Integral $int_0^infty frac{ln x}{(pi^2+ln^2 x)(1+x)^2} frac{dx}{sqrt x}$












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I have stumbled upon the following integral:$$I=int_0^infty frac{ln x}{(pi^2+ln^2 x)(1+x)^2} frac{dx}{sqrt x}=-frac{pi}{24}$$
Although I could solve it, I am not quite comfortable with the way I did it.



But first I will show the way. We can substitute $ln x rightarrow t $ which gives:
$$I=int_{-infty}^infty frac{t}{pi^2+t^2}frac{e^{frac{t}{2}}}{(1+e^t)^2}dtoverset{t=-x}=int_{-infty}^infty frac{-x}{pi^2+x^2}frac{e^{-frac{x}{2}}}{(1+e^{-x})^2}dx$$
Also adding the two integral from above and simplify some of it yields:
$$2I= int_{-infty}^infty frac{x}{pi^2+x^2}left(frac{e^{frac{x}{2}}}{(1+e^x)^2}-frac{e^{-frac{x}{2}}}{(1+e^{-x})^2}right)dx$$
$$Rightarrow I=-frac{1}{4} int_{-infty}^infty frac{x}{pi^2+x^2}frac{sinh left(frac{x}{2}right)}{cosh ^2left(frac{x}{2}right)}dx$$
And now a round of IBP gives:
$$I=frac12 int_{-infty}^infty left(frac{x^2-pi^2}{(x^2+pi^2)^2}right)left(frac{1}{cosh left(frac{x}{2}right)}right)dx$$
Using the Plancherel theorem the integral simplifies to: $$I=int_0^infty left(sqrt{frac{pi}{2}}xleft(-e^{-pi x}right)right)left(sqrt{2pi}frac{1}{cosh(pi x)}right)dxoverset{pi xrightarrow x}=-frac{1}{pi}int_0^infty frac{x}{cosh( x)}e^{- x}dx$$
We also have the following Laplace tranform for:$$f(t)=frac{t}{cosh( t)}rightarrow F(s)=frac18left(psi_1left(frac{s+1}{4}right)-psi_1left(frac{s+3}{4}right)right)$$
Where $displaystyle{psi_1(z)=sum_{n=0}^infty frac{1}{(z+n)^2}},$ is the trigamma function.
$$Rightarrow I=-frac{1}{pi}F(s=1)=-frac{1}{pi}cdot frac18left(psi_1left(frac{1}{2}right)-psi_1 (1)right)=-frac{1}{pi}cdot frac18left(frac{pi^2}{2}-frac{pi^2}{6}right)=-frac{pi}{24}$$
Have I done anything wrong, or can it be improved?
I have to admit that I mostly used wolfram when applying Plancherel theorem and Laplace transform which I'm not comfortable with, but I didn't find an alternative method myself.




For this question I would like to see a different proof that doesn't rely on that theorem.




Probably not needed, but I should mention that my contour integration knowledge is pretty low. Also maybe there is a conexion with this integral, but I didn't find any.










share|cite|improve this question











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    20












    $begingroup$


    I have stumbled upon the following integral:$$I=int_0^infty frac{ln x}{(pi^2+ln^2 x)(1+x)^2} frac{dx}{sqrt x}=-frac{pi}{24}$$
    Although I could solve it, I am not quite comfortable with the way I did it.



    But first I will show the way. We can substitute $ln x rightarrow t $ which gives:
    $$I=int_{-infty}^infty frac{t}{pi^2+t^2}frac{e^{frac{t}{2}}}{(1+e^t)^2}dtoverset{t=-x}=int_{-infty}^infty frac{-x}{pi^2+x^2}frac{e^{-frac{x}{2}}}{(1+e^{-x})^2}dx$$
    Also adding the two integral from above and simplify some of it yields:
    $$2I= int_{-infty}^infty frac{x}{pi^2+x^2}left(frac{e^{frac{x}{2}}}{(1+e^x)^2}-frac{e^{-frac{x}{2}}}{(1+e^{-x})^2}right)dx$$
    $$Rightarrow I=-frac{1}{4} int_{-infty}^infty frac{x}{pi^2+x^2}frac{sinh left(frac{x}{2}right)}{cosh ^2left(frac{x}{2}right)}dx$$
    And now a round of IBP gives:
    $$I=frac12 int_{-infty}^infty left(frac{x^2-pi^2}{(x^2+pi^2)^2}right)left(frac{1}{cosh left(frac{x}{2}right)}right)dx$$
    Using the Plancherel theorem the integral simplifies to: $$I=int_0^infty left(sqrt{frac{pi}{2}}xleft(-e^{-pi x}right)right)left(sqrt{2pi}frac{1}{cosh(pi x)}right)dxoverset{pi xrightarrow x}=-frac{1}{pi}int_0^infty frac{x}{cosh( x)}e^{- x}dx$$
    We also have the following Laplace tranform for:$$f(t)=frac{t}{cosh( t)}rightarrow F(s)=frac18left(psi_1left(frac{s+1}{4}right)-psi_1left(frac{s+3}{4}right)right)$$
    Where $displaystyle{psi_1(z)=sum_{n=0}^infty frac{1}{(z+n)^2}},$ is the trigamma function.
    $$Rightarrow I=-frac{1}{pi}F(s=1)=-frac{1}{pi}cdot frac18left(psi_1left(frac{1}{2}right)-psi_1 (1)right)=-frac{1}{pi}cdot frac18left(frac{pi^2}{2}-frac{pi^2}{6}right)=-frac{pi}{24}$$
    Have I done anything wrong, or can it be improved?
    I have to admit that I mostly used wolfram when applying Plancherel theorem and Laplace transform which I'm not comfortable with, but I didn't find an alternative method myself.




    For this question I would like to see a different proof that doesn't rely on that theorem.




    Probably not needed, but I should mention that my contour integration knowledge is pretty low. Also maybe there is a conexion with this integral, but I didn't find any.










    share|cite|improve this question











    $endgroup$















      20












      20








      20


      17



      $begingroup$


      I have stumbled upon the following integral:$$I=int_0^infty frac{ln x}{(pi^2+ln^2 x)(1+x)^2} frac{dx}{sqrt x}=-frac{pi}{24}$$
      Although I could solve it, I am not quite comfortable with the way I did it.



      But first I will show the way. We can substitute $ln x rightarrow t $ which gives:
      $$I=int_{-infty}^infty frac{t}{pi^2+t^2}frac{e^{frac{t}{2}}}{(1+e^t)^2}dtoverset{t=-x}=int_{-infty}^infty frac{-x}{pi^2+x^2}frac{e^{-frac{x}{2}}}{(1+e^{-x})^2}dx$$
      Also adding the two integral from above and simplify some of it yields:
      $$2I= int_{-infty}^infty frac{x}{pi^2+x^2}left(frac{e^{frac{x}{2}}}{(1+e^x)^2}-frac{e^{-frac{x}{2}}}{(1+e^{-x})^2}right)dx$$
      $$Rightarrow I=-frac{1}{4} int_{-infty}^infty frac{x}{pi^2+x^2}frac{sinh left(frac{x}{2}right)}{cosh ^2left(frac{x}{2}right)}dx$$
      And now a round of IBP gives:
      $$I=frac12 int_{-infty}^infty left(frac{x^2-pi^2}{(x^2+pi^2)^2}right)left(frac{1}{cosh left(frac{x}{2}right)}right)dx$$
      Using the Plancherel theorem the integral simplifies to: $$I=int_0^infty left(sqrt{frac{pi}{2}}xleft(-e^{-pi x}right)right)left(sqrt{2pi}frac{1}{cosh(pi x)}right)dxoverset{pi xrightarrow x}=-frac{1}{pi}int_0^infty frac{x}{cosh( x)}e^{- x}dx$$
      We also have the following Laplace tranform for:$$f(t)=frac{t}{cosh( t)}rightarrow F(s)=frac18left(psi_1left(frac{s+1}{4}right)-psi_1left(frac{s+3}{4}right)right)$$
      Where $displaystyle{psi_1(z)=sum_{n=0}^infty frac{1}{(z+n)^2}},$ is the trigamma function.
      $$Rightarrow I=-frac{1}{pi}F(s=1)=-frac{1}{pi}cdot frac18left(psi_1left(frac{1}{2}right)-psi_1 (1)right)=-frac{1}{pi}cdot frac18left(frac{pi^2}{2}-frac{pi^2}{6}right)=-frac{pi}{24}$$
      Have I done anything wrong, or can it be improved?
      I have to admit that I mostly used wolfram when applying Plancherel theorem and Laplace transform which I'm not comfortable with, but I didn't find an alternative method myself.




