Integral of $ln(tanh(x))$
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I'd like a hint toward how I could evaluate this definite integral. I'm aware it's likely to be non elementary and I haven't found a way to evaluate it yet:$$int_0^infty ln(tanh(x)),,mathrm{d}x$$
If you're curious where this came from, I was looking at an integral involving $ln(sin(x))$ and I thought of this one.
Thanks.
calculus integration hyperbolic-functions
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add a comment |
$begingroup$
I'd like a hint toward how I could evaluate this definite integral. I'm aware it's likely to be non elementary and I haven't found a way to evaluate it yet:$$int_0^infty ln(tanh(x)),,mathrm{d}x$$
If you're curious where this came from, I was looking at an integral involving $ln(sin(x))$ and I thought of this one.
Thanks.
calculus integration hyperbolic-functions
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2
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Related: math.stackexchange.com/questions/2182256/integral-of-ln-cosh-x Did you find a solution for $int_0^{infty}log(sinh(x))$? Mathematica gives $int_0^{infty}log(tanh(x))=-frac{pi^2}{8}$ so there might be a nice way to solve this one
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– Václav Mordvinov
Dec 5 '18 at 20:13
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I believe that both $int_0^infty{lnleft(sinhleft(xright)right)},,mathrm{d}x$ and $int_0^infty{lnleft(coshleft(xright)right)},,mathrm{d}x$ diverge but that result for tanh seems promising for a really satisfying solution.
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– Sheepe
Dec 5 '18 at 20:19
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In the light of some of the solutions below, why not now consider a generalization of your original integral: $int_0^infty left(ln(tanh(x)) right)^n, dx$ where n is a positive integer.
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– James Arathoon
Dec 8 '18 at 17:32
add a comment |
$begingroup$
I'd like a hint toward how I could evaluate this definite integral. I'm aware it's likely to be non elementary and I haven't found a way to evaluate it yet:$$int_0^infty ln(tanh(x)),,mathrm{d}x$$
If you're curious where this came from, I was looking at an integral involving $ln(sin(x))$ and I thought of this one.
Thanks.
calculus integration hyperbolic-functions
$endgroup$
I'd like a hint toward how I could evaluate this definite integral. I'm aware it's likely to be non elementary and I haven't found a way to evaluate it yet:$$int_0^infty ln(tanh(x)),,mathrm{d}x$$
If you're curious where this came from, I was looking at an integral involving $ln(sin(x))$ and I thought of this one.
Thanks.
calculus integration hyperbolic-functions
calculus integration hyperbolic-functions
edited Dec 5 '18 at 20:08
amWhy
1
1
asked Dec 5 '18 at 20:04
SheepeSheepe
433
433
2
$begingroup$
Related: math.stackexchange.com/questions/2182256/integral-of-ln-cosh-x Did you find a solution for $int_0^{infty}log(sinh(x))$? Mathematica gives $int_0^{infty}log(tanh(x))=-frac{pi^2}{8}$ so there might be a nice way to solve this one
$endgroup$
– Václav Mordvinov
Dec 5 '18 at 20:13
$begingroup$
I believe that both $int_0^infty{lnleft(sinhleft(xright)right)},,mathrm{d}x$ and $int_0^infty{lnleft(coshleft(xright)right)},,mathrm{d}x$ diverge but that result for tanh seems promising for a really satisfying solution.
$endgroup$
– Sheepe
Dec 5 '18 at 20:19
$begingroup$
In the light of some of the solutions below, why not now consider a generalization of your original integral: $int_0^infty left(ln(tanh(x)) right)^n, dx$ where n is a positive integer.
$endgroup$
– James Arathoon
Dec 8 '18 at 17:32
add a comment |
2
$begingroup$
Related: math.stackexchange.com/questions/2182256/integral-of-ln-cosh-x Did you find a solution for $int_0^{infty}log(sinh(x))$? Mathematica gives $int_0^{infty}log(tanh(x))=-frac{pi^2}{8}$ so there might be a nice way to solve this one
$endgroup$
– Václav Mordvinov
Dec 5 '18 at 20:13
$begingroup$
I believe that both $int_0^infty{lnleft(sinhleft(xright)right)},,mathrm{d}x$ and $int_0^infty{lnleft(coshleft(xright)right)},,mathrm{d}x$ diverge but that result for tanh seems promising for a really satisfying solution.
$endgroup$
– Sheepe
Dec 5 '18 at 20:19
$begingroup$
In the light of some of the solutions below, why not now consider a generalization of your original integral: $int_0^infty left(ln(tanh(x)) right)^n, dx$ where n is a positive integer.
