Proving $Imoperatorname{Li}_2(sqrt i(sqrt 2-1))=frac34G+frac18piln(sqrt2-1)$












3












$begingroup$


$newcommand{Li}{operatorname{Li}_2}$




I found, numerically, that $$ImLi(sqrt i(sqrt 2-1))=frac34G+frac18piln(sqrt2-1).$$
How can we prove it?




My attempt of proving this equation:
Using identity $$Li(x)=int_0^1frac{x}{xt-1}ln tdt,$$
we can deduce$$begin{align}ImLi(sqrt i(sqrt 2-1))&=frac1{2i}int_0^1left(frac{sqrt i(sqrt2-1)}{sqrt i(sqrt2-1)t-1}-frac{sqrt {-i}(sqrt2-1)}{sqrt {-i}(sqrt2-1)t-1}right)ln tdt\
&=int_0^1frac{2-sqrt{2}}{left(4 sqrt{2}-6right) t^2-2 left(sqrt{2}-2right) t-2}ln tdt\
&=int_0^{2-sqrt2}-frac{1}{u^2-2u+2}lnfrac u{2-sqrt 2}du\
&=frac18piln(2-sqrt2)-int_{-1}^{1-sqrt2}frac{ln(v+1)}{v^2+1}dv\
&=frac18piln(2-sqrt2)-int_{pi/8}^{pi/4}ln(1-tan x)dx\
&=frac18piln(sqrt2-1)-int_{pi/8}^{pi/4}lnsec x+lnsinleft(fracpi4-xright)dx\
end{align}$$
I have no idea how to deal with the log-trig integral.






calculus definite-integrals polylogarithm







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$endgroup$












  • $begingroup$
    Try reducing the lower limit of your integral to zero and then use the identity ; $$G = int_0^{π/4}log cot theta dtheta $$
    $endgroup$
    – Awe Kumar Jha
    Dec 7 '18 at 11:49












  • $begingroup$
    $logsin x$ and $logcos x$ have well-known Fourier series, and you just have to perform a termwise integration of them.
    $endgroup$
    – Jack D'Aurizio
    Dec 7 '18 at 17:06










  • $begingroup$
    What is $sqrt{i}$?
    $endgroup$
    – FDP
    Dec 7 '18 at 18:14










  • $begingroup$
    @FDP $sqrt i= e^{pi i/4}=frac{1+i}{sqrt 2}$.
    $endgroup$
    – Kemono Chen
    Dec 8 '18 at 0:55










  • $begingroup$
    why not begin{align}sqrt i= -e^{pi i/4}?end{align} if $x_0^2=a$ then $(-x_0)^2=a$.
    $endgroup$
    – FDP
    Dec 8 '18 at 11:30


















3












$begingroup$


$newcommand{Li}{operatorname{Li}_2}$




I found, numerically, that $$ImLi(sqrt i(sqrt 2-1))=frac34G+frac18piln(sqrt2-1).$$
How can we prove it?




My attempt of proving this equation:
Using identity $$Li(x)=int_0^1frac{x}{xt-1}ln tdt,$$
we can deduce$$begin{align}ImLi(sqrt i(sqrt 2-1))&=frac1{2i}int_0^1left(frac{sqrt i(sqrt2-1)}{sqrt i(sqrt2-1)t-1}-frac{sqrt {-i}(sqrt2-1)}{sqrt {-i}(sqrt2-1)t-1}right)ln tdt\
&=int_0^1frac{2-sqrt{2}}{left(4 sqrt{2}-6right) t^2-2 left(sqrt{2}-2right) t-2}ln tdt\
&=int_0^{2-sqrt2}-frac{1}{u^2-2u+2}lnfrac u{2-sqrt 2}du\
&=frac18piln(2-sqrt2)-int_{-1}^{1-sqrt2}frac{ln(v+1)}{v^2+1}dv\
&=frac18piln(2-sqrt2)-int_{pi/8}^{pi/4}ln(1-tan x)dx\
&=frac18piln(sqrt2-1)-int_{pi/8}^{pi/4}lnsec x+lnsinleft(fracpi4-xright)dx\
end{align}$$
I have no idea how to deal with the log-trig integral.