      For this question I would like to see a different proof that doesn't rely on that theorem.




      Probably not needed, but I should mention that my contour integration knowledge is pretty low. Also maybe there is a conexion with this integral, but I didn't find any.










      share|cite|improve this question











      $endgroup$




      I have stumbled upon the following integral:$$I=int_0^infty frac{ln x}{(pi^2+ln^2 x)(1+x)^2} frac{dx}{sqrt x}=-frac{pi}{24}$$
      Although I could solve it, I am not quite comfortable with the way I did it.



      But first I will show the way. We can substitute $ln x rightarrow t $ which gives:
      $$I=int_{-infty}^infty frac{t}{pi^2+t^2}frac{e^{frac{t}{2}}}{(1+e^t)^2}dtoverset{t=-x}=int_{-infty}^infty frac{-x}{pi^2+x^2}frac{e^{-frac{x}{2}}}{(1+e^{-x})^2}dx$$
      Also adding the two integral from above and simplify some of it yields:
      $$2I= int_{-infty}^infty frac{x}{pi^2+x^2}left(frac{e^{frac{x}{2}}}{(1+e^x)^2}-frac{e^{-frac{x}{2}}}{(1+e^{-x})^2}right)dx$$
      $$Rightarrow I=-frac{1}{4} int_{-infty}^infty frac{x}{pi^2+x^2}frac{sinh left(frac{x}{2}right)}{cosh ^2left(frac{x}{2}right)}dx$$
      And now a round of IBP gives:
      $$I=frac12 int_{-infty}^infty left(frac{x^2-pi^2}{(x^2+pi^2)^2}right)left(frac{1}{cosh left(frac{x}{2}right)}right)dx$$
      Using the Plancherel theorem the integral simplifies to: $$I=int_0^infty left(sqrt{frac{pi}{2}}xleft(-e^{-pi x}right)right)left(sqrt{2pi}frac{1}{cosh(pi x)}right)dxoverset{pi xrightarrow x}=-frac{1}{pi}int_0^infty frac{x}{cosh( x)}e^{- x}dx$$
      We also have the following Laplace tranform for:$$f(t)=frac{t}{cosh( t)}rightarrow F(s)=frac18left(psi_1left(frac{s+1}{4}right)-psi_1left(frac{s+3}{4}right)right)$$
      Where $displaystyle{psi_1(z)=sum_{n=0}^infty frac{1}{(z+n)^2}},$ is the trigamma function.
      $$Rightarrow I=-frac{1}{pi}F(s=1)=-frac{1}{pi}cdot frac18left(psi_1left(frac{1}{2}right)-psi_1 (1)right)=-frac{1}{pi}cdot frac18left(frac{pi^2}{2}-frac{pi^2}{6}right)=-frac{pi}{24}$$
      Have I done anything wrong, or can it be improved?
      I have to admit that I mostly used wolfram when applying Plancherel theorem and Laplace transform which I'm not comfortable with, but I didn't find an alternative method myself.




      For this question I would like to see a different proof that doesn't rely on that theorem.




      Probably not needed, but I should mention that my contour integration knowledge is pretty low. Also maybe there is a conexion with this integral, but I didn't find any.







      integration definite-integrals alternative-proof






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      share|cite|improve this question













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      edited Jan 1 at 19:26







      Zacky

















      asked Jan 1 at 17:30









      ZackyZacky

      7,76511062




      7,76511062






















          3 Answers
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          9












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          From the identity



          $$Imint_0^infty e^{-(pi-it)x},dx=frac t{pi^2+t^2}$$



          we see that it suffices to compute the imaginary part of the integral



          $$int_0^infty dxint_{-infty}^infty dt;
          frac{e^{alpha t}}{(1+e^t)^2}e^{-pi x}$$



          where $alpha=1/2+ix$. Now, the integral with respect to $t$ is easy by means of the substitution $u=e^t$ and using the beta function. We thus get



          $$int_0^infty pileft(frac12-ixright)frac{e^{-pi x}}{cosh(pi x)},dx.$$



          Taking the imaginary part we see that the problem boils down to compute the integral



          $$int_0^infty frac{x e^{-pi x}}{cosh(pi x)},dx
          =2int_0^infty frac{x}{1+e^{2pi x}},dx$$



          which, after the substitution $v=2pi x$, reduces to the integral representation of the eta function $eta(2)$. Also notice that taking the real part we obtain the evaluation



          $$int_0^inftyfrac1{(pi^2+log^2 x)(1+x)^2} frac{dx}{sqrt x}=
          frac{log2}{2pi}.$$



          This method generalises to other integrals such as



          begin{align*}
          int_0^infty frac{1}{(pi^2+ln^2 x)(1+x)^3} frac{dx}{sqrt x}
          &=frac{3log (2)}{8 pi }-frac{3 zeta (3)}{16 pi ^3}\
          int_0^infty frac{ln x}{(pi^2+ln^2 x)(1+x)^3} frac{dx}{sqrt x}
          &=-frac{pi }{24}\
          int_0^infty frac{1}{(pi^2+ln^2 x)(1+x)^4} frac{dx}{sqrt x}
          &=frac{5 log (2)}{16 pi }-frac{9 zeta (3)}{32 pi ^3}\
          int_0^infty frac{ln x}{(pi^2+ln^2 x)(1+x)^4} frac{dx}{sqrt x}
          &=-frac{223 pi }{5760}\
          int_0^infty frac{1}{(pi^2+ln^2 x)(1+x)^5} frac{dx}{sqrt x}
          &=-frac{43 zeta (3)}{128 pi ^3}+frac{15 zeta (5)}{256 pi ^5}+frac{35 log (2)}{128
          pi }\
          int_0^infty frac{ln x}{(pi^2+ln^2 x)(1+x)^5} frac{dx}{sqrt x}
          &=-frac{103 pi }{2880}\
          end{align*}



          Incidentally, since the integrals $int_0^infty frac{ln x}{(pi^2+ln^2 x)(1+x)^k} frac{dx}{sqrt x}$ both yield the same value for $k=2,3$, we also deduce



          $$int_0^infty frac{sqrt xln x}{(pi^2+ln^2 x)(1+x)^3};dx=0.$$



          This, however, should not be surprising due to the symmetry $xmapsto1/x$.






          share|cite|improve this answer











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          • $begingroup$
            At first sight this looks quite impressive! Give me some time to try to understand it better.
            $endgroup$
            – Zacky
            Jan 1 at 19:13






          • 1




            $begingroup$
            Thanks! I hope you enjoy completing the steps, but tell me if you need more details.
            $endgroup$
            – diech
            Jan 1 at 19:18










          • $begingroup$
            This is very beautiful. But I get here: $$int_0^infty pileft(frac12color{red}+ixright)frac{e^{-pi x}}{cosh(pi x)},dx.$$ Am I wrong?
            $endgroup$
            – Zacky
            Jan 1 at 22:50












          • $begingroup$
            @Zacky Are you sure? I've checked my calculations and I am still getting $-ix$. In any case, your integral is negative, so the imaginary part should also be negative, shouldn't it?
            $endgroup$
            – diech
            Jan 2 at 9:34










          • $begingroup$
            Well, I have $$int_{-infty}^infty frac{e^{at}}{(1+e^t)^2}dtoverset{t=ln u}=int_0^infty frac{u^{a-1}}{(1+u)^2}du=B(a,2-a)=aGamma(a)Gamma(1-a)$$And since $a=frac12color{red}+ix$ and by the reflection formula it gives that.
            $endgroup$
            – Zacky
            Jan 2 at 11:30



















          7












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          That $pi^2+log^2(x)$ makes me think to the integral representation for Gregory coefficients:



          $$ int_{0}^{+infty}frac{dx}{(1+x)^n (pi^2+log^2 x)} = frac{1}{n!}left[frac{d^n}{dx^n}frac{z}{log(1-z)}right]_{z=0}=[z^n]frac{z}{log(1-z)} tag{1}$$
          which can be seen as a consequence of the Lagrange-Buhrmann inversion theorem. We just need to insert a factor $frac{log x}{sqrt{x}}$ in the integrand function appearing in the LHS, so let's go back to the residue theorem.