$endgroup$
– James Arathoon
Dec 8 '18 at 17:32
2
2
$begingroup$
Related: math.stackexchange.com/questions/2182256/integral-of-ln-cosh-x Did you find a solution for $int_0^{infty}log(sinh(x))$? Mathematica gives $int_0^{infty}log(tanh(x))=-frac{pi^2}{8}$ so there might be a nice way to solve this one
$endgroup$
– Václav Mordvinov
Dec 5 '18 at 20:13
$begingroup$
Related: math.stackexchange.com/questions/2182256/integral-of-ln-cosh-x Did you find a solution for $int_0^{infty}log(sinh(x))$? Mathematica gives $int_0^{infty}log(tanh(x))=-frac{pi^2}{8}$ so there might be a nice way to solve this one
$endgroup$
– Václav Mordvinov
Dec 5 '18 at 20:13
$begingroup$
I believe that both $int_0^infty{lnleft(sinhleft(xright)right)},,mathrm{d}x$ and $int_0^infty{lnleft(coshleft(xright)right)},,mathrm{d}x$ diverge but that result for tanh seems promising for a really satisfying solution.
$endgroup$
– Sheepe
Dec 5 '18 at 20:19
$begingroup$
I believe that both $int_0^infty{lnleft(sinhleft(xright)right)},,mathrm{d}x$ and $int_0^infty{lnleft(coshleft(xright)right)},,mathrm{d}x$ diverge but that result for tanh seems promising for a really satisfying solution.
$endgroup$
– Sheepe
Dec 5 '18 at 20:19
$begingroup$
In the light of some of the solutions below, why not now consider a generalization of your original integral: $int_0^infty left(ln(tanh(x)) right)^n, dx$ where n is a positive integer.
$endgroup$
– James Arathoon
Dec 8 '18 at 17:32
$begingroup$
In the light of some of the solutions below, why not now consider a generalization of your original integral: $int_0^infty left(ln(tanh(x)) right)^n, dx$ where n is a positive integer.
$endgroup$
– James Arathoon
Dec 8 '18 at 17:32
add a comment |
4 Answers
4
active
oldest
votes
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By applying the definition of the hyperbolic tangent function in terms of the exponential we obtain
$$begin{align}
int_0^{infty}log(tanh(x))dx&=int_0^{infty}logleft(frac{e^x-e^{-x}}{e^x+e^{-x}}right)dx \
&=int_0^{infty}logleft(frac{1-e^{-2x}}{1+e^{-2x}}right)dx\
&=int_0^{infty}logleft(1-e^{-2x}right)-logleft(1+e^{-2x}right)dx
end{align}$$
Now by expanding the logarithm as a series $($!$)$ we further get
$$begin{align}
int_0^{infty}logleft(1-e^{-2x}right)-logleft(1+e^{-2x}right)dx&=int_0^{infty}-sum_{n=1}^{infty}frac1{n}e^{-2nx}+sum_{n=1}^{infty}frac{(-1)^n}{n}e^{-2nx}dx\
&=-sum_{n=1}^{infty}left[-frac{e^{-2nx}}{2n^2}right]_0^{infty}+sum_{n=1}^{infty}left[(-1)^{n+1}frac{e^{-2nx}}{2n^2}right]_0^{infty}\
&=-sum_{n=1}^{infty}frac1{2n^2}-sum_{n=1}^{infty}frac{(-1)^{n+1}}{2n^2}\
&=-frac12zeta(2)-frac12eta(2)\
&=-frac12frac{pi^2}6-frac12frac{pi^2}{12}\
&=-frac{pi^2}8
end{align}$$
Which is the desired result. $zeta(s)$ denotes the Riemann Zeta Function and $eta(s)$ the Dirichlet Eta Function respectively for which the values are known.
Anyway this solution is kind of unsteady since I cannot really justify the validity of the power series of the logarithmic functions $($which is normally restricted to $|x|<1$ $)$ nor the termwise integration. Nevertheless it leads to the right solution.
EDIT:
As illustrated by ComplexYetTrivial within the comments the validity of the series expansion of the logarithm is guaranteed due the the fact that $e^{-2x}<1$ for all $x>0$ from where the series converges. Whereas the termwise integration is justified by the monotone dominate convergence theorem. Thus my proposed solution seems to be entirely fine.
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Absolutely amazing, thank you very much.
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– Sheepe
Dec 5 '18 at 20:40
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@Sheepe Thank you. Howsoever I am still unsure about the convergence radius of the logarithmic series expansion which is dismissed withtin this solution^^
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– mrtaurho
Dec 5 '18 at 20:41
3
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Since $mathrm{e}^{-2 x} < 1$ for $x>0$, the series expansion is valid. Termwise integration is justified by the monotone/dominated convergence theorem, so the solution is perfectly fine!
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– ComplexYetTrivial
Dec 5 '18 at 20:42
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I did get up until the series substitution in my own attempts, but out of my own concerns of the same idea I couldn't see how to continue but it seems to work out nevertheless.
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– Sheepe
Dec 5 '18 at 20:43
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@ComplexYetTrivial This completes my evaluation and clears my doubts! Thank you! Firstly I was sceptical about my own attempt but out of pure curiosity I finished the evaluation and indeed it turned out to work :)
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– mrtaurho
Dec 5 '18 at 20:50
add a comment |
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Through the substitution $x=text{arctanh}(t)$ we have $I=int_{0}^{+infty}logtanh x,dx = int_{0}^{1}frac{log t}{1-t^2},dt$.