calculus definite-integrals polylogarithm







share|cite|improve this question









$endgroup$












  • $begingroup$
    Try reducing the lower limit of your integral to zero and then use the identity ; $$G = int_0^{π/4}log cot theta dtheta $$
    $endgroup$
    – Awe Kumar Jha
    Dec 7 '18 at 11:49












  • $begingroup$
    $logsin x$ and $logcos x$ have well-known Fourier series, and you just have to perform a termwise integration of them.
    $endgroup$
    – Jack D'Aurizio
    Dec 7 '18 at 17:06










  • $begingroup$
    What is $sqrt{i}$?
    $endgroup$
    – FDP
    Dec 7 '18 at 18:14










  • $begingroup$
    @FDP $sqrt i= e^{pi i/4}=frac{1+i}{sqrt 2}$.
    $endgroup$
    – Kemono Chen
    Dec 8 '18 at 0:55










  • $begingroup$
    why not begin{align}sqrt i= -e^{pi i/4}?end{align} if $x_0^2=a$ then $(-x_0)^2=a$.
    $endgroup$
    – FDP
    Dec 8 '18 at 11:30
















3












3








3


0



$begingroup$


$newcommand{Li}{operatorname{Li}_2}$




I found, numerically, that $$ImLi(sqrt i(sqrt 2-1))=frac34G+frac18piln(sqrt2-1).$$
How can we prove it?




My attempt of proving this equation:
Using identity $$Li(x)=int_0^1frac{x}{xt-1}ln tdt,$$
we can deduce$$begin{align}ImLi(sqrt i(sqrt 2-1))&=frac1{2i}int_0^1left(frac{sqrt i(sqrt2-1)}{sqrt i(sqrt2-1)t-1}-frac{sqrt {-i}(sqrt2-1)}{sqrt {-i}(sqrt2-1)t-1}right)ln tdt\
&=int_0^1frac{2-sqrt{2}}{left(4 sqrt{2}-6right) t^2-2 left(sqrt{2}-2right) t-2}ln tdt\
&=int_0^{2-sqrt2}-frac{1}{u^2-2u+2}lnfrac u{2-sqrt 2}du\
&=frac18piln(2-sqrt2)-int_{-1}^{1-sqrt2}frac{ln(v+1)}{v^2+1}dv\
&=frac18piln(2-sqrt2)-int_{pi/8}^{pi/4}ln(1-tan x)dx\
&=frac18piln(sqrt2-1)-int_{pi/8}^{pi/4}lnsec x+lnsinleft(fracpi4-xright)dx\
end{align}$$
I have no idea how to deal with the log-trig integral.






calculus definite-integrals polylogarithm







share|cite|improve this question









$endgroup$




$newcommand{Li}{operatorname{Li}_2}$




I found, numerically, that $$ImLi(sqrt i(sqrt 2-1))=frac34G+frac18piln(sqrt2-1).$$
How can we prove it?




My attempt of proving this equation:
Using identity $$Li(x)=int_0^1frac{x}{xt-1}ln tdt,$$
we can deduce$$begin{align}ImLi(sqrt i(sqrt 2-1))&=frac1{2i}int_0^1left(frac{sqrt i(sqrt2-1)}{sqrt i(sqrt2-1)t-1}-frac{sqrt {-i}(sqrt2-1)}{sqrt {-i}(sqrt2-1)t-1}right)ln tdt\
&=int_0^1frac{2-sqrt{2}}{left(4 sqrt{2}-6right) t^2-2 left(sqrt{2}-2right) t-2}ln tdt\
&=int_0^{2-sqrt2}-frac{1}{u^2-2u+2}lnfrac u{2-sqrt 2}du\
&=frac18piln(2-sqrt2)-int_{-1}^{1-sqrt2}frac{ln(v+1)}{v^2+1}dv\
&=frac18piln(2-sqrt2)-int_{pi/8}^{pi/4}ln(1-tan x)dx\
&=frac18piln(sqrt2-1)-int_{pi/8}^{pi/4}lnsec x+lnsinleft(fracpi4-xright)dx\
end{align}$$
I have no idea how to deal with the log-trig integral.