          $$ int_{0}^{+infty}frac{log(x),dx}{sqrt{x}(1+x)^2(pi^2+log^2 x)}=int_{-infty}^{+infty}frac{t e^{-t/2}}{(2coshfrac{t}{2})^2 (pi^2+t^2)},dt$$
          equals
          $$ -frac{1}{4}int_{mathbb{R}}frac{tsinhfrac{t}{2}}{(t^2+pi^2)cosh^2frac{t}{2}},dt=-int_{mathbb{R}}frac{tsinh t}{(4t^2+pi^2)cosh^2 t},dt. $$
          The meromorphic function $frac{sinh t}{cosh^2 t}=-frac{d}{dt}left(frac{1}{cosh t}right)$ only has double poles with residue zero, hence all the mass of the last integral comes from the singularity at $frac{pi i}{2}$ and from the behaviour at infinity. The residue theorem grants
          $$ frac{1}{cosh x}=sum_{ngeq 0}(-1)^n frac{pi(2n+1)}{frac{pi^2}{4}(2n+1)^2+x^2} $$
          and
          $$ frac{sinh x}{cosh^2 x} = sum_{ngeq 0}(-1)^n frac{2pi(2n+1)x}{(frac{pi^2}{4}(2n+1)^2+x^2)^2}.$$
          Since
          $$ int_{mathbb{R}}frac{2pi(2n+1)x^2}{(frac{pi^2}{4}(2n+1)^2+x^2)^2 (pi^2+4x^2)},dx = frac{1}{2pi(n+1)^2}$$
          our integral equals $-frac{1}{2pi}eta(2)=color{red}{-frac{pi}{24}}$ by the dominated convergence theorem, allowing to switch $int_{mathbb{R}}$ and $sum_{ngeq 0}$. $frac{1}{24}$ is also the coefficient of $z^3$ in $frac{z}{log(1-z)}$, but so far I have not found a direct way to relate the original integral to the $n=3$ instance of $(1)$.






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          • 2




            $begingroup$
            Thank you for the answer, I thought too that it can be related to the Gregory coefficients, also I should mention that I learnt about the Plancherel theorem from your answer here:math.stackexchange.com/a/2891545/515527
            $endgroup$
            – Zacky
            Jan 1 at 19:24








          • 1




            $begingroup$
            @Zachary: this is Herglotz trick for $frac{1}{cosh}$, followed by termwise differentiation. We have $frac{1}{cosh z} = sum_{xi} frac{R_{xi}}{z-xi}$ where $xi$ are the (simple) zeroes of $cosh$ and $R_xi$ is the residue at $z=xi$ of $frac{1}{cosh}$.
            $endgroup$
            – Jack D'Aurizio
            Jan 1 at 19:50






          • 1




            $begingroup$
            @Zachary: the trick to get my representation is to couple the contribution provided by the poles at $(2n+1)frac{pi i}{2}$ for $n=Ninmathbb{N}$ and $n=-(N+1)$.This cancels the imaginary part and leaves a series on $ngeq 0$.
            $endgroup$
            – Jack D'Aurizio
            Jan 2 at 0:51






          • 1




            $begingroup$
            @Zachary: that's my fault, I wrote wrong constants, now fixing.
            $endgroup$
            – Jack D'Aurizio
            Jan 2 at 6:36






          • 1




            $begingroup$
            Alright. Can I just say that the coupling of poles method is genius. Before doing that, I had an absolute mess of an infinite series, but thanks to your suggestion I managed to nicely simplify the sum. I really appreciate your answers here and I learn a lot from them. I especially enjoyed this one: math.stackexchange.com/questions/2529614/… :) Now, I'm always ready to use the Laplace "Integration by parts" methods.
            $endgroup$
            – Zachary
            Jan 2 at 7:53



















          4












          $begingroup$

          This is not a full answer, but one does not need the full Laplace transform of $x/cosh x$, as the integral can be done by expanding $1/cosh x$ into a geometric series



          $$begin{aligned} I = int_{0}^{infty}frac{xe^{-x}}{cosh x},mathrm{d}x &= 2int_{0}^{infty}frac{xe^{-x}}{e^{x}}frac{mathrm{d}x}{1+e^{-2x}} = 2sum_{n=0}^{infty}(-1)^{n}int_{0}^{infty}xe^{-(2+2n)x},mathrm{d}x \
          &= 2sum_{n=0}^{infty}frac{(-1)^{n}}{4(1+n)^{2}}int_{0}^{infty}ue^{-u},mathrm{d}u = frac{1}{2}sum_{n=0}^{infty}frac{(-1)^{n}}{(1+n)^{2}} \ &= frac{1}{2}sum_{n=1}^{infty}frac{(-1)^{n-1}}{n^{2}} = frac{eta(2)}{2} = frac{pi^{2}}{24}end{aligned}$$



          where $eta(s)$ is the Dirichlet eta function.






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            3 Answers
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            3 Answers
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            active

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            active

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            9












            $begingroup$

            From the identity



            $$Imint_0^infty e^{-(pi-it)x},dx=frac t{pi^2+t^2}$$



            we see that it suffices to compute the imaginary part of the integral



            $$int_0^infty dxint_{-infty}^infty dt;
            frac{e^{alpha t}}{(1+e^t)^2}e^{-pi x}$$



            where $alpha=1/2+ix$. Now, the integral with respect to $t$ is easy by means of the substitution $u=e^t$ and using the beta function. We thus get



            $$int_0^infty pileft(frac12-ixright)frac{e^{-pi x}}{cosh(pi x)},dx.$$



            Taking the imaginary part we see that the problem boils down to compute the integral



            $$int_0^infty frac{x e^{-pi x}}{cosh(pi x)},dx
            =2int_0^infty frac{x}{1+e^{2pi x}},dx$$



            which, after the substitution $v=2pi x$, reduces to the integral representation of the eta function $eta(2)$. Also notice that taking the real part we obtain the evaluation



            $$int_0^inftyfrac1{(pi^2+log^2 x)(1+x)^2} frac{dx}{sqrt x}=
            frac{log2}{2pi}.$$



            This method generalises to other integrals such as



            begin{align*}
            int_0^infty frac{1}{(pi^2+ln^2 x)(1+x)^3} frac{dx}{sqrt x}
            &=frac{3log (2)}{8 pi }-frac{3 zeta (3)}{16 pi ^3}\
            int_0^infty frac{ln x}{(pi^2+ln^2 x)(1+x)^3} frac{dx}{sqrt x}
            &=-frac{pi }{24}\
            int_0^infty frac{1}{(pi^2+ln^2 x)(1+x)^4} frac{dx}{sqrt x}
            &=frac{5 log (2)}{16 pi }-frac{9 zeta (3)}{32 pi ^3}\
            int_0^infty frac{ln x}{(pi^2+ln^2 x)(1+x)^4} frac{dx}{sqrt x}
            &=-frac{223 pi }{5760}\
            int_0^infty frac{1}{(pi^2+ln^2 x)(1+x)^5} frac{dx}{sqrt x}
            &=-frac{43 zeta (3)}{128 pi ^3}+frac{15 zeta (5)}{256 pi ^5}+frac{35 log (2)}{128
            pi }\
            int_0^infty frac{ln x}{(pi^2+ln^2 x)(1+x)^5} frac{dx}{sqrt x}
            &=-frac{103 pi }{2880}\
            end{align*}



            Incidentally, since the integrals $int_0^infty frac{ln x}{(pi^2+ln^2 x)(1+x)^k} frac{dx}{sqrt x}$ both yield the same value for $k=2,3$, we also deduce



            $$int_0^infty frac{sqrt xln x}{(pi^2+ln^2 x)(1+x)^3};dx=0.$$



            This, however, should not be surprising due to the symmetry $xmapsto1/x$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              At first sight this looks quite impressive! Give me some time to try to understand it better.
              $endgroup$
              – Zacky
              Jan 1 at 19:13






            • 1




              $begingroup$
              Thanks! I hope you enjoy completing the steps, but tell me if you need more details.
              $endgroup$
              – diech
              Jan 1 at 19:18