Since $int_{0}^{1}t^{2n}log(t),dt = -frac{1}{(2n+1)^2}$ we have
$$ I = -sum_{ngeq 0}frac{1}{(2n+1)^2}=-left[zeta(2)-frac{1}{4}zeta(2)right]=-frac{3}{4}cdotfrac{pi^2}{6}=color{red}{-frac{pi^2}{8}}. $$
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add a comment |
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Here is a slight variation on a theme.
Making use of the result $tanh^2 x = 1 - mbox{sech}^2 x$, we can write the integral as
$$I = frac{1}{2} int_0^infty ln (1 - text{sech}^2 x) , dx.$$
Setting $text{sech}^2 x mapsto x$ gives
$$I = frac{1}{4} int_0^1 frac{ln (1 - x)}{x sqrt{1 - x}} , dx.$$
There are many ways to evaluate this integral. One way is by enforcing a substitution of $x mapsto 1 - x^2$. Doing so we arrive at
$$I = int_0^1 frac{ln x}{1 - x^2} , dx,$$
which is exactly the same point @Jack D'Aurizio arrived at in his solution.
Departing from Jack, we now employ a self-similar substitution of $u = dfrac{1 - x}{1 + x}$.
Thus
$$I = frac{1}{2} int_0^1 frac{1}{u} ln left (frac{1 - u}{1 + u} right ) , du = frac{1}{2} int_0^1 frac{ln (1 - u)}{u} , du - frac{1}{2} int_0^1 frac{ln (1 + u)}{u} , du.$$
In the second of these integrals let $u mapsto -u$
begin{align*}
I &= frac{1}{2} int_0^1 frac{ln (1 - u)}{u} , du - frac{1}{2} int_0^{-1} frac{ln (1 - u)}{u} , du\
&= -frac{1}{2} text{Li}_2 (1) + frac{1}{2} text{Li}_2 (-1)\
&= -frac{1}{2} cdot frac{pi^2}{6} + frac{1}{2} cdot -frac{pi^2}{12}\
&= -frac{pi^2}{8},
end{align*}
where use of the dilogarithm function has been made.
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add a comment |
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Another option: $$lntanh x=-2operatorname{artanh}exp -2x=-2sum_{kge 0}frac{exp -(4k+2)x}{2k+1},$$so$$int_0^inftylntanh x dx=-sum_{kge 0}frac{1}{(2k+1)^2}=-frac{pi^2}{8}.$$
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add a comment |
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4 Answers
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4 Answers
4
active
oldest
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active
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$begingroup$
By applying the definition of the hyperbolic tangent function in terms of the exponential we obtain
$$begin{align}
int_0^{infty}log(tanh(x))dx&=int_0^{infty}logleft(frac{e^x-e^{-x}}{e^x+e^{-x}}right)dx \
&=int_0^{infty}logleft(frac{1-e^{-2x}}{1+e^{-2x}}right)dx\
&=int_0^{infty}logleft(1-e^{-2x}right)-logleft(1+e^{-2x}right)dx
end{align}$$
Now by expanding the logarithm as a series $($!$)$ we further get
$$begin{align}
int_0^{infty}logleft(1-e^{-2x}right)-logleft(1+e^{-2x}right)dx&=int_0^{infty}-sum_{n=1}^{infty}frac1{n}e^{-2nx}+sum_{n=1}^{infty}frac{(-1)^n}{n}e^{-2nx}dx\
&=-sum_{n=1}^{infty}left[-frac{e^{-2nx}}{2n^2}right]_0^{infty}+sum_{n=1}^{infty}left[(-1)^{n+1}frac{e^{-2nx}}{2n^2}right]_0^{infty}\
&=-sum_{n=1}^{infty}frac1{2n^2}-sum_{n=1}^{infty}frac{(-1)^{n+1}}{2n^2}\
&=-frac12zeta(2)-frac12eta(2)\
&=-frac12frac{pi^2}6-frac12frac{pi^2}{12}\
&=-frac{pi^2}8
end{align}$$
Which is the desired result. $zeta(s)$ denotes the Riemann Zeta Function and $eta(s)$ the Dirichlet Eta Function respectively for which the values are known.
Anyway this solution is kind of unsteady since I cannot really justify the validity of the power series of the logarithmic functions $($which is normally restricted to $|x|<1$ $)$ nor the termwise integration. Nevertheless it leads to the right solution.
EDIT:
As illustrated by ComplexYetTrivial within the comments the validity of the series expansion of the logarithm is guaranteed due the the fact that $e^{-2x}<1$ for all $x>0$ from where the series converges. Whereas the termwise integration is justified by the monotone dominate convergence theorem. Thus my proposed solution seems to be entirely fine.
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$begingroup$
Absolutely amazing, thank you very much.
$endgroup$
– Sheepe
Dec 5 '18 at 20:40
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@Sheepe Thank you. Howsoever I am still unsure about the convergence radius of the logarithmic series expansion which is dismissed withtin this solution^^
$endgroup$
– mrtaurho
Dec 5 '18 at 20:41
3
$begingroup$
Since $mathrm{e}^{-2 x} < 1$ for $x>0$, the series expansion is valid. Termwise integration is justified by the monotone/dominated convergence theorem, so the solution is perfectly fine!