calculus definite-integrals polylogarithm




calculus definite-integrals polylogarithm






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share|cite|improve this question











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asked Dec 7 '18 at 11:17









Kemono ChenKemono Chen

3,1941844




3,1941844












  • $begingroup$
    Try reducing the lower limit of your integral to zero and then use the identity ; $$G = int_0^{π/4}log cot theta dtheta $$
    $endgroup$
    – Awe Kumar Jha
    Dec 7 '18 at 11:49












  • $begingroup$
    $logsin x$ and $logcos x$ have well-known Fourier series, and you just have to perform a termwise integration of them.
    $endgroup$
    – Jack D'Aurizio
    Dec 7 '18 at 17:06










  • $begingroup$
    What is $sqrt{i}$?
    $endgroup$
    – FDP
    Dec 7 '18 at 18:14










  • $begingroup$
    @FDP $sqrt i= e^{pi i/4}=frac{1+i}{sqrt 2}$.
    $endgroup$
    – Kemono Chen
    Dec 8 '18 at 0:55










  • $begingroup$
    why not begin{align}sqrt i= -e^{pi i/4}?end{align} if $x_0^2=a$ then $(-x_0)^2=a$.
    $endgroup$
    – FDP
    Dec 8 '18 at 11:30




















  • $begingroup$
    Try reducing the lower limit of your integral to zero and then use the identity ; $$G = int_0^{π/4}log cot theta dtheta $$
    $endgroup$
    – Awe Kumar Jha
    Dec 7 '18 at 11:49












  • $begingroup$
    $logsin x$ and $logcos x$ have well-known Fourier series, and you just have to perform a termwise integration of them.
    $endgroup$
    – Jack D'Aurizio
    Dec 7 '18 at 17:06










  • $begingroup$
    What is $sqrt{i}$?
    $endgroup$
    – FDP
    Dec 7 '18 at 18:14










  • $begingroup$
    @FDP $sqrt i= e^{pi i/4}=frac{1+i}{sqrt 2}$.
    $endgroup$
    – Kemono Chen
    Dec 8 '18 at 0:55










  • $begingroup$
    why not begin{align}sqrt i= -e^{pi i/4}?end{align} if $x_0^2=a$ then $(-x_0)^2=a$.
    $endgroup$
    – FDP
    Dec 8 '18 at 11:30


















$begingroup$
Try reducing the lower limit of your integral to zero and then use the identity ; $$G = int_0^{π/4}log cot theta dtheta $$
$endgroup$
– Awe Kumar Jha
Dec 7 '18 at 11:49






$begingroup$
Try reducing the lower limit of your integral to zero and then use the identity ; $$G = int_0^{π/4}log cot theta dtheta $$
$endgroup$
– Awe Kumar Jha
Dec 7 '18 at 11:49














$begingroup$
$logsin x$ and $logcos x$ have well-known Fourier series, and you just have to perform a termwise integration of them.
$endgroup$
– Jack D'Aurizio
Dec 7 '18 at 17:06




$begingroup$
$logsin x$ and $logcos x$ have well-known Fourier series, and you just have to perform a termwise integration of them.
$endgroup$
– Jack D'Aurizio
Dec 7 '18 at 17:06












$begingroup$
What is $sqrt{i}$?
$endgroup$
– FDP
Dec 7 '18 at 18:14




$begingroup$
What is $sqrt{i}$?
$endgroup$
– FDP
Dec 7 '18 at 18:14












$begingroup$
@FDP $sqrt i= e^{pi i/4}=frac{1+i}{sqrt 2}$.
$endgroup$
– Kemono Chen
Dec 8 '18 at 0:55




$begingroup$
@FDP $sqrt i= e^{pi i/4}=frac{1+i}{sqrt 2}$.
$endgroup$
– Kemono Chen
Dec 8 '18 at 0:55












$begingroup$
why not begin{align}sqrt i= -e^{pi i/4}?end{align} if $x_0^2=a$ then $(-x_0)^2=a$.
$endgroup$
– FDP
Dec 8 '18 at 11:30