            • $begingroup$
              This is very beautiful. But I get here: $$int_0^infty pileft(frac12color{red}+ixright)frac{e^{-pi x}}{cosh(pi x)},dx.$$ Am I wrong?
              $endgroup$
              – Zacky
              Jan 1 at 22:50












            • $begingroup$
              @Zacky Are you sure? I've checked my calculations and I am still getting $-ix$. In any case, your integral is negative, so the imaginary part should also be negative, shouldn't it?
              $endgroup$
              – diech
              Jan 2 at 9:34










            • $begingroup$
              Well, I have $$int_{-infty}^infty frac{e^{at}}{(1+e^t)^2}dtoverset{t=ln u}=int_0^infty frac{u^{a-1}}{(1+u)^2}du=B(a,2-a)=aGamma(a)Gamma(1-a)$$And since $a=frac12color{red}+ix$ and by the reflection formula it gives that.
              $endgroup$
              – Zacky
              Jan 2 at 11:30
















            9












            $begingroup$

            From the identity



            $$Imint_0^infty e^{-(pi-it)x},dx=frac t{pi^2+t^2}$$



            we see that it suffices to compute the imaginary part of the integral



            $$int_0^infty dxint_{-infty}^infty dt;
            frac{e^{alpha t}}{(1+e^t)^2}e^{-pi x}$$



            where $alpha=1/2+ix$. Now, the integral with respect to $t$ is easy by means of the substitution $u=e^t$ and using the beta function. We thus get



            $$int_0^infty pileft(frac12-ixright)frac{e^{-pi x}}{cosh(pi x)},dx.$$



            Taking the imaginary part we see that the problem boils down to compute the integral



            $$int_0^infty frac{x e^{-pi x}}{cosh(pi x)},dx
            =2int_0^infty frac{x}{1+e^{2pi x}},dx$$



            which, after the substitution $v=2pi x$, reduces to the integral representation of the eta function $eta(2)$. Also notice that taking the real part we obtain the evaluation



            $$int_0^inftyfrac1{(pi^2+log^2 x)(1+x)^2} frac{dx}{sqrt x}=
            frac{log2}{2pi}.$$



            This method generalises to other integrals such as



            begin{align*}
            int_0^infty frac{1}{(pi^2+ln^2 x)(1+x)^3} frac{dx}{sqrt x}
            &=frac{3log (2)}{8 pi }-frac{3 zeta (3)}{16 pi ^3}\
            int_0^infty frac{ln x}{(pi^2+ln^2 x)(1+x)^3} frac{dx}{sqrt x}
            &=-frac{pi }{24}\
            int_0^infty frac{1}{(pi^2+ln^2 x)(1+x)^4} frac{dx}{sqrt x}
            &=frac{5 log (2)}{16 pi }-frac{9 zeta (3)}{32 pi ^3}\
            int_0^infty frac{ln x}{(pi^2+ln^2 x)(1+x)^4} frac{dx}{sqrt x}
            &=-frac{223 pi }{5760}\
            int_0^infty frac{1}{(pi^2+ln^2 x)(1+x)^5} frac{dx}{sqrt x}
            &=-frac{43 zeta (3)}{128 pi ^3}+frac{15 zeta (5)}{256 pi ^5}+frac{35 log (2)}{128
            pi }\
            int_0^infty frac{ln x}{(pi^2+ln^2 x)(1+x)^5} frac{dx}{sqrt x}
            &=-frac{103 pi }{2880}\
            end{align*}



            Incidentally, since the integrals $int_0^infty frac{ln x}{(pi^2+ln^2 x)(1+x)^k} frac{dx}{sqrt x}$ both yield the same value for $k=2,3$, we also deduce



            $$int_0^infty frac{sqrt xln x}{(pi^2+ln^2 x)(1+x)^3};dx=0.$$



            This, however, should not be surprising due to the symmetry $xmapsto1/x$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              At first sight this looks quite impressive! Give me some time to try to understand it better.
              $endgroup$
              – Zacky
              Jan 1 at 19:13






            • 1




              $begingroup$
              Thanks! I hope you enjoy completing the steps, but tell me if you need more details.
              $endgroup$
              – diech
              Jan 1 at 19:18










            • $begingroup$
              This is very beautiful. But I get here: $$int_0^infty pileft(frac12color{red}+ixright)frac{e^{-pi x}}{cosh(pi x)},dx.$$ Am I wrong?
              $endgroup$
              – Zacky
              Jan 1 at 22:50












            • $begingroup$
              @Zacky Are you sure? I've checked my calculations and I am still getting $-ix$. In any case, your integral is negative, so the imaginary part should also be negative, shouldn't it?
              $endgroup$
              – diech
              Jan 2 at 9:34










            • $begingroup$
              Well, I have $$int_{-infty}^infty frac{e^{at}}{(1+e^t)^2}dtoverset{t=ln u}=int_0^infty frac{u^{a-1}}{(1+u)^2}du=B(a,2-a)=aGamma(a)Gamma(1-a)$$And since $a=frac12color{red}+ix$ and by the reflection formula it gives that.
              $endgroup$
              – Zacky
              Jan 2 at 11:30














            9












            9








            9





            $begingroup$

            From the identity



            $$Imint_0^infty e^{-(pi-it)x},dx=frac t{pi^2+t^2}$$



            we see that it suffices to compute the imaginary part of the integral



            $$int_0^infty dxint_{-infty}^infty dt;
            frac{e^{alpha t}}{(1+e^t)^2}e^{-pi x}$$



            where $alpha=1/2+ix$. Now, the integral with respect to $t$ is easy by means of the substitution $u=e^t$ and using the beta function. We thus get



            $$int_0^infty pileft(frac12-ixright)frac{e^{-pi x}}{cosh(pi x)},dx.$$



            Taking the imaginary part we see that the problem boils down to compute the integral



            $$int_0^infty frac{x e^{-pi x}}{cosh(pi x)},dx
            =2int_0^infty frac{x}{1+e^{2pi x}},dx$$



            which, after the substitution $v=2pi x$, reduces to the integral representation of the eta function $eta(2)$. Also notice that taking the real part we obtain the evaluation



            $$int_0^inftyfrac1{(pi^2+log^2 x)(1+x)^2} frac{dx}{sqrt x}=
            frac{log2}{2pi}.$$



            This method generalises to other integrals such as



            begin{align*}
            int_0^infty frac{1}{(pi^2+ln^2 x)(1+x)^3} frac{dx}{sqrt x}
            &=frac{3log (2)}{8 pi }-frac{3 zeta (3)}{16 pi ^3}\
            int_0^infty frac{ln x}{(pi^2+ln^2 x)(1+x)^3} frac{dx}{sqrt x}
            &=-frac{pi }{24}\
            int_0^infty frac{1}{(pi^2+ln^2 x)(1+x)^4} frac{dx}{sqrt x}
            &=frac{5 log (2)}{16 pi }-frac{9 zeta (3)}{32 pi ^3}\
            int_0^infty frac{ln x}{(pi^2+ln^2 x)(1+x)^4} frac{dx}{sqrt x}
            &=-frac{223 pi }{5760}\
            int_0^infty frac{1}{(pi^2+ln^2 x)(1+x)^5} frac{dx}{sqrt x}
            &=-frac{43 zeta (3)}{128 pi ^3}+frac{15 zeta (5)}{256 pi ^5}+frac{35 log (2)}{128
            pi }\
            int_0^infty frac{ln x}{(pi^2+ln^2 x)(1+x)^5} frac{dx}{sqrt x}
            &=-frac{103 pi }{2880}\
            end{align*}



            Incidentally, since the integrals $int_0^infty frac{ln x}{(pi^2+ln^2 x)(1+x)^k} frac{dx}{sqrt x}$ both yield the same value for $k=2,3$, we also deduce



            $$int_0^infty frac{sqrt xln x}{(pi^2+ln^2 x)(1+x)^3};dx=0.$$



            This, however, should not be surprising due to the symmetry $xmapsto1/x$.