$endgroup$
– ComplexYetTrivial
Dec 5 '18 at 20:42
$begingroup$
I did get up until the series substitution in my own attempts, but out of my own concerns of the same idea I couldn't see how to continue but it seems to work out nevertheless.
$endgroup$
– Sheepe
Dec 5 '18 at 20:43
$begingroup$
@ComplexYetTrivial This completes my evaluation and clears my doubts! Thank you! Firstly I was sceptical about my own attempt but out of pure curiosity I finished the evaluation and indeed it turned out to work :)
$endgroup$
– mrtaurho
Dec 5 '18 at 20:50
add a comment |
$begingroup$
By applying the definition of the hyperbolic tangent function in terms of the exponential we obtain
$$begin{align}
int_0^{infty}log(tanh(x))dx&=int_0^{infty}logleft(frac{e^x-e^{-x}}{e^x+e^{-x}}right)dx \
&=int_0^{infty}logleft(frac{1-e^{-2x}}{1+e^{-2x}}right)dx\
&=int_0^{infty}logleft(1-e^{-2x}right)-logleft(1+e^{-2x}right)dx
end{align}$$
Now by expanding the logarithm as a series $($!$)$ we further get
$$begin{align}
int_0^{infty}logleft(1-e^{-2x}right)-logleft(1+e^{-2x}right)dx&=int_0^{infty}-sum_{n=1}^{infty}frac1{n}e^{-2nx}+sum_{n=1}^{infty}frac{(-1)^n}{n}e^{-2nx}dx\
&=-sum_{n=1}^{infty}left[-frac{e^{-2nx}}{2n^2}right]_0^{infty}+sum_{n=1}^{infty}left[(-1)^{n+1}frac{e^{-2nx}}{2n^2}right]_0^{infty}\
&=-sum_{n=1}^{infty}frac1{2n^2}-sum_{n=1}^{infty}frac{(-1)^{n+1}}{2n^2}\
&=-frac12zeta(2)-frac12eta(2)\
&=-frac12frac{pi^2}6-frac12frac{pi^2}{12}\
&=-frac{pi^2}8
end{align}$$
Which is the desired result. $zeta(s)$ denotes the Riemann Zeta Function and $eta(s)$ the Dirichlet Eta Function respectively for which the values are known.
Anyway this solution is kind of unsteady since I cannot really justify the validity of the power series of the logarithmic functions $($which is normally restricted to $|x|<1$ $)$ nor the termwise integration. Nevertheless it leads to the right solution.
EDIT:
As illustrated by ComplexYetTrivial within the comments the validity of the series expansion of the logarithm is guaranteed due the the fact that $e^{-2x}<1$ for all $x>0$ from where the series converges. Whereas the termwise integration is justified by the monotone dominate convergence theorem. Thus my proposed solution seems to be entirely fine.
$endgroup$
$begingroup$
Absolutely amazing, thank you very much.
$endgroup$
– Sheepe
Dec 5 '18 at 20:40
$begingroup$
@Sheepe Thank you. Howsoever I am still unsure about the convergence radius of the logarithmic series expansion which is dismissed withtin this solution^^
$endgroup$
– mrtaurho
Dec 5 '18 at 20:41
3
$begingroup$
Since $mathrm{e}^{-2 x} < 1$ for $x>0$, the series expansion is valid. Termwise integration is justified by the monotone/dominated convergence theorem, so the solution is perfectly fine!
$endgroup$
– ComplexYetTrivial
Dec 5 '18 at 20:42
$begingroup$
I did get up until the series substitution in my own attempts, but out of my own concerns of the same idea I couldn't see how to continue but it seems to work out nevertheless.
$endgroup$
– Sheepe
Dec 5 '18 at 20:43
$begingroup$
@ComplexYetTrivial This completes my evaluation and clears my doubts! Thank you! Firstly I was sceptical about my own attempt but out of pure curiosity I finished the evaluation and indeed it turned out to work :)
$endgroup$
– mrtaurho
Dec 5 '18 at 20:50
add a comment |
$begingroup$
By applying the definition of the hyperbolic tangent function in terms of the exponential we obtain
$$begin{align}
int_0^{infty}log(tanh(x))dx&=int_0^{infty}logleft(frac{e^x-e^{-x}}{e^x+e^{-x}}right)dx \
&=int_0^{infty}logleft(frac{1-e^{-2x}}{1+e^{-2x}}right)dx\
&=int_0^{infty}logleft(1-e^{-2x}right)-logleft(1+e^{-2x}right)dx
end{align}$$
Now by expanding the logarithm as a series $($!$)$ we further get
$$begin{align}
int_0^{infty}logleft(1-e^{-2x}right)-logleft(1+e^{-2x}right)dx&=int_0^{infty}-sum_{n=1}^{infty}frac1{n}e^{-2nx}+sum_{n=1}^{infty}frac{(-1)^n}{n}e^{-2nx}dx\
&=-sum_{n=1}^{infty}left[-frac{e^{-2nx}}{2n^2}right]_0^{infty}+sum_{n=1}^{infty}left[(-1)^{n+1}frac{e^{-2nx}}{2n^2}right]_0^{infty}\
&=-sum_{n=1}^{infty}frac1{2n^2}-sum_{n=1}^{infty}frac{(-1)^{n+1}}{2n^2}\
&=-frac12zeta(2)-frac12eta(2)\
&=-frac12frac{pi^2}6-frac12frac{pi^2}{12}\
&=-frac{pi^2}8
end{align}$$
Which is the desired result. $zeta(s)$ denotes the Riemann Zeta Function and $eta(s)$ the Dirichlet Eta Function respectively for which the values are known.