$begingroup$
why not begin{align}sqrt i= -e^{pi i/4}?end{align} if $x_0^2=a$ then $(-x_0)^2=a$.
$endgroup$
– FDP
Dec 8 '18 at 11:30












1 Answer
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3












$begingroup$

$newcommand{Cl}{operatorname{Cl}}$In order to dodge the extensive usage of the Fourier Series Expansions I will use the Clausen Function $Cl_2(z)$ to shorten things up; nevertheless the result will remain the same as one could get going all along the long way. Anyway, the only difficulty that remains after your attempt is the evaluation of the following integral




$$mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}xtag1$$




First of all I will do a bit of reshaping to actually apply useful formulae involving the Clausen Function. Therefore split up the first integral and enforce the substitution $x+fracpi4mapsto x$ within the second integral to get



$$begin{align*}
mathfrak{I}&=-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x\
&=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{pi/8}^{pi/4}logleft(cos x+frac pi4right)mathrm{d}x\
&=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{3pi/8}^{pi/2}log(cos x)mathrm{d}x\
&=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_0^{pi/2}log(cos x)mathrm{d}x+int_0^{3pi/8}log(cos x)mathrm{d}x
end{align*}$$



Now it is time to apply the first useful formula of the Clausen Function, namely




$$int_0^t log(cos x)mathrm{d}x~=~frac12Cl_2(pi-2t)-tlog(2)tag2$$




Formula $(2)$ can be shown rather easy be utilizing the well-known Fourier Series Expansion of $log(cos x)$ combined with the series representation of the $Cl_2(z)$ function. However, with this knowledge we can rewrite the integrals from above in terms of the Clausen Function to get



$$begin{align*}
mathfrak{I}&=frac12Cl_2left(pi-2fracpi4right)-frac12Cl_2left(pi-2fracpi8right)-frac12Cl_2left(pi-2fracpi2right)+frac12Cl_2left(pi-2frac{3pi}8right)\
&~~~underbrace{-fracpi4log(2)+fracpi8log(2)+fracpi2log(2)-frac{3pi}8log(2)}_{=0}\
&=frac12left[Cl_2left(fracpi2right)-Cl_2left(0right)+Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)right]
end{align*}$$



We are almost done! It is time to throw some more important formulae in: firstly the already mentioned series representation and secondly the so-called Duplication Formula




$$begin{align*}
Cl_2(z)~&=~sum_{n=1}^infty frac{sin(nz)}{n^2}tag3\
Cl_2(2z)~&=~2Cl_2(z)-2Cl_2(pi-z)tag4
end{align*}$$




From $(3)$ we can direcetly conclude that $Cl_2left(fracpi2right)=G$ and that $Cl_2(0)=0$ where $G$ denotes Catalan's Constant. Using the $(4)$ with $z=fracpi4$ we get a representation for the other terms from above. Putting this all together gives us the final value



$$begin{align*}
mathfrak{I}&=frac12left[underbrace{Cl_2left(fracpi2right)}_{=G}-underbrace{Cl_2left(0right)}_{=0}+underbrace{Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)}_{=frac G2}right]\
&=frac12left[G+frac G2right]
end{align*}$$




$$therefore~mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x~=~frac34 G$$




I recommend to study the Clausen Function hence it reduces the number of caculations needed for linear logarithmo-trigonometric integrals tremendously. If you are feeling uncomfortable with a part of the proof let me know and I will try to clear your doubts.






share|cite|improve this answer











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    1 Answer
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    1 Answer
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    oldest

    votes






    active

    oldest

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    3












    $begingroup$

    $newcommand{Cl}{operatorname{Cl}}$In order to dodge the extensive usage of the Fourier Series Expansions I will use the Clausen Function $Cl_2(z)$ to shorten things up; nevertheless the result will remain the same as one could get going all along the long way. Anyway, the only difficulty that remains after your attempt is the evaluation of the following integral




    $$mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}xtag1$$




    First of all I will do a bit of reshaping to actually apply useful formulae involving the Clausen Function. Therefore split up the first integral and enforce the substitution $x+fracpi4mapsto x$ within the second integral to get



    $$begin{align*}
    mathfrak{I}&=-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x\
    &=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{pi/8}^{pi/4}logleft(cos x+frac pi4right)mathrm{d}x\
    &=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{3pi/8}^{pi/2}log(cos x)mathrm{d}x\
    &=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_0^{pi/2}log(cos x)mathrm{d}x+int_0^{3pi/8}log(cos x)mathrm{d}x
    end{align*}$$