            share|cite|improve this answer











            $endgroup$



            From the identity



            $$Imint_0^infty e^{-(pi-it)x},dx=frac t{pi^2+t^2}$$



            we see that it suffices to compute the imaginary part of the integral



            $$int_0^infty dxint_{-infty}^infty dt;
            frac{e^{alpha t}}{(1+e^t)^2}e^{-pi x}$$



            where $alpha=1/2+ix$. Now, the integral with respect to $t$ is easy by means of the substitution $u=e^t$ and using the beta function. We thus get



            $$int_0^infty pileft(frac12-ixright)frac{e^{-pi x}}{cosh(pi x)},dx.$$



            Taking the imaginary part we see that the problem boils down to compute the integral



            $$int_0^infty frac{x e^{-pi x}}{cosh(pi x)},dx
            =2int_0^infty frac{x}{1+e^{2pi x}},dx$$



            which, after the substitution $v=2pi x$, reduces to the integral representation of the eta function $eta(2)$. Also notice that taking the real part we obtain the evaluation



            $$int_0^inftyfrac1{(pi^2+log^2 x)(1+x)^2} frac{dx}{sqrt x}=
            frac{log2}{2pi}.$$



            This method generalises to other integrals such as



            begin{align*}
            int_0^infty frac{1}{(pi^2+ln^2 x)(1+x)^3} frac{dx}{sqrt x}
            &=frac{3log (2)}{8 pi }-frac{3 zeta (3)}{16 pi ^3}\
            int_0^infty frac{ln x}{(pi^2+ln^2 x)(1+x)^3} frac{dx}{sqrt x}
            &=-frac{pi }{24}\
            int_0^infty frac{1}{(pi^2+ln^2 x)(1+x)^4} frac{dx}{sqrt x}
            &=frac{5 log (2)}{16 pi }-frac{9 zeta (3)}{32 pi ^3}\
            int_0^infty frac{ln x}{(pi^2+ln^2 x)(1+x)^4} frac{dx}{sqrt x}
            &=-frac{223 pi }{5760}\
            int_0^infty frac{1}{(pi^2+ln^2 x)(1+x)^5} frac{dx}{sqrt x}
            &=-frac{43 zeta (3)}{128 pi ^3}+frac{15 zeta (5)}{256 pi ^5}+frac{35 log (2)}{128
            pi }\
            int_0^infty frac{ln x}{(pi^2+ln^2 x)(1+x)^5} frac{dx}{sqrt x}
            &=-frac{103 pi }{2880}\
            end{align*}



            Incidentally, since the integrals $int_0^infty frac{ln x}{(pi^2+ln^2 x)(1+x)^k} frac{dx}{sqrt x}$ both yield the same value for $k=2,3$, we also deduce



            $$int_0^infty frac{sqrt xln x}{(pi^2+ln^2 x)(1+x)^3};dx=0.$$



            This, however, should not be surprising due to the symmetry $xmapsto1/x$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 3 at 19:34

























            answered Jan 1 at 19:06









            diechdiech

            1063




            1063












            • $begingroup$
              At first sight this looks quite impressive! Give me some time to try to understand it better.
              $endgroup$
              – Zacky
              Jan 1 at 19:13






            • 1




              $begingroup$
              Thanks! I hope you enjoy completing the steps, but tell me if you need more details.
              $endgroup$
              – diech
              Jan 1 at 19:18










            • $begingroup$
              This is very beautiful. But I get here: $$int_0^infty pileft(frac12color{red}+ixright)frac{e^{-pi x}}{cosh(pi x)},dx.$$ Am I wrong?
              $endgroup$
              – Zacky
              Jan 1 at 22:50












            • $begingroup$
              @Zacky Are you sure? I've checked my calculations and I am still getting $-ix$. In any case, your integral is negative, so the imaginary part should also be negative, shouldn't it?
              $endgroup$
              – diech
              Jan 2 at 9:34










            • $begingroup$
              Well, I have $$int_{-infty}^infty frac{e^{at}}{(1+e^t)^2}dtoverset{t=ln u}=int_0^infty frac{u^{a-1}}{(1+u)^2}du=B(a,2-a)=aGamma(a)Gamma(1-a)$$And since $a=frac12color{red}+ix$ and by the reflection formula it gives that.
              $endgroup$
              – Zacky
              Jan 2 at 11:30


















            • $begingroup$
              At first sight this looks quite impressive! Give me some time to try to understand it better.
              $endgroup$
              – Zacky
              Jan 1 at 19:13






            • 1




              $begingroup$
              Thanks! I hope you enjoy completing the steps, but tell me if you need more details.
              $endgroup$
              – diech
              Jan 1 at 19:18










            • $begingroup$
              This is very beautiful. But I get here: $$int_0^infty pileft(frac12color{red}+ixright)frac{e^{-pi x}}{cosh(pi x)},dx.$$ Am I wrong?
              $endgroup$
              – Zacky
              Jan 1 at 22:50












            • $begingroup$
              @Zacky Are you sure? I've checked my calculations and I am still getting $-ix$. In any case, your integral is negative, so the imaginary part should also be negative, shouldn't it?
              $endgroup$
              – diech
              Jan 2 at 9:34










            • $begingroup$
              Well, I have $$int_{-infty}^infty frac{e^{at}}{(1+e^t)^2}dtoverset{t=ln u}=int_0^infty frac{u^{a-1}}{(1+u)^2}du=B(a,2-a)=aGamma(a)Gamma(1-a)$$And since $a=frac12color{red}+ix$ and by the reflection formula it gives that.
              $endgroup$
              – Zacky
              Jan 2 at 11:30
















            $begingroup$
            At first sight this looks quite impressive! Give me some time to try to understand it better.
            $endgroup$
            – Zacky
            Jan 1 at 19:13




            $begingroup$
            At first sight this looks quite impressive! Give me some time to try to understand it better.
            $endgroup$
            – Zacky
            Jan 1 at 19:13




            1




            1




            $begingroup$
            Thanks! I hope you enjoy completing the steps, but tell me if you need more details.
            $endgroup$
            – diech
            Jan 1 at 19:18




            $begingroup$
            Thanks! I hope you enjoy completing the steps, but tell me if you need more details.
            $endgroup$
            – diech
            Jan 1 at 19:18












            $begingroup$
            This is very beautiful. But I get here: $$int_0^infty pileft(frac12color{red}+ixright)frac{e^{-pi x}}{cosh(pi x)},dx.$$ Am I wrong?
            $endgroup$
            – Zacky
            Jan 1 at 22:50






            $begingroup$
            This is very beautiful. But I get here: $$int_0^infty pileft(frac12color{red}+ixright)frac{e^{-pi x}}{cosh(pi x)},dx.$$ Am I wrong?
            $endgroup$
            – Zacky
            Jan 1 at 22:50














            $begingroup$
            @Zacky Are you sure? I've checked my calculations and I am still getting $-ix$. In any case, your integral is negative, so the imaginary part should also be negative, shouldn't it?
            $endgroup$
            – diech
            Jan 2 at 9:34




            $begingroup$
            @Zacky Are you sure? I've checked my calculations and I am still getting $-ix$. In any case, your integral is negative, so the imaginary part should also be negative, shouldn't it?
            $endgroup$
            – diech
            Jan 2 at 9:34












            $begingroup$
            Well, I have $$int_{-infty}^infty frac{e^{at}}{(1+e^t)^2}dtoverset{t=ln u}=int_0^infty frac{u^{a-1}}{(1+u)^2}du=B(a,2-a)=aGamma(a)Gamma(1-a)$$And since $a=frac12color{red}+ix$ and by the reflection formula it gives that.
            $endgroup$
            – Zacky
            Jan 2 at 11:30




            $begingroup$
            Well, I have $$int_{-infty}^infty frac{e^{at}}{(1+e^t)^2}dtoverset{t=ln u}=int_0^infty frac{u^{a-1}}{(1+u)^2}du=B(a,2-a)=aGamma(a)Gamma(1-a)$$And since $a=frac12color{red}+ix$ and by the reflection formula it gives that.
            $endgroup$
            – Zacky
            Jan 2 at 11:30











            7












            $begingroup$

            That $pi^2+log^2(x)$ makes me think to the integral representation for Gregory coefficients:



            $$ int_{0}^{+infty}frac{dx}{(1+x)^n (pi^2+log^2 x)} = frac{1}{n!}left[frac{d^n}{dx^n}frac{z}{log(1-z)}right]_{z=0}=[z^n]frac{z}{log(1-z)} tag{1}$$
            which can be seen as a consequence of the Lagrange-Buhrmann inversion theorem. We just need to insert a factor $frac{log x}{sqrt{x}}$ in the integrand function appearing in the LHS, so let's go back to the residue theorem.



