1).
      $endgroup$
      – Markus Scheuer
      Mar 28 at 21:46










    • $begingroup$
      @MarkusScheuer Thanks, you are welcome!
      $endgroup$
      – Yuri Negometyanov
      Mar 28 at 22:16














    • 1




      $begingroup$
      Very nice presentation, already before the alternate approach (+1).
      $endgroup$
      – Markus Scheuer
      Mar 28 at 21:46










    • $begingroup$
      @MarkusScheuer Thanks, you are welcome!
      $endgroup$
      – Yuri Negometyanov
      Mar 28 at 22:16








    1




    1




    $begingroup$
    Very nice presentation, already before the alternate approach (+1).
    $endgroup$
    – Markus Scheuer
    Mar 28 at 21:46




    $begingroup$
    Very nice presentation, already before the alternate approach (+1).
    $endgroup$
    – Markus Scheuer
    Mar 28 at 21:46












    $begingroup$
    @MarkusScheuer Thanks, you are welcome!
    $endgroup$
    – Yuri Negometyanov
    Mar 28 at 22:16




    $begingroup$
    @MarkusScheuer Thanks, you are welcome!
    $endgroup$
    – Yuri Negometyanov
    Mar 28 at 22:16











    2












    $begingroup$

    I did it!



    I actually have no idea whether or not this works, but this is how I did it.



    $ninBbb N$



    Define the sequence ${r_k^{(n)}}_{k=1}^{k=n}$ such that
    $$x^n+1=prod_{k=1}^{n}big(x-r^{(n)}_{k}big)$$
    We then know that $$r_k^{(n)}=expbigg[frac{ipi}{n}(2k-1)bigg]$$
    Then we define
    $$S_n={r_k^{(n)}:kin[1,n]capBbb N}$$
    So we have that
    $$frac1{x^n+1}=prod_{rin S_n}frac1{x-r}=prod_{k=1}^nfrac1{x-r_k^{(n)}}$$
    Then we assume that we can write
    $$prod_{rin S_n}frac1{x-r}=sum_{rin S_n}frac{b(r)}{x-r}$$
    Multiplying both sides by $prod_{ain S_n}(x-a)$,
    $$1=sum_{rin S_n}b(r)prod_{ain S_n\ aneq r}(x-a)$$
    So for any $omegain S_n$,
    $$1=b(omega)prod_{ain S_n\ aneq omega}(omega-a)$$
    $$b(omega)=prod_{ain S_n\ aneq omega}frac1{omega-a}$$
    $$b(r_k^{(n)})=prod_{p=1\ pneq k}^nfrac1{r_k^{(n)}-r_p^{(n)}}$$
    So we know that
    $$I_n=int_0^1frac{mathrm{d}x}{1+x^n}=sum_{k=1}^{n}b(r_k^{(n)})int_0^1frac{mathrm{d}x}{x-r_k^{(n)}}$$
    $$I_n=sum_{k=1}^{n}b(r_k^{(n)})logbigg|frac{r_k^{(n)}-1}{r_k^{(n)}}bigg|$$
    $$I_n=sum_{k=1}^{n}logbigg|frac{r_k^{(n)}-1}{r_k^{(n)}}bigg|prod_{p=1\ pneq k}^nfrac1{r_k^{(n)}-r_p^{(n)}}$$
    So we have
    $$int_0^1log(1+x^n)mathrm{d}x=log2-n+nsum_{k=1}^{n}logbigg|frac{r_k^{(n)}-1}{r_k^{(n)}}bigg|prod_{p=1\ pneq k}^nfrac1{r_k^{(n)}-r_p^{(n)}}$$
    along with a plethora of other identities...






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Cheers @clathratus - I will have to have a look over in detail, but on the surface it looks good. Thanks heaps!
      $endgroup$
      – user150203
      Dec 31 '18 at 6:16










    • $begingroup$
      I say this not as a critique of your post, but of my own... do I have defend the set $r_j$ as having size $n$ ?? i.e. that the $c_j$ are distinct...
      $endgroup$
      – user150203
      Dec 31 '18 at 9:47










    • $begingroup$
      Well any polynomial with degree $n$ is going to have $n$ distinct roots, so it would make sense that $|S_n|=n$
      $endgroup$
      – clathratus
      Dec 31 '18 at 20:28
















    2












    $begingroup$

    I did it!