Anyway this solution is kind of unsteady since I cannot really justify the validity of the power series of the logarithmic functions $($which is normally restricted to $|x|<1$ $)$ nor the termwise integration. Nevertheless it leads to the right solution.
EDIT:
As illustrated by ComplexYetTrivial within the comments the validity of the series expansion of the logarithm is guaranteed due the the fact that $e^{-2x}<1$ for all $x>0$ from where the series converges. Whereas the termwise integration is justified by the monotone dominate convergence theorem. Thus my proposed solution seems to be entirely fine.
$endgroup$
By applying the definition of the hyperbolic tangent function in terms of the exponential we obtain
$$begin{align}
int_0^{infty}log(tanh(x))dx&=int_0^{infty}logleft(frac{e^x-e^{-x}}{e^x+e^{-x}}right)dx \
&=int_0^{infty}logleft(frac{1-e^{-2x}}{1+e^{-2x}}right)dx\
&=int_0^{infty}logleft(1-e^{-2x}right)-logleft(1+e^{-2x}right)dx
end{align}$$
Now by expanding the logarithm as a series $($!$)$ we further get
$$begin{align}
int_0^{infty}logleft(1-e^{-2x}right)-logleft(1+e^{-2x}right)dx&=int_0^{infty}-sum_{n=1}^{infty}frac1{n}e^{-2nx}+sum_{n=1}^{infty}frac{(-1)^n}{n}e^{-2nx}dx\
&=-sum_{n=1}^{infty}left[-frac{e^{-2nx}}{2n^2}right]_0^{infty}+sum_{n=1}^{infty}left[(-1)^{n+1}frac{e^{-2nx}}{2n^2}right]_0^{infty}\
&=-sum_{n=1}^{infty}frac1{2n^2}-sum_{n=1}^{infty}frac{(-1)^{n+1}}{2n^2}\
&=-frac12zeta(2)-frac12eta(2)\
&=-frac12frac{pi^2}6-frac12frac{pi^2}{12}\
&=-frac{pi^2}8
end{align}$$
Which is the desired result. $zeta(s)$ denotes the Riemann Zeta Function and $eta(s)$ the Dirichlet Eta Function respectively for which the values are known.
Anyway this solution is kind of unsteady since I cannot really justify the validity of the power series of the logarithmic functions $($which is normally restricted to $|x|<1$ $)$ nor the termwise integration. Nevertheless it leads to the right solution.
EDIT:
As illustrated by ComplexYetTrivial within the comments the validity of the series expansion of the logarithm is guaranteed due the the fact that $e^{-2x}<1$ for all $x>0$ from where the series converges. Whereas the termwise integration is justified by the monotone dominate convergence theorem. Thus my proposed solution seems to be entirely fine.
edited Dec 6 '18 at 17:06
answered Dec 5 '18 at 20:37
mrtaurhomrtaurho
5,65051540
5,65051540
$begingroup$
Absolutely amazing, thank you very much.
$endgroup$
– Sheepe
Dec 5 '18 at 20:40
$begingroup$
@Sheepe Thank you. Howsoever I am still unsure about the convergence radius of the logarithmic series expansion which is dismissed withtin this solution^^
$endgroup$
– mrtaurho
Dec 5 '18 at 20:41
3
$begingroup$
Since $mathrm{e}^{-2 x} < 1$ for $x>0$, the series expansion is valid. Termwise integration is justified by the monotone/dominated convergence theorem, so the solution is perfectly fine!
$endgroup$
– ComplexYetTrivial
Dec 5 '18 at 20:42
$begingroup$
I did get up until the series substitution in my own attempts, but out of my own concerns of the same idea I couldn't see how to continue but it seems to work out nevertheless.
$endgroup$
– Sheepe
Dec 5 '18 at 20:43
$begingroup$
@ComplexYetTrivial This completes my evaluation and clears my doubts! Thank you! Firstly I was sceptical about my own attempt but out of pure curiosity I finished the evaluation and indeed it turned out to work :)
$endgroup$
– mrtaurho
Dec 5 '18 at 20:50
add a comment |
$begingroup$
Absolutely amazing, thank you very much.