            $$ int_{0}^{+infty}frac{log(x),dx}{sqrt{x}(1+x)^2(pi^2+log^2 x)}=int_{-infty}^{+infty}frac{t e^{-t/2}}{(2coshfrac{t}{2})^2 (pi^2+t^2)},dt$$
            equals
            $$ -frac{1}{4}int_{mathbb{R}}frac{tsinhfrac{t}{2}}{(t^2+pi^2)cosh^2frac{t}{2}},dt=-int_{mathbb{R}}frac{tsinh t}{(4t^2+pi^2)cosh^2 t},dt. $$
            The meromorphic function $frac{sinh t}{cosh^2 t}=-frac{d}{dt}left(frac{1}{cosh t}right)$ only has double poles with residue zero, hence all the mass of the last integral comes from the singularity at $frac{pi i}{2}$ and from the behaviour at infinity. The residue theorem grants
            $$ frac{1}{cosh x}=sum_{ngeq 0}(-1)^n frac{pi(2n+1)}{frac{pi^2}{4}(2n+1)^2+x^2} $$
            and
            $$ frac{sinh x}{cosh^2 x} = sum_{ngeq 0}(-1)^n frac{2pi(2n+1)x}{(frac{pi^2}{4}(2n+1)^2+x^2)^2}.$$
            Since
            $$ int_{mathbb{R}}frac{2pi(2n+1)x^2}{(frac{pi^2}{4}(2n+1)^2+x^2)^2 (pi^2+4x^2)},dx = frac{1}{2pi(n+1)^2}$$
            our integral equals $-frac{1}{2pi}eta(2)=color{red}{-frac{pi}{24}}$ by the dominated convergence theorem, allowing to switch $int_{mathbb{R}}$ and $sum_{ngeq 0}$. $frac{1}{24}$ is also the coefficient of $z^3$ in $frac{z}{log(1-z)}$, but so far I have not found a direct way to relate the original integral to the $n=3$ instance of $(1)$.






            share|cite|improve this answer











            $endgroup$









            • 2




              $begingroup$
              Thank you for the answer, I thought too that it can be related to the Gregory coefficients, also I should mention that I learnt about the Plancherel theorem from your answer here:math.stackexchange.com/a/2891545/515527
              $endgroup$
              – Zacky
              Jan 1 at 19:24








            • 1




              $begingroup$
              @Zachary: this is Herglotz trick for $frac{1}{cosh}$, followed by termwise differentiation. We have $frac{1}{cosh z} = sum_{xi} frac{R_{xi}}{z-xi}$ where $xi$ are the (simple) zeroes of $cosh$ and $R_xi$ is the residue at $z=xi$ of $frac{1}{cosh}$.
              $endgroup$
              – Jack D'Aurizio
              Jan 1 at 19:50






            • 1




              $begingroup$
              @Zachary: the trick to get my representation is to couple the contribution provided by the poles at $(2n+1)frac{pi i}{2}$ for $n=Ninmathbb{N}$ and $n=-(N+1)$.This cancels the imaginary part and leaves a series on $ngeq 0$.
              $endgroup$
              – Jack D'Aurizio
              Jan 2 at 0:51






            • 1




              $begingroup$
              @Zachary: that's my fault, I wrote wrong constants, now fixing.
              $endgroup$
              – Jack D'Aurizio
              Jan 2 at 6:36






            • 1




              $begingroup$
              Alright. Can I just say that the coupling of poles method is genius. Before doing that, I had an absolute mess of an infinite series, but thanks to your suggestion I managed to nicely simplify the sum. I really appreciate your answers here and I learn a lot from them. I especially enjoyed this one: math.stackexchange.com/questions/2529614/… :) Now, I'm always ready to use the Laplace "Integration by parts" methods.
              $endgroup$
              – Zachary
              Jan 2 at 7:53
















            7












            $begingroup$

            That $pi^2+log^2(x)$ makes me think to the integral representation for Gregory coefficients:



            $$ int_{0}^{+infty}frac{dx}{(1+x)^n (pi^2+log^2 x)} = frac{1}{n!}left[frac{d^n}{dx^n}frac{z}{log(1-z)}right]_{z=0}=[z^n]frac{z}{log(1-z)} tag{1}$$
            which can be seen as a consequence of the Lagrange-Buhrmann inversion theorem. We just need to insert a factor $frac{log x}{sqrt{x}}$ in the integrand function appearing in the LHS, so let's go back to the residue theorem.



            $$ int_{0}^{+infty}frac{log(x),dx}{sqrt{x}(1+x)^2(pi^2+log^2 x)}=int_{-infty}^{+infty}frac{t e^{-t/2}}{(2coshfrac{t}{2})^2 (pi^2+t^2)},dt$$
            equals
            $$ -frac{1}{4}int_{mathbb{R}}frac{tsinhfrac{t}{2}}{(t^2+pi^2)cosh^2frac{t}{2}},dt=-int_{mathbb{R}}frac{tsinh t}{(4t^2+pi^2)cosh^2 t},dt. $$
            The meromorphic function $frac{sinh t}{cosh^2 t}=-frac{d}{dt}left(frac{1}{cosh t}right)$ only has double poles with residue zero, hence all the mass of the last integral comes from the singularity at $frac{pi i}{2}$ and from the behaviour at infinity. The residue theorem grants
            $$ frac{1}{cosh x}=sum_{ngeq 0}(-1)^n frac{pi(2n+1)}{frac{pi^2}{4}(2n+1)^2+x^2} $$
            and
            $$ frac{sinh x}{cosh^2 x} = sum_{ngeq 0}(-1)^n frac{2pi(2n+1)x}{(frac{pi^2}{4}(2n+1)^2+x^2)^2}.$$
            Since
            $$ int_{mathbb{R}}frac{2pi(2n+1)x^2}{(frac{pi^2}{4}(2n+1)^2+x^2)^2 (pi^2+4x^2)},dx = frac{1}{2pi(n+1)^2}$$
            our integral equals $-frac{1}{2pi}eta(2)=color{red}{-frac{pi}{24}}$ by the dominated convergence theorem, allowing to switch $int_{mathbb{R}}$ and $sum_{ngeq 0}$. $frac{1}{24}$ is also the coefficient of $z^3$ in $frac{z}{log(1-z)}$, but so far I have not found a direct way to relate the original integral to the $n=3$ instance of $(1)$.






            share|cite|improve this answer











            $endgroup$









            • 2




              $begingroup$
              Thank you for the answer, I thought too that it can be related to the Gregory coefficients, also I should mention that I learnt about the Plancherel theorem from your answer here:math.stackexchange.com/a/2891545/515527
              $endgroup$
              – Zacky
              Jan 1 at 19:24








            • 1




              $begingroup$
              @Zachary: this is Herglotz trick for $frac{1}{cosh}$, followed by termwise differentiation. We have $frac{1}{cosh z} = sum_{xi} frac{R_{xi}}{z-xi}$ where $xi$ are the (simple) zeroes of $cosh$ and $R_xi$ is the residue at $z=xi$ of $frac{1}{cosh}$.
              $endgroup$
              – Jack D'Aurizio
              Jan 1 at 19:50






            • 1




              $begingroup$
              @Zachary: the trick to get my representation is to couple the contribution provided by the poles at $(2n+1)frac{pi i}{2}$ for $n=Ninmathbb{N}$ and $n=-(N+1)$.This cancels the imaginary part and leaves a series on $ngeq 0$.
              $endgroup$
              – Jack D'Aurizio
              Jan 2 at 0:51






            • 1




              $begingroup$
              @Zachary: that's my fault, I wrote wrong constants, now fixing.
              $endgroup$
              – Jack D'Aurizio
              Jan 2 at 6:36






            • 1




              $begingroup$
              Alright. Can I just say that the coupling of poles method is genius. Before doing that, I had an absolute mess of an infinite series, but thanks to your suggestion I managed to nicely simplify the sum. I really appreciate your answers here and I learn a lot from them. I especially enjoyed this one: math.stackexchange.com/questions/2529614/… :) Now, I'm always ready to use the Laplace "Integration by parts" methods.
              $endgroup$
              – Zachary
              Jan 2 at 7:53