    I actually have no idea whether or not this works, but this is how I did it.



    $ninBbb N$



    Define the sequence ${r_k^{(n)}}_{k=1}^{k=n}$ such that
    $$x^n+1=prod_{k=1}^{n}big(x-r^{(n)}_{k}big)$$
    We then know that $$r_k^{(n)}=expbigg[frac{ipi}{n}(2k-1)bigg]$$
    Then we define
    $$S_n={r_k^{(n)}:kin[1,n]capBbb N}$$
    So we have that
    $$frac1{x^n+1}=prod_{rin S_n}frac1{x-r}=prod_{k=1}^nfrac1{x-r_k^{(n)}}$$
    Then we assume that we can write
    $$prod_{rin S_n}frac1{x-r}=sum_{rin S_n}frac{b(r)}{x-r}$$
    Multiplying both sides by $prod_{ain S_n}(x-a)$,
    $$1=sum_{rin S_n}b(r)prod_{ain S_n\ aneq r}(x-a)$$
    So for any $omegain S_n$,
    $$1=b(omega)prod_{ain S_n\ aneq omega}(omega-a)$$
    $$b(omega)=prod_{ain S_n\ aneq omega}frac1{omega-a}$$
    $$b(r_k^{(n)})=prod_{p=1\ pneq k}^nfrac1{r_k^{(n)}-r_p^{(n)}}$$
    So we know that
    $$I_n=int_0^1frac{mathrm{d}x}{1+x^n}=sum_{k=1}^{n}b(r_k^{(n)})int_0^1frac{mathrm{d}x}{x-r_k^{(n)}}$$
    $$I_n=sum_{k=1}^{n}b(r_k^{(n)})logbigg|frac{r_k^{(n)}-1}{r_k^{(n)}}bigg|$$
    $$I_n=sum_{k=1}^{n}logbigg|frac{r_k^{(n)}-1}{r_k^{(n)}}bigg|prod_{p=1\ pneq k}^nfrac1{r_k^{(n)}-r_p^{(n)}}$$
    So we have
    $$int_0^1log(1+x^n)mathrm{d}x=log2-n+nsum_{k=1}^{n}logbigg|frac{r_k^{(n)}-1}{r_k^{(n)}}bigg|prod_{p=1\ pneq k}^nfrac1{r_k^{(n)}-r_p^{(n)}}$$
    along with a plethora of other identities...






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Cheers @clathratus - I will have to have a look over in detail, but on the surface it looks good. Thanks heaps!
      $endgroup$
      – user150203
      Dec 31 '18 at 6:16










    • $begingroup$
      I say this not as a critique of your post, but of my own... do I have defend the set $r_j$ as having size $n$ ?? i.e. that the $c_j$ are distinct...
      $endgroup$
      – user150203
      Dec 31 '18 at 9:47










    • $begingroup$
      Well any polynomial with degree $n$ is going to have $n$ distinct roots, so it would make sense that $|S_n|=n$
      $endgroup$
      – clathratus
      Dec 31 '18 at 20:28














    2












    2








    2





    $begingroup$

    I did it!



    I actually have no idea whether or not this works, but this is how I did it.