$endgroup$
– Sheepe
Dec 5 '18 at 20:40
$begingroup$
@Sheepe Thank you. Howsoever I am still unsure about the convergence radius of the logarithmic series expansion which is dismissed withtin this solution^^
$endgroup$
– mrtaurho
Dec 5 '18 at 20:41
3
$begingroup$
Since $mathrm{e}^{-2 x} < 1$ for $x>0$, the series expansion is valid. Termwise integration is justified by the monotone/dominated convergence theorem, so the solution is perfectly fine!
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– ComplexYetTrivial
Dec 5 '18 at 20:42
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I did get up until the series substitution in my own attempts, but out of my own concerns of the same idea I couldn't see how to continue but it seems to work out nevertheless.
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– Sheepe
Dec 5 '18 at 20:43
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@ComplexYetTrivial This completes my evaluation and clears my doubts! Thank you! Firstly I was sceptical about my own attempt but out of pure curiosity I finished the evaluation and indeed it turned out to work :)
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– mrtaurho
Dec 5 '18 at 20:50
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Absolutely amazing, thank you very much.
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– Sheepe
Dec 5 '18 at 20:40
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Absolutely amazing, thank you very much.
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– Sheepe
Dec 5 '18 at 20:40
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@Sheepe Thank you. Howsoever I am still unsure about the convergence radius of the logarithmic series expansion which is dismissed withtin this solution
^^
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– mrtaurho
Dec 5 '18 at 20:41
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@Sheepe Thank you. Howsoever I am still unsure about the convergence radius of the logarithmic series expansion which is dismissed withtin this solution
^^
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– mrtaurho
Dec 5 '18 at 20:41
3
3
$begingroup$
Since $mathrm{e}^{-2 x} < 1$ for $x>0$, the series expansion is valid. Termwise integration is justified by the monotone/dominated convergence theorem, so the solution is perfectly fine!
$endgroup$
– ComplexYetTrivial
Dec 5 '18 at 20:42
$begingroup$
Since $mathrm{e}^{-2 x} < 1$ for $x>0$, the series expansion is valid. Termwise integration is justified by the monotone/dominated convergence theorem, so the solution is perfectly fine!
$endgroup$
– ComplexYetTrivial
Dec 5 '18 at 20:42
$begingroup$
I did get up until the series substitution in my own attempts, but out of my own concerns of the same idea I couldn't see how to continue but it seems to work out nevertheless.
$endgroup$
– Sheepe
Dec 5 '18 at 20:43
$begingroup$
I did get up until the series substitution in my own attempts, but out of my own concerns of the same idea I couldn't see how to continue but it seems to work out nevertheless.
$endgroup$
– Sheepe
Dec 5 '18 at 20:43
$begingroup$
@ComplexYetTrivial This completes my evaluation and clears my doubts! Thank you! Firstly I was sceptical about my own attempt but out of pure curiosity I finished the evaluation and indeed it turned out to work :)
$endgroup$
– mrtaurho
Dec 5 '18 at 20:50
$begingroup$
@ComplexYetTrivial This completes my evaluation and clears my doubts! Thank you! Firstly I was sceptical about my own attempt but out of pure curiosity I finished the evaluation and indeed it turned out to work :)
$endgroup$
– mrtaurho
Dec 5 '18 at 20:50
add a comment |
$begingroup$
Through the substitution $x=text{arctanh}(t)$ we have $I=int_{0}^{+infty}logtanh x,dx = int_{0}^{1}frac{log t}{1-t^2},dt$.
Since $int_{0}^{1}t^{2n}log(t),dt = -frac{1}{(2n+1)^2}$ we have
$$ I = -sum_{ngeq 0}frac{1}{(2n+1)^2}=-left[zeta(2)-frac{1}{4}zeta(2)right]=-frac{3}{4}cdotfrac{pi^2}{6}=color{red}{-frac{pi^2}{8}}. $$
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add a comment |
$begingroup$
Through the substitution $x=text{arctanh}(t)$ we have $I=int_{0}^{+infty}logtanh x,dx = int_{0}^{1}frac{log t}{1-t^2},dt$.
Since $int_{0}^{1}t^{2n}log(t),dt = -frac{1}{(2n+1)^2}$ we have
$$ I = -sum_{ngeq 0}frac{1}{(2n+1)^2}=-left[zeta(2)-frac{1}{4}zeta(2)right]=-frac{3}{4}cdotfrac{pi^2}{6}=color{red}{-frac{pi^2}{8}}. $$
$endgroup$
add a comment |
$begingroup$
Through the substitution $x=text{arctanh}(t)$ we have $I=int_{0}^{+infty}logtanh x,dx = int_{0}^{1}frac{log t}{1-t^2},dt$.
Since $int_{0}^{1}t^{2n}log(t),dt = -frac{1}{(2n+1)^2}$ we have
$$ I = -sum_{ngeq 0}frac{1}{(2n+1)^2}=-left[zeta(2)-frac{1}{4}zeta(2)right]=-frac{3}{4}cdotfrac{pi^2}{6}=color{red}{-frac{pi^2}{8}}. $$
$endgroup$
Through the substitution $x=text{arctanh}(t)$ we have $I=int_{0}^{+infty}logtanh x,dx = int_{0}^{1}frac{log t}{1-t^2},dt$.