    Now it is time to apply the first useful formula of the Clausen Function, namely




    $$int_0^t log(cos x)mathrm{d}x~=~frac12Cl_2(pi-2t)-tlog(2)tag2$$




    Formula $(2)$ can be shown rather easy be utilizing the well-known Fourier Series Expansion of $log(cos x)$ combined with the series representation of the $Cl_2(z)$ function. However, with this knowledge we can rewrite the integrals from above in terms of the Clausen Function to get



    $$begin{align*}
    mathfrak{I}&=frac12Cl_2left(pi-2fracpi4right)-frac12Cl_2left(pi-2fracpi8right)-frac12Cl_2left(pi-2fracpi2right)+frac12Cl_2left(pi-2frac{3pi}8right)\
    &~~~underbrace{-fracpi4log(2)+fracpi8log(2)+fracpi2log(2)-frac{3pi}8log(2)}_{=0}\
    &=frac12left[Cl_2left(fracpi2right)-Cl_2left(0right)+Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)right]
    end{align*}$$



    We are almost done! It is time to throw some more important formulae in: firstly the already mentioned series representation and secondly the so-called Duplication Formula




    $$begin{align*}
    Cl_2(z)~&=~sum_{n=1}^infty frac{sin(nz)}{n^2}tag3\
    Cl_2(2z)~&=~2Cl_2(z)-2Cl_2(pi-z)tag4
    end{align*}$$




    From $(3)$ we can direcetly conclude that $Cl_2left(fracpi2right)=G$ and that $Cl_2(0)=0$ where $G$ denotes Catalan's Constant. Using the $(4)$ with $z=fracpi4$ we get a representation for the other terms from above. Putting this all together gives us the final value



    $$begin{align*}
    mathfrak{I}&=frac12left[underbrace{Cl_2left(fracpi2right)}_{=G}-underbrace{Cl_2left(0right)}_{=0}+underbrace{Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)}_{=frac G2}right]\
    &=frac12left[G+frac G2right]
    end{align*}$$




    $$therefore~mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x~=~frac34 G$$




    I recommend to study the Clausen Function hence it reduces the number of caculations needed for linear logarithmo-trigonometric integrals tremendously. If you are feeling uncomfortable with a part of the proof let me know and I will try to clear your doubts.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      $newcommand{Cl}{operatorname{Cl}}$In order to dodge the extensive usage of the Fourier Series Expansions I will use the Clausen Function $Cl_2(z)$ to shorten things up; nevertheless the result will remain the same as one could get going all along the long way. Anyway, the only difficulty that remains after your attempt is the evaluation of the following integral




      $$mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}xtag1$$




      First of all I will do a bit of reshaping to actually apply useful formulae involving the Clausen Function. Therefore split up the first integral and enforce the substitution $x+fracpi4mapsto x$ within the second integral to get



      $$begin{align*}
      mathfrak{I}&=-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x\
      &=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{pi/8}^{pi/4}logleft(cos x+frac pi4right)mathrm{d}x\
      &=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{3pi/8}^{pi/2}log(cos x)mathrm{d}x\
      &=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_0^{pi/2}log(cos x)mathrm{d}x+int_0^{3pi/8}log(cos x)mathrm{d}x
      end{align*}$$



      Now it is time to apply the first useful formula of the Clausen Function, namely




      $$int_0^t log(cos x)mathrm{d}x~=~frac12Cl_2(pi-2t)-tlog(2)tag2$$




      Formula $(2)$ can be shown rather easy be utilizing the well-known Fourier Series Expansion of $log(cos x)$ combined with the series representation of the $Cl_2(z)$ function. However, with this knowledge we can rewrite the integrals from above in terms of the Clausen Function to get



      $$begin{align*}
      mathfrak{I}&=frac12Cl_2left(pi-2fracpi4right)-frac12Cl_2left(pi-2fracpi8right)-frac12Cl_2left(pi-2fracpi2right)+frac12Cl_2left(pi-2frac{3pi}8right)\
      &~~~underbrace{-fracpi4log(2)+fracpi8log(2)+fracpi2log(2)-frac{3pi}8log(2)}_{=0}\
      &=frac12left[Cl_2left(fracpi2right)-Cl_2left(0right)+Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)right]
      end{align*}$$