            7












            7








            7





            $begingroup$

            That $pi^2+log^2(x)$ makes me think to the integral representation for Gregory coefficients:



            $$ int_{0}^{+infty}frac{dx}{(1+x)^n (pi^2+log^2 x)} = frac{1}{n!}left[frac{d^n}{dx^n}frac{z}{log(1-z)}right]_{z=0}=[z^n]frac{z}{log(1-z)} tag{1}$$
            which can be seen as a consequence of the Lagrange-Buhrmann inversion theorem. We just need to insert a factor $frac{log x}{sqrt{x}}$ in the integrand function appearing in the LHS, so let's go back to the residue theorem.



            $$ int_{0}^{+infty}frac{log(x),dx}{sqrt{x}(1+x)^2(pi^2+log^2 x)}=int_{-infty}^{+infty}frac{t e^{-t/2}}{(2coshfrac{t}{2})^2 (pi^2+t^2)},dt$$
            equals
            $$ -frac{1}{4}int_{mathbb{R}}frac{tsinhfrac{t}{2}}{(t^2+pi^2)cosh^2frac{t}{2}},dt=-int_{mathbb{R}}frac{tsinh t}{(4t^2+pi^2)cosh^2 t},dt. $$
            The meromorphic function $frac{sinh t}{cosh^2 t}=-frac{d}{dt}left(frac{1}{cosh t}right)$ only has double poles with residue zero, hence all the mass of the last integral comes from the singularity at $frac{pi i}{2}$ and from the behaviour at infinity. The residue theorem grants
            $$ frac{1}{cosh x}=sum_{ngeq 0}(-1)^n frac{pi(2n+1)}{frac{pi^2}{4}(2n+1)^2+x^2} $$
            and
            $$ frac{sinh x}{cosh^2 x} = sum_{ngeq 0}(-1)^n frac{2pi(2n+1)x}{(frac{pi^2}{4}(2n+1)^2+x^2)^2}.$$
            Since
            $$ int_{mathbb{R}}frac{2pi(2n+1)x^2}{(frac{pi^2}{4}(2n+1)^2+x^2)^2 (pi^2+4x^2)},dx = frac{1}{2pi(n+1)^2}$$
            our integral equals $-frac{1}{2pi}eta(2)=color{red}{-frac{pi}{24}}$ by the dominated convergence theorem, allowing to switch $int_{mathbb{R}}$ and $sum_{ngeq 0}$. $frac{1}{24}$ is also the coefficient of $z^3$ in $frac{z}{log(1-z)}$, but so far I have not found a direct way to relate the original integral to the $n=3$ instance of $(1)$.






            share|cite|improve this answer











            $endgroup$



            That $pi^2+log^2(x)$ makes me think to the integral representation for Gregory coefficients:



            $$ int_{0}^{+infty}frac{dx}{(1+x)^n (pi^2+log^2 x)} = frac{1}{n!}left[frac{d^n}{dx^n}frac{z}{log(1-z)}right]_{z=0}=[z^n]frac{z}{log(1-z)} tag{1}$$
            which can be seen as a consequence of the Lagrange-Buhrmann inversion theorem. We just need to insert a factor $frac{log x}{sqrt{x}}$ in the integrand function appearing in the LHS, so let's go back to the residue theorem.



            $$ int_{0}^{+infty}frac{log(x),dx}{sqrt{x}(1+x)^2(pi^2+log^2 x)}=int_{-infty}^{+infty}frac{t e^{-t/2}}{(2coshfrac{t}{2})^2 (pi^2+t^2)},dt$$
            equals
            $$ -frac{1}{4}int_{mathbb{R}}frac{tsinhfrac{t}{2}}{(t^2+pi^2)cosh^2frac{t}{2}},dt=-int_{mathbb{R}}frac{tsinh t}{(4t^2+pi^2)cosh^2 t},dt. $$
            The meromorphic function $frac{sinh t}{cosh^2 t}=-frac{d}{dt}left(frac{1}{cosh t}right)$ only has double poles with residue zero, hence all the mass of the last integral comes from the singularity at $frac{pi i}{2}$ and from the behaviour at infinity. The residue theorem grants
            $$ frac{1}{cosh x}=sum_{ngeq 0}(-1)^n frac{pi(2n+1)}{frac{pi^2}{4}(2n+1)^2+x^2} $$
            and
            $$ frac{sinh x}{cosh^2 x} = sum_{ngeq 0}(-1)^n frac{2pi(2n+1)x}{(frac{pi^2}{4}(2n+1)^2+x^2)^2}.$$
            Since
            $$ int_{mathbb{R}}frac{2pi(2n+1)x^2}{(frac{pi^2}{4}(2n+1)^2+x^2)^2 (pi^2+4x^2)},dx = frac{1}{2pi(n+1)^2}$$
            our integral equals $-frac{1}{2pi}eta(2)=color{red}{-frac{pi}{24}}$ by the dominated convergence theorem, allowing to switch $int_{mathbb{R}}$ and $sum_{ngeq 0}$. $frac{1}{24}$ is also the coefficient of $z^3$ in $frac{z}{log(1-z)}$, but so far I have not found a direct way to relate the original integral to the $n=3$ instance of $(1)$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 2 at 6:40

























            answered Jan 1 at 19:04









            Jack D'AurizioJack D'Aurizio

            292k33284672




            292k33284672








            • 2




              $begingroup$
              Thank you for the answer, I thought too that it can be related to the Gregory coefficients, also I should mention that I learnt about the Plancherel theorem from your answer here:math.stackexchange.com/a/2891545/515527
              $endgroup$
              – Zacky
              Jan 1 at 19:24








            • 1




              $begingroup$
              @Zachary: this is Herglotz trick for $frac{1}{cosh}$, followed by termwise differentiation. We have $frac{1}{cosh z} = sum_{xi} frac{R_{xi}}{z-xi}$ where $xi$ are the (simple) zeroes of $cosh$ and $R_xi$ is the residue at $z=xi$ of $frac{1}{cosh}$.
              $endgroup$
              – Jack D'Aurizio
              Jan 1 at 19:50






            • 1




              $begingroup$
              @Zachary: the trick to get my representation is to couple the contribution provided by the poles at $(2n+1)frac{pi i}{2}$ for $n=Ninmathbb{N}$ and $n=-(N+1)$.This cancels the imaginary part and leaves a series on $ngeq 0$.
              $endgroup$
              – Jack D'Aurizio
              Jan 2 at 0:51






            • 1




              $begingroup$
              @Zachary: that's my fault, I wrote wrong constants, now fixing.
              $endgroup$
              – Jack D'Aurizio
              Jan 2 at 6:36






            • 1




              $begingroup$
              Alright. Can I just say that the coupling of poles method is genius. Before doing that, I had an absolute mess of an infinite series, but thanks to your suggestion I managed to nicely simplify the sum. I really appreciate your answers here and I learn a lot from them. I especially enjoyed this one: math.stackexchange.com/questions/2529614/… :) Now, I'm always ready to use the Laplace "Integration by parts" methods.
              $endgroup$
              – Zachary
              Jan 2 at 7:53














            • 2




              $begingroup$
              Thank you for the answer, I thought too that it can be related to the Gregory coefficients, also I should mention that I learnt about the Plancherel theorem from your answer here:math.stackexchange.com/a/2891545/515527
              $endgroup$
              – Zacky
              Jan 1 at 19:24








            • 1




              $begingroup$
              @Zachary: this is Herglotz trick for $frac{1}{cosh}$, followed by termwise differentiation. We have $frac{1}{cosh z} = sum_{xi} frac{R_{xi}}{z-xi}$ where $xi$ are the (simple) zeroes of $cosh$ and $R_xi$ is the residue at $z=xi$ of $frac{1}{cosh}$.
              $endgroup$
              – Jack D'Aurizio
              Jan 1 at 19:50






            • 1




              $begingroup$
              @Zachary: the trick to get my representation is to couple the contribution provided by the poles at $(2n+1)frac{pi i}{2}$ for $n=Ninmathbb{N}$ and $n=-(N+1)$.This cancels the imaginary part and leaves a series on $ngeq 0$.
              $endgroup$
              – Jack D'Aurizio
              Jan 2 at 0:51






            • 1




              $begingroup$
              @Zachary: that's my fault, I wrote wrong constants, now fixing.
              $endgroup$
              – Jack D'Aurizio
              Jan 2 at 6:36