    $ninBbb N$



    Define the sequence ${r_k^{(n)}}_{k=1}^{k=n}$ such that
    $$x^n+1=prod_{k=1}^{n}big(x-r^{(n)}_{k}big)$$
    We then know that $$r_k^{(n)}=expbigg[frac{ipi}{n}(2k-1)bigg]$$
    Then we define
    $$S_n={r_k^{(n)}:kin[1,n]capBbb N}$$
    So we have that
    $$frac1{x^n+1}=prod_{rin S_n}frac1{x-r}=prod_{k=1}^nfrac1{x-r_k^{(n)}}$$
    Then we assume that we can write
    $$prod_{rin S_n}frac1{x-r}=sum_{rin S_n}frac{b(r)}{x-r}$$
    Multiplying both sides by $prod_{ain S_n}(x-a)$,
    $$1=sum_{rin S_n}b(r)prod_{ain S_n\ aneq r}(x-a)$$
    So for any $omegain S_n$,
    $$1=b(omega)prod_{ain S_n\ aneq omega}(omega-a)$$
    $$b(omega)=prod_{ain S_n\ aneq omega}frac1{omega-a}$$
    $$b(r_k^{(n)})=prod_{p=1\ pneq k}^nfrac1{r_k^{(n)}-r_p^{(n)}}$$
    So we know that
    $$I_n=int_0^1frac{mathrm{d}x}{1+x^n}=sum_{k=1}^{n}b(r_k^{(n)})int_0^1frac{mathrm{d}x}{x-r_k^{(n)}}$$
    $$I_n=sum_{k=1}^{n}b(r_k^{(n)})logbigg|frac{r_k^{(n)}-1}{r_k^{(n)}}bigg|$$
    $$I_n=sum_{k=1}^{n}logbigg|frac{r_k^{(n)}-1}{r_k^{(n)}}bigg|prod_{p=1\ pneq k}^nfrac1{r_k^{(n)}-r_p^{(n)}}$$
    So we have
    $$int_0^1log(1+x^n)mathrm{d}x=log2-n+nsum_{k=1}^{n}logbigg|frac{r_k^{(n)}-1}{r_k^{(n)}}bigg|prod_{p=1\ pneq k}^nfrac1{r_k^{(n)}-r_p^{(n)}}$$
    along with a plethora of other identities...






    share|cite|improve this answer











    $endgroup$



    I did it!



    I actually have no idea whether or not this works, but this is how I did it.



    $ninBbb N$



    Define the sequence ${r_k^{(n)}}_{k=1}^{k=n}$ such that
    $$x^n+1=prod_{k=1}^{n}big(x-r^{(n)}_{k}big)$$
    We then know that $$r_k^{(n)}=expbigg[frac{ipi}{n}(2k-1)bigg]$$
    Then we define
    $$S_n={r_k^{(n)}:kin[1,n]capBbb N}$$
    So we have that
    $$frac1{x^n+1}=prod_{rin S_n}frac1{x-r}=prod_{k=1}^nfrac1{x-r_k^{(n)}}$$
    Then we assume that we can write
    $$prod_{rin S_n}frac1{x-r}=sum_{rin S_n}frac{b(r)}{x-r}$$
    Multiplying both sides by $prod_{ain S_n}(x-a)$,
    $$1=sum_{rin S_n}b(r)prod_{ain S_n\ aneq r}(x-a)$$
    So for any $omegain S_n$,
    $$1=b(omega)prod_{ain S_n\ aneq omega}(omega-a)$$
    $$b(omega)=prod_{ain S_n\ aneq omega}frac1{omega-a}$$
    $$b(r_k^{(n)})=prod_{p=1\ pneq k}^nfrac1{r_k^{(n)}-r_p^{(n)}}$$
    So we know that
    $$I_n=int_0^1frac{mathrm{d}x}{1+x^n}=sum_{k=1}^{n}b(r_k^{(n)})int_0^1frac{mathrm{d}x}{x-r_k^{(n)}}$$
    $$I_n=sum_{k=1}^{n}b(r_k^{(n)})logbigg|frac{r_k^{(n)}-1}{r_k^{(n)}}bigg|$$
    $$I_n=sum_{k=1}^{n}logbigg|frac{r_k^{(n)}-1}{r_k^{(n)}}bigg|prod_{p=1\ pneq k}^nfrac1{r_k^{(n)}-r_p^{(n)}}$$
    So we have
    $$int_0^1log(1+x^n)mathrm{d}x=log2-n+nsum_{k=1}^{n}logbigg|frac{r_k^{(n)}-1}{r_k^{(n)}}bigg|prod_{p=1\ pneq k}^nfrac1{r_k^{(n)}-r_p^{(n)}}$$
    along with a plethora of other identities...







