Integral of $ln(tanh(x))$












8












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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.










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$endgroup$








  • 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
















8












$begingroup$


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.










share|cite|improve this question











$endgroup$








  • 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














8












8








8


0



$begingroup$


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.










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








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














  • 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










4 Answers
4






active

oldest

votes


















9












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















8












$begingroup$

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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$endgroup$





















    2












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      Another option: $$lntanh x=-2operatorname{artanh}exp -2x=-2sum_{kge 0}frac{exp -(4k+2)x}{2k+1},$$so$$int_0^inftylntanh x dx=-sum_{kge 0}frac{1}{(2k+1)^2}=-frac{pi^2}{8}.$$






      share|cite|improve this answer









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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        9












        $begingroup$

        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.






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          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
















        9












        $begingroup$

        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.






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          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














        9












        9








        9





        $begingroup$

        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.






        share|cite|improve this answer











        $endgroup$



        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        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


















        • $begingroup$
          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
















        $begingroup$
        Absolutely amazing, thank you very much.
        $endgroup$
        – Sheepe
        Dec 5 '18 at 20:40




        $begingroup$
        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




        $begingroup$
        @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




        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











        8












        $begingroup$

        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



        $$ I = -sum_{ngeq 0}frac{1}{(2n+1)^2}=-left[zeta(2)-frac{1}{4}zeta(2)right]=-frac{3}{4}cdotfrac{pi^2}{6}=color{red}{-frac{pi^2}{8}}. $$






        share|cite|improve this answer









        $endgroup$


















          8












          $begingroup$

          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



          $$ I = -sum_{ngeq 0}frac{1}{(2n+1)^2}=-left[zeta(2)-frac{1}{4}zeta(2)right]=-frac{3}{4}cdotfrac{pi^2}{6}=color{red}{-frac{pi^2}{8}}. $$






          share|cite|improve this answer









          $endgroup$
















            8












            8








            8





            $begingroup$

            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



            $$ I = -sum_{ngeq 0}frac{1}{(2n+1)^2}=-left[zeta(2)-frac{1}{4}zeta(2)right]=-frac{3}{4}cdotfrac{pi^2}{6}=color{red}{-frac{pi^2}{8}}. $$






            share|cite|improve this answer









            $endgroup$



            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



            $$ I = -sum_{ngeq 0}frac{1}{(2n+1)^2}=-left[zeta(2)-frac{1}{4}zeta(2)right]=-frac{3}{4}cdotfrac{pi^2}{6}=color{red}{-frac{pi^2}{8}}. $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 5 '18 at 21:30









            Jack D'AurizioJack D'Aurizio

            290k33283665




            290k33283665























                2












                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  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.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    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.






                    share|cite|improve this answer









                    $endgroup$



                    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.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 6 '18 at 3:25









                    omegadotomegadot

                    6,3672828




                    6,3672828























                        2












                        $begingroup$

                        Another option: $$lntanh x=-2operatorname{artanh}exp -2x=-2sum_{kge 0}frac{exp -(4k+2)x}{2k+1},$$so$$int_0^inftylntanh x dx=-sum_{kge 0}frac{1}{(2k+1)^2}=-frac{pi^2}{8}.$$






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          Another option: $$lntanh x=-2operatorname{artanh}exp -2x=-2sum_{kge 0}frac{exp -(4k+2)x}{2k+1},$$so$$int_0^inftylntanh x dx=-sum_{kge 0}frac{1}{(2k+1)^2}=-frac{pi^2}{8}.$$






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            Another option: $$lntanh x=-2operatorname{artanh}exp -2x=-2sum_{kge 0}frac{exp -(4k+2)x}{2k+1},$$so$$int_0^inftylntanh x dx=-sum_{kge 0}frac{1}{(2k+1)^2}=-frac{pi^2}{8}.$$






                            share|cite|improve this answer









                            $endgroup$



                            Another option: $$lntanh x=-2operatorname{artanh}exp -2x=-2sum_{kge 0}frac{exp -(4k+2)x}{2k+1},$$so$$int_0^inftylntanh x dx=-sum_{kge 0}frac{1}{(2k+1)^2}=-frac{pi^2}{8}.$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 6 '18 at 17:24









                            J.G.J.G.

                            28.7k22845




                            28.7k22845






























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