      We are almost done! It is time to throw some more important formulae in: firstly the already mentioned series representation and secondly the so-called Duplication Formula




      $$begin{align*}
      Cl_2(z)~&=~sum_{n=1}^infty frac{sin(nz)}{n^2}tag3\
      Cl_2(2z)~&=~2Cl_2(z)-2Cl_2(pi-z)tag4
      end{align*}$$




      From $(3)$ we can direcetly conclude that $Cl_2left(fracpi2right)=G$ and that $Cl_2(0)=0$ where $G$ denotes Catalan's Constant. Using the $(4)$ with $z=fracpi4$ we get a representation for the other terms from above. Putting this all together gives us the final value



      $$begin{align*}
      mathfrak{I}&=frac12left[underbrace{Cl_2left(fracpi2right)}_{=G}-underbrace{Cl_2left(0right)}_{=0}+underbrace{Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)}_{=frac G2}right]\
      &=frac12left[G+frac G2right]
      end{align*}$$




      $$therefore~mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x~=~frac34 G$$




      I recommend to study the Clausen Function hence it reduces the number of caculations needed for linear logarithmo-trigonometric integrals tremendously. If you are feeling uncomfortable with a part of the proof let me know and I will try to clear your doubts.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        $newcommand{Cl}{operatorname{Cl}}$In order to dodge the extensive usage of the Fourier Series Expansions I will use the Clausen Function $Cl_2(z)$ to shorten things up; nevertheless the result will remain the same as one could get going all along the long way. Anyway, the only difficulty that remains after your attempt is the evaluation of the following integral




        $$mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}xtag1$$




        First of all I will do a bit of reshaping to actually apply useful formulae involving the Clausen Function. Therefore split up the first integral and enforce the substitution $x+fracpi4mapsto x$ within the second integral to get



        $$begin{align*}
        mathfrak{I}&=-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x\
        &=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{pi/8}^{pi/4}logleft(cos x+frac pi4right)mathrm{d}x\
        &=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{3pi/8}^{pi/2}log(cos x)mathrm{d}x\
        &=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_0^{pi/2}log(cos x)mathrm{d}x+int_0^{3pi/8}log(cos x)mathrm{d}x
        end{align*}$$



        Now it is time to apply the first useful formula of the Clausen Function, namely




        $$int_0^t log(cos x)mathrm{d}x~=~frac12Cl_2(pi-2t)-tlog(2)tag2$$




        Formula $(2)$ can be shown rather easy be utilizing the well-known Fourier Series Expansion of $log(cos x)$ combined with the series representation of the $Cl_2(z)$ function. However, with this knowledge we can rewrite the integrals from above in terms of the Clausen Function to get



        $$begin{align*}
        mathfrak{I}&=frac12Cl_2left(pi-2fracpi4right)-frac12Cl_2left(pi-2fracpi8right)-frac12Cl_2left(pi-2fracpi2right)+frac12Cl_2left(pi-2frac{3pi}8right)\
        &~~~underbrace{-fracpi4log(2)+fracpi8log(2)+fracpi2log(2)-frac{3pi}8log(2)}_{=0}\
        &=frac12left[Cl_2left(fracpi2right)-Cl_2left(0right)+Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)right]
        end{align*}$$



        We are almost done! It is time to throw some more important formulae in: firstly the already mentioned series representation and secondly the so-called Duplication Formula




        $$begin{align*}
        Cl_2(z)~&=~sum_{n=1}^infty frac{sin(nz)}{n^2}tag3\
        Cl_2(2z)~&=~2Cl_2(z)-2Cl_2(pi-z)tag4
        end{align*}$$