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 11 at 21:56

























    answered Dec 31 '18 at 5:52









    clathratusclathratus

    5,1441439




    5,1441439








    • 1




      $begingroup$
      Cheers @clathratus - I will have to have a look over in detail, but on the surface it looks good. Thanks heaps!
      $endgroup$
      – user150203
      Dec 31 '18 at 6:16










    • $begingroup$
      I say this not as a critique of your post, but of my own... do I have defend the set $r_j$ as having size $n$ ?? i.e. that the $c_j$ are distinct...
      $endgroup$
      – user150203
      Dec 31 '18 at 9:47










    • $begingroup$
      Well any polynomial with degree $n$ is going to have $n$ distinct roots, so it would make sense that $|S_n|=n$
      $endgroup$
      – clathratus
      Dec 31 '18 at 20:28














    • 1




      $begingroup$
      Cheers @clathratus - I will have to have a look over in detail, but on the surface it looks good. Thanks heaps!
      $endgroup$
      – user150203
      Dec 31 '18 at 6:16










    • $begingroup$
      I say this not as a critique of your post, but of my own... do I have defend the set $r_j$ as having size $n$ ?? i.e. that the $c_j$ are distinct...
      $endgroup$
      – user150203
      Dec 31 '18 at 9:47










    • $begingroup$
      Well any polynomial with degree $n$ is going to have $n$ distinct roots, so it would make sense that $|S_n|=n$
      $endgroup$
      – clathratus
      Dec 31 '18 at 20:28








    1




    1




    $begingroup$
    Cheers @clathratus - I will have to have a look over in detail, but on the surface it looks good. Thanks heaps!
    $endgroup$
    – user150203
    Dec 31 '18 at 6:16




    $begingroup$
    Cheers @clathratus - I will have to have a look over in detail, but on the surface it looks good. Thanks heaps!
    $endgroup$
    – user150203
    Dec 31 '18 at 6:16












    $begingroup$
    I say this not as a critique of your post, but of my own... do I have defend the set $r_j$ as having size $n$ ?? i.e. that the $c_j$ are distinct...
    $endgroup$
    – user150203
    Dec 31 '18 at 9:47




    $begingroup$
    I say this not as a critique of your post, but of my own... do I have defend the set $r_j$ as having size $n$ ?? i.e. that the $c_j$ are distinct...
    $endgroup$
    – user150203
    Dec 31 '18 at 9:47












    $begingroup$
    Well any polynomial with degree $n$ is going to have $n$ distinct roots, so it would make sense that $|S_n|=n$
    $endgroup$
    – clathratus
    Dec 31 '18 at 20:28




    $begingroup$
    Well any polynomial with degree $n$ is going to have $n$ distinct roots, so it would make sense that $|S_n|=n$
    $endgroup$
    – clathratus
    Dec 31 '18 at 20:28











    1












    $begingroup$

    Here's another, quicker, method (I also don't know if this one works)



    Using the same $r_k^{(n)}$ as last time, we apply the $logprod_{i}a_i=sum_ilog a_i$ property to see that
    $$log(1+x^n)=logprod_{k=1}^{n}(x-r_k^{(n)})=sum_{k=1}^{n}log(x-r_k^{(n)})$$
    So
    $$I_n=int_0^1log(1+x^n)mathrm dx=sum_{k=1}^{n}int_0^1log(x-r_k^{(n)})mathrm dx$$
    This last integral boils down to
    $$begin{align}
    int_0^1log(x-a)mathrm dx=&alogfrac{a}{1+a}+log(1-a)-1\
    =&logfrac{a^a(1-a)}{e(1+a)^a}
    end{align}$$
    So
    $$I_n=sum_{rin S_n}logfrac{r^r(1-r)}{e(1+r)^r}$$
    And you know how I love product representations, so we again use $logprod_{i}a_i=sum_ilog a_i$ to see that
    $$
    I_n=logprod_{rin S_n}frac{r^r(1-r)}{e(1+r)^r}\
    prod_{rin S_n}frac{r^r(1-r)}{(1+r)^r}=exp(n+I_n)
    $$
    Which I just think is really neat.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Here's another, quicker, method (I also don't know if this one works)