Since $int_{0}^{1}t^{2n}log(t),dt = -frac{1}{(2n+1)^2}$ we have
$$ I = -sum_{ngeq 0}frac{1}{(2n+1)^2}=-left[zeta(2)-frac{1}{4}zeta(2)right]=-frac{3}{4}cdotfrac{pi^2}{6}=color{red}{-frac{pi^2}{8}}. $$
answered Dec 5 '18 at 21:30
Jack D'AurizioJack D'Aurizio
290k33283665
290k33283665
add a comment |
add a comment |
$begingroup$
Here is a slight variation on a theme.
Making use of the result $tanh^2 x = 1 - mbox{sech}^2 x$, we can write the integral as
$$I = frac{1}{2} int_0^infty ln (1 - text{sech}^2 x) , dx.$$
Setting $text{sech}^2 x mapsto x$ gives
$$I = frac{1}{4} int_0^1 frac{ln (1 - x)}{x sqrt{1 - x}} , dx.$$
There are many ways to evaluate this integral. One way is by enforcing a substitution of $x mapsto 1 - x^2$. Doing so we arrive at
$$I = int_0^1 frac{ln x}{1 - x^2} , dx,$$
which is exactly the same point @Jack D'Aurizio arrived at in his solution.
Departing from Jack, we now employ a self-similar substitution of $u = dfrac{1 - x}{1 + x}$.
Thus
$$I = frac{1}{2} int_0^1 frac{1}{u} ln left (frac{1 - u}{1 + u} right ) , du = frac{1}{2} int_0^1 frac{ln (1 - u)}{u} , du - frac{1}{2} int_0^1 frac{ln (1 + u)}{u} , du.$$
In the second of these integrals let $u mapsto -u$
begin{align*}
I &= frac{1}{2} int_0^1 frac{ln (1 - u)}{u} , du - frac{1}{2} int_0^{-1} frac{ln (1 - u)}{u} , du\
&= -frac{1}{2} text{Li}_2 (1) + frac{1}{2} text{Li}_2 (-1)\
&= -frac{1}{2} cdot frac{pi^2}{6} + frac{1}{2} cdot -frac{pi^2}{12}\
&= -frac{pi^2}{8},
end{align*}
where use of the dilogarithm function has been made.
$endgroup$
add a comment |
$begingroup$
Here is a slight variation on a theme.
Making use of the result $tanh^2 x = 1 - mbox{sech}^2 x$, we can write the integral as
$$I = frac{1}{2} int_0^infty ln (1 - text{sech}^2 x) , dx.$$
Setting $text{sech}^2 x mapsto x$ gives
$$I = frac{1}{4} int_0^1 frac{ln (1 - x)}{x sqrt{1 - x}} , dx.$$
There are many ways to evaluate this integral. One way is by enforcing a substitution of $x mapsto 1 - x^2$. Doing so we arrive at
$$I = int_0^1 frac{ln x}{1 - x^2} , dx,$$
which is exactly the same point @Jack D'Aurizio arrived at in his solution.
Departing from Jack, we now employ a self-similar substitution of $u = dfrac{1 - x}{1 + x}$.
Thus
$$I = frac{1}{2} int_0^1 frac{1}{u} ln left (frac{1 - u}{1 + u} right ) , du = frac{1}{2} int_0^1 frac{ln (1 - u)}{u} , du - frac{1}{2} int_0^1 frac{ln (1 + u)}{u} , du.$$
In the second of these integrals let $u mapsto -u$
begin{align*}
I &= frac{1}{2} int_0^1 frac{ln (1 - u)}{u} , du - frac{1}{2} int_0^{-1} frac{ln (1 - u)}{u} , du\
&= -frac{1}{2} text{Li}_2 (1) + frac{1}{2} text{Li}_2 (-1)\
&= -frac{1}{2} cdot frac{pi^2}{6} + frac{1}{2} cdot -frac{pi^2}{12}\
&= -frac{pi^2}{8},
end{align*}
where use of the dilogarithm function has been made.
$endgroup$
add a comment |
$begingroup$
Here is a slight variation on a theme.
Making use of the result $tanh^2 x = 1 - mbox{sech}^2 x$, we can write the integral as
$$I = frac{1}{2} int_0^infty ln (1 - text{sech}^2 x) , dx.$$
Setting $text{sech}^2 x mapsto x$ gives
$$I = frac{1}{4} int_0^1 frac{ln (1 - x)}{x sqrt{1 - x}} , dx.$$
There are many ways to evaluate this integral. One way is by enforcing a substitution of $x mapsto 1 - x^2$. Doing so we arrive at
$$I = int_0^1 frac{ln x}{1 - x^2} , dx,$$
which is exactly the same point @Jack D'Aurizio arrived at in his solution.
Departing from Jack, we now employ a self-similar substitution of $u = dfrac{1 - x}{1 + x}$.