        From $(3)$ we can direcetly conclude that $Cl_2left(fracpi2right)=G$ and that $Cl_2(0)=0$ where $G$ denotes Catalan's Constant. Using the $(4)$ with $z=fracpi4$ we get a representation for the other terms from above. Putting this all together gives us the final value



        $$begin{align*}
        mathfrak{I}&=frac12left[underbrace{Cl_2left(fracpi2right)}_{=G}-underbrace{Cl_2left(0right)}_{=0}+underbrace{Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)}_{=frac G2}right]\
        &=frac12left[G+frac G2right]
        end{align*}$$




        $$therefore~mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x~=~frac34 G$$




        I recommend to study the Clausen Function hence it reduces the number of caculations needed for linear logarithmo-trigonometric integrals tremendously. If you are feeling uncomfortable with a part of the proof let me know and I will try to clear your doubts.






        share|cite|improve this answer











        $endgroup$



        $newcommand{Cl}{operatorname{Cl}}$In order to dodge the extensive usage of the Fourier Series Expansions I will use the Clausen Function $Cl_2(z)$ to shorten things up; nevertheless the result will remain the same as one could get going all along the long way. Anyway, the only difficulty that remains after your attempt is the evaluation of the following integral




        $$mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}xtag1$$




        First of all I will do a bit of reshaping to actually apply useful formulae involving the Clausen Function. Therefore split up the first integral and enforce the substitution $x+fracpi4mapsto x$ within the second integral to get



        $$begin{align*}
        mathfrak{I}&=-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x\
        &=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{pi/8}^{pi/4}logleft(cos x+frac pi4right)mathrm{d}x\
        &=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{3pi/8}^{pi/2}log(cos x)mathrm{d}x\
        &=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_0^{pi/2}log(cos x)mathrm{d}x+int_0^{3pi/8}log(cos x)mathrm{d}x
        end{align*}$$



        Now it is time to apply the first useful formula of the Clausen Function, namely




        $$int_0^t log(cos x)mathrm{d}x~=~frac12Cl_2(pi-2t)-tlog(2)tag2$$




        Formula $(2)$ can be shown rather easy be utilizing the well-known Fourier Series Expansion of $log(cos x)$ combined with the series representation of the $Cl_2(z)$ function. However, with this knowledge we can rewrite the integrals from above in terms of the Clausen Function to get



        $$begin{align*}
        mathfrak{I}&=frac12Cl_2left(pi-2fracpi4right)-frac12Cl_2left(pi-2fracpi8right)-frac12Cl_2left(pi-2fracpi2right)+frac12Cl_2left(pi-2frac{3pi}8right)\
        &~~~underbrace{-fracpi4log(2)+fracpi8log(2)+fracpi2log(2)-frac{3pi}8log(2)}_{=0}\
        &=frac12left[Cl_2left(fracpi2right)-Cl_2left(0right)+Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)right]
        end{align*}$$



        We are almost done! It is time to throw some more important formulae in: firstly the already mentioned series representation and secondly the so-called Duplication Formula




        $$begin{align*}
        Cl_2(z)~&=~sum_{n=1}^infty frac{sin(nz)}{n^2}tag3\
        Cl_2(2z)~&=~2Cl_2(z)-2Cl_2(pi-z)tag4
        end{align*}$$




        From $(3)$ we can direcetly conclude that $Cl_2left(fracpi2right)=G$ and that $Cl_2(0)=0$ where $G$ denotes Catalan's Constant. Using the $(4)$ with $z=fracpi4$ we get a representation for the other terms from above. Putting this all together gives us the final value



        $$begin{align*}
        mathfrak{I}&=frac12left[underbrace{Cl_2left(fracpi2right)}_{=G}-underbrace{Cl_2left(0right)}_{=0}+underbrace{Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)}_{=frac G2}right]\
        &=frac12left[G+frac G2right]
        end{align*}$$




        $$therefore~mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x~=~frac34 G$$




        I recommend to study the Clausen Function hence it reduces the number of caculations needed for linear logarithmo-trigonometric integrals tremendously. If you are feeling uncomfortable with a part of the proof let me know and I will try to clear your doubts.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 28 at 14:28

























        answered Jan 4 at 19:21









        mrtaurhomrtaurho

        5,75551540




        5,75551540






























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