      Using the same $r_k^{(n)}$ as last time, we apply the $logprod_{i}a_i=sum_ilog a_i$ property to see that
      $$log(1+x^n)=logprod_{k=1}^{n}(x-r_k^{(n)})=sum_{k=1}^{n}log(x-r_k^{(n)})$$
      So
      $$I_n=int_0^1log(1+x^n)mathrm dx=sum_{k=1}^{n}int_0^1log(x-r_k^{(n)})mathrm dx$$
      This last integral boils down to
      $$begin{align}
      int_0^1log(x-a)mathrm dx=&alogfrac{a}{1+a}+log(1-a)-1\
      =&logfrac{a^a(1-a)}{e(1+a)^a}
      end{align}$$
      So
      $$I_n=sum_{rin S_n}logfrac{r^r(1-r)}{e(1+r)^r}$$
      And you know how I love product representations, so we again use $logprod_{i}a_i=sum_ilog a_i$ to see that
      $$
      I_n=logprod_{rin S_n}frac{r^r(1-r)}{e(1+r)^r}\
      prod_{rin S_n}frac{r^r(1-r)}{(1+r)^r}=exp(n+I_n)
      $$
      Which I just think is really neat.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Here's another, quicker, method (I also don't know if this one works)



        Using the same $r_k^{(n)}$ as last time, we apply the $logprod_{i}a_i=sum_ilog a_i$ property to see that
        $$log(1+x^n)=logprod_{k=1}^{n}(x-r_k^{(n)})=sum_{k=1}^{n}log(x-r_k^{(n)})$$
        So
        $$I_n=int_0^1log(1+x^n)mathrm dx=sum_{k=1}^{n}int_0^1log(x-r_k^{(n)})mathrm dx$$
        This last integral boils down to
        $$begin{align}
        int_0^1log(x-a)mathrm dx=&alogfrac{a}{1+a}+log(1-a)-1\
        =&logfrac{a^a(1-a)}{e(1+a)^a}
        end{align}$$
        So
        $$I_n=sum_{rin S_n}logfrac{r^r(1-r)}{e(1+r)^r}$$
        And you know how I love product representations, so we again use $logprod_{i}a_i=sum_ilog a_i$ to see that
        $$
        I_n=logprod_{rin S_n}frac{r^r(1-r)}{e(1+r)^r}\
        prod_{rin S_n}frac{r^r(1-r)}{(1+r)^r}=exp(n+I_n)
        $$
        Which I just think is really neat.






        share|cite|improve this answer









        $endgroup$



        Here's another, quicker, method (I also don't know if this one works)



        Using the same $r_k^{(n)}$ as last time, we apply the $logprod_{i}a_i=sum_ilog a_i$ property to see that
        $$log(1+x^n)=logprod_{k=1}^{n}(x-r_k^{(n)})=sum_{k=1}^{n}log(x-r_k^{(n)})$$
        So
        $$I_n=int_0^1log(1+x^n)mathrm dx=sum_{k=1}^{n}int_0^1log(x-r_k^{(n)})mathrm dx$$
        This last integral boils down to
        $$begin{align}
        int_0^1log(x-a)mathrm dx=&alogfrac{a}{1+a}+log(1-a)-1\
        =&logfrac{a^a(1-a)}{e(1+a)^a}
        end{align}$$
        So
        $$I_n=sum_{rin S_n}logfrac{r^r(1-r)}{e(1+r)^r}$$
        And you know how I love product representations, so we again use $logprod_{i}a_i=sum_ilog a_i$ to see that
        $$
        I_n=logprod_{rin S_n}frac{r^r(1-r)}{e(1+r)^r}\
        prod_{rin S_n}frac{r^r(1-r)}{(1+r)^r}=exp(n+I_n)
        $$
        Which I just think is really neat.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 3 at 23:58









        clathratusclathratus

        5,1441439




        5,1441439






























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