Thus
$$I = frac{1}{2} int_0^1 frac{1}{u} ln left (frac{1 - u}{1 + u} right ) , du = frac{1}{2} int_0^1 frac{ln (1 - u)}{u} , du - frac{1}{2} int_0^1 frac{ln (1 + u)}{u} , du.$$
In the second of these integrals let $u mapsto -u$
begin{align*}
I &= frac{1}{2} int_0^1 frac{ln (1 - u)}{u} , du - frac{1}{2} int_0^{-1} frac{ln (1 - u)}{u} , du\
&= -frac{1}{2} text{Li}_2 (1) + frac{1}{2} text{Li}_2 (-1)\
&= -frac{1}{2} cdot frac{pi^2}{6} + frac{1}{2} cdot -frac{pi^2}{12}\
&= -frac{pi^2}{8},
end{align*}
where use of the dilogarithm function has been made.
$endgroup$
Here is a slight variation on a theme.
Making use of the result $tanh^2 x = 1 - mbox{sech}^2 x$, we can write the integral as
$$I = frac{1}{2} int_0^infty ln (1 - text{sech}^2 x) , dx.$$
Setting $text{sech}^2 x mapsto x$ gives
$$I = frac{1}{4} int_0^1 frac{ln (1 - x)}{x sqrt{1 - x}} , dx.$$
There are many ways to evaluate this integral. One way is by enforcing a substitution of $x mapsto 1 - x^2$. Doing so we arrive at
$$I = int_0^1 frac{ln x}{1 - x^2} , dx,$$
which is exactly the same point @Jack D'Aurizio arrived at in his solution.
Departing from Jack, we now employ a self-similar substitution of $u = dfrac{1 - x}{1 + x}$.
Thus
$$I = frac{1}{2} int_0^1 frac{1}{u} ln left (frac{1 - u}{1 + u} right ) , du = frac{1}{2} int_0^1 frac{ln (1 - u)}{u} , du - frac{1}{2} int_0^1 frac{ln (1 + u)}{u} , du.$$
In the second of these integrals let $u mapsto -u$
begin{align*}
I &= frac{1}{2} int_0^1 frac{ln (1 - u)}{u} , du - frac{1}{2} int_0^{-1} frac{ln (1 - u)}{u} , du\
&= -frac{1}{2} text{Li}_2 (1) + frac{1}{2} text{Li}_2 (-1)\
&= -frac{1}{2} cdot frac{pi^2}{6} + frac{1}{2} cdot -frac{pi^2}{12}\
&= -frac{pi^2}{8},
end{align*}
where use of the dilogarithm function has been made.
answered Dec 6 '18 at 3:25
omegadotomegadot
6,3672828
6,3672828
add a comment |
add a comment |
$begingroup$
Another option: $$lntanh x=-2operatorname{artanh}exp -2x=-2sum_{kge 0}frac{exp -(4k+2)x}{2k+1},$$so$$int_0^inftylntanh x dx=-sum_{kge 0}frac{1}{(2k+1)^2}=-frac{pi^2}{8}.$$
$endgroup$
add a comment |
$begingroup$
Another option: $$lntanh x=-2operatorname{artanh}exp -2x=-2sum_{kge 0}frac{exp -(4k+2)x}{2k+1},$$so$$int_0^inftylntanh x dx=-sum_{kge 0}frac{1}{(2k+1)^2}=-frac{pi^2}{8}.$$
$endgroup$
add a comment |
$begingroup$
Another option: $$lntanh x=-2operatorname{artanh}exp -2x=-2sum_{kge 0}frac{exp -(4k+2)x}{2k+1},$$so$$int_0^inftylntanh x dx=-sum_{kge 0}frac{1}{(2k+1)^2}=-frac{pi^2}{8}.$$
$endgroup$
Another option: $$lntanh x=-2operatorname{artanh}exp -2x=-2sum_{kge 0}frac{exp -(4k+2)x}{2k+1},$$so$$int_0^inftylntanh x dx=-sum_{kge 0}frac{1}{(2k+1)^2}=-frac{pi^2}{8}.$$
answered Dec 6 '18 at 17:24
J.G.J.G.
28.7k22845
28.7k22845
add a comment |
add a comment |
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$begingroup$
Related: math.stackexchange.com/questions/2182256/integral-of-ln-cosh-x Did you find a solution for $int_0^{infty}log(sinh(x))$? Mathematica gives $int_0^{infty}log(tanh(x))=-frac{pi^2}{8}$ so there might be a nice way to solve this one
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– Václav Mordvinov
Dec 5 '18 at 20:13
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I believe that both $int_0^infty{lnleft(sinhleft(xright)right)},,mathrm{d}x$ and $int_0^infty{lnleft(coshleft(xright)right)},,mathrm{d}x$ diverge but that result for tanh seems promising for a really satisfying solution.
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– Sheepe
Dec 5 '18 at 20:19
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In the light of some of the solutions below, why not now consider a generalization of your original integral: $int_0^infty left(ln(tanh(x)) right)^n, dx$ where n is a positive integer.
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– James Arathoon
Dec 8 '18 at 17:32