            • 1




              $begingroup$
              Alright. Can I just say that the coupling of poles method is genius. Before doing that, I had an absolute mess of an infinite series, but thanks to your suggestion I managed to nicely simplify the sum. I really appreciate your answers here and I learn a lot from them. I especially enjoyed this one: math.stackexchange.com/questions/2529614/… :) Now, I'm always ready to use the Laplace "Integration by parts" methods.
              $endgroup$
              – Zachary
              Jan 2 at 7:53








            2




            2




            $begingroup$
            Thank you for the answer, I thought too that it can be related to the Gregory coefficients, also I should mention that I learnt about the Plancherel theorem from your answer here:math.stackexchange.com/a/2891545/515527
            $endgroup$
            – Zacky
            Jan 1 at 19:24






            $begingroup$
            Thank you for the answer, I thought too that it can be related to the Gregory coefficients, also I should mention that I learnt about the Plancherel theorem from your answer here:math.stackexchange.com/a/2891545/515527
            $endgroup$
            – Zacky
            Jan 1 at 19:24






            1




            1




            $begingroup$
            @Zachary: this is Herglotz trick for $frac{1}{cosh}$, followed by termwise differentiation. We have $frac{1}{cosh z} = sum_{xi} frac{R_{xi}}{z-xi}$ where $xi$ are the (simple) zeroes of $cosh$ and $R_xi$ is the residue at $z=xi$ of $frac{1}{cosh}$.
            $endgroup$
            – Jack D'Aurizio
            Jan 1 at 19:50




            $begingroup$
            @Zachary: this is Herglotz trick for $frac{1}{cosh}$, followed by termwise differentiation. We have $frac{1}{cosh z} = sum_{xi} frac{R_{xi}}{z-xi}$ where $xi$ are the (simple) zeroes of $cosh$ and $R_xi$ is the residue at $z=xi$ of $frac{1}{cosh}$.
            $endgroup$
            – Jack D'Aurizio
            Jan 1 at 19:50




            1




            1




            $begingroup$
            @Zachary: the trick to get my representation is to couple the contribution provided by the poles at $(2n+1)frac{pi i}{2}$ for $n=Ninmathbb{N}$ and $n=-(N+1)$.This cancels the imaginary part and leaves a series on $ngeq 0$.
            $endgroup$
            – Jack D'Aurizio
            Jan 2 at 0:51




            $begingroup$
            @Zachary: the trick to get my representation is to couple the contribution provided by the poles at $(2n+1)frac{pi i}{2}$ for $n=Ninmathbb{N}$ and $n=-(N+1)$.This cancels the imaginary part and leaves a series on $ngeq 0$.
            $endgroup$
            – Jack D'Aurizio
            Jan 2 at 0:51




            1




            1




            $begingroup$
            @Zachary: that's my fault, I wrote wrong constants, now fixing.
            $endgroup$
            – Jack D'Aurizio
            Jan 2 at 6:36




            $begingroup$
            @Zachary: that's my fault, I wrote wrong constants, now fixing.
            $endgroup$
            – Jack D'Aurizio
            Jan 2 at 6:36




            1




            1




            $begingroup$
            Alright. Can I just say that the coupling of poles method is genius. Before doing that, I had an absolute mess of an infinite series, but thanks to your suggestion I managed to nicely simplify the sum. I really appreciate your answers here and I learn a lot from them. I especially enjoyed this one: math.stackexchange.com/questions/2529614/… :) Now, I'm always ready to use the Laplace "Integration by parts" methods.
            $endgroup$
            – Zachary
            Jan 2 at 7:53




            $begingroup$
            Alright. Can I just say that the coupling of poles method is genius. Before doing that, I had an absolute mess of an infinite series, but thanks to your suggestion I managed to nicely simplify the sum. I really appreciate your answers here and I learn a lot from them. I especially enjoyed this one: math.stackexchange.com/questions/2529614/… :) Now, I'm always ready to use the Laplace "Integration by parts" methods.
            $endgroup$
            – Zachary
            Jan 2 at 7:53











            4












            $begingroup$

            This is not a full answer, but one does not need the full Laplace transform of $x/cosh x$, as the integral can be done by expanding $1/cosh x$ into a geometric series



            $$begin{aligned} I = int_{0}^{infty}frac{xe^{-x}}{cosh x},mathrm{d}x &= 2int_{0}^{infty}frac{xe^{-x}}{e^{x}}frac{mathrm{d}x}{1+e^{-2x}} = 2sum_{n=0}^{infty}(-1)^{n}int_{0}^{infty}xe^{-(2+2n)x},mathrm{d}x \
            &= 2sum_{n=0}^{infty}frac{(-1)^{n}}{4(1+n)^{2}}int_{0}^{infty}ue^{-u},mathrm{d}u = frac{1}{2}sum_{n=0}^{infty}frac{(-1)^{n}}{(1+n)^{2}} \ &= frac{1}{2}sum_{n=1}^{infty}frac{(-1)^{n-1}}{n^{2}} = frac{eta(2)}{2} = frac{pi^{2}}{24}end{aligned}$$



            where $eta(s)$ is the Dirichlet eta function.






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              This is not a full answer, but one does not need the full Laplace transform of $x/cosh x$, as the integral can be done by expanding $1/cosh x$ into a geometric series



              $$begin{aligned} I = int_{0}^{infty}frac{xe^{-x}}{cosh x},mathrm{d}x &= 2int_{0}^{infty}frac{xe^{-x}}{e^{x}}frac{mathrm{d}x}{1+e^{-2x}} = 2sum_{n=0}^{infty}(-1)^{n}int_{0}^{infty}xe^{-(2+2n)x},mathrm{d}x \
              &= 2sum_{n=0}^{infty}frac{(-1)^{n}}{4(1+n)^{2}}int_{0}^{infty}ue^{-u},mathrm{d}u = frac{1}{2}sum_{n=0}^{infty}frac{(-1)^{n}}{(1+n)^{2}} \ &= frac{1}{2}sum_{n=1}^{infty}frac{(-1)^{n-1}}{n^{2}} = frac{eta(2)}{2} = frac{pi^{2}}{24}end{aligned}$$



              where $eta(s)$ is the Dirichlet eta function.






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                This is not a full answer, but one does not need the full Laplace transform of $x/cosh x$, as the integral can be done by expanding $1/cosh x$ into a geometric series



                $$begin{aligned} I = int_{0}^{infty}frac{xe^{-x}}{cosh x},mathrm{d}x &= 2int_{0}^{infty}frac{xe^{-x}}{e^{x}}frac{mathrm{d}x}{1+e^{-2x}} = 2sum_{n=0}^{infty}(-1)^{n}int_{0}^{infty}xe^{-(2+2n)x},mathrm{d}x \
                &= 2sum_{n=0}^{infty}frac{(-1)^{n}}{4(1+n)^{2}}int_{0}^{infty}ue^{-u},mathrm{d}u = frac{1}{2}sum_{n=0}^{infty}frac{(-1)^{n}}{(1+n)^{2}} \ &= frac{1}{2}sum_{n=1}^{infty}frac{(-1)^{n-1}}{n^{2}} = frac{eta(2)}{2} = frac{pi^{2}}{24}end{aligned}$$



                where $eta(s)$ is the Dirichlet eta function.






                share|cite|improve this answer









                $endgroup$



                This is not a full answer, but one does not need the full Laplace transform of $x/cosh x$, as the integral can be done by expanding $1/cosh x$ into a geometric series



                $$begin{aligned} I = int_{0}^{infty}frac{xe^{-x}}{cosh x},mathrm{d}x &= 2int_{0}^{infty}frac{xe^{-x}}{e^{x}}frac{mathrm{d}x}{1+e^{-2x}} = 2sum_{n=0}^{infty}(-1)^{n}int_{0}^{infty}xe^{-(2+2n)x},mathrm{d}x \
                &= 2sum_{n=0}^{infty}frac{(-1)^{n}}{4(1+n)^{2}}int_{0}^{infty}ue^{-u},mathrm{d}u = frac{1}{2}sum_{n=0}^{infty}frac{(-1)^{n}}{(1+n)^{2}} \ &= frac{1}{2}sum_{n=1}^{infty}frac{(-1)^{n-1}}{n^{2}} = frac{eta(2)}{2} = frac{pi^{2}}{24}end{aligned}$$



                where $eta(s)$ is the Dirichlet eta function.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 1 at 19:05









                IninterrompueIninterrompue

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