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Here is a question that naturally arose in the study of some specific integrals. I'm curious if for such integrals are known nice real analysis tools for calculating them (including here all possible sources
in literature that are publicaly available
). At some point I'll add my real analysis solution.
It's a question for the informative purpose rather than finding solutions, the solution is optional.

Prove that

$$\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6}. $$

Here is a supplementary question

$$\int_{-1}^1 \frac{\log(1-x)}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{4}+\frac{\gamma }{6}+\frac{\log (2)}{6}-2 \log (A) $$

where $A$ is Glaisher–Kinkelin constant.

for the passionates of integrals, series and limits.

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    $\begingroup$ In terms of complex analysis tools, making the substitution $u = \text{arctanh}(x)$ converts it into $$ \int_{-\infty}^{\infty} \frac{\text{sech}^{2}(u)}{\pi^{2}+4u^{2}} \, du$$ which can be evaluated by summing the residues in the upper half of the complex plane. $\endgroup$ – Random Variable Jul 29 '15 at 23:34
  • $\begingroup$ @RandomVariable Do you have any idea about using complex analysis for the second integral. Your above approach doesn't seem to work in this case $\endgroup$ – tired Jul 30 '15 at 11:47
  • $\begingroup$ @tired No, not really. $\endgroup$ – Random Variable Jul 30 '15 at 15:13
  • $\begingroup$ Would it be possible to provide some information about the $A$ that appears in the value for the supplementary integral? $\endgroup$ – robjohn Aug 1 '15 at 15:56
  • $\begingroup$ @robjohn I updated the post. $\endgroup$ – user 1357113 Aug 1 '15 at 16:17
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Approach 1:

For the first integral \begin{align} 2\int^1_{0}\frac{{\rm d}x}{\pi^2+\ln^2\left(\frac{1-x}{1+x}\right)} &=-\frac{4}{\pi}\mathrm{Im}\int^1_0\frac{{\rm d}x}{(\ln{x}+\pi i)(1+x)^2}\tag1\\ &=-\frac{4}{\pi}\mathrm{Im}\int^1_{-1}\frac{{\rm d}x}{\ln{x}(1-x)^2}\tag2\\ &=\frac{4}{\pi}\mathrm{Im}\left[\int^\pi_0\frac{ie^{i\phi}}{i\phi(1-e^{i\phi})^2}\ {\rm d}\phi-\frac{\pi i}{2}\operatorname*{Res}_{z=1}\frac{1}{\ln{z}(z-1)^2}\right]\tag3\\ &=\frac{4}{\pi}\mathrm{Im}\left[\int^\pi_0\frac{{\rm d}\phi}{-4\phi\sin^2\left(\frac{\phi}{2}\right)}-\frac{\pi i}{2}\operatorname*{Res}_{z=1}\frac{1}{\ln{z}(z-1)^2}\right]\\ &=\frac{4}{\pi}\left(-\frac{\pi}{2}\right)\operatorname*{Res}_{z=1}\left[\frac{1}{(z-1)^3}+\frac{1}{2(z-1)^2}\color{red}{-}\frac{\color{red}{1}}{\color{red}{12}(z-1)}+\mathcal{O}(1)\right]\tag4\\ &=\frac{1}{6} \end{align} Explanation:
$(1)$: Substitute $x\mapsto \dfrac{1-x}{1+x}$.
$(2)$: Substitute $x\mapsto -x$ and extend the integration interval to $[-1,1]$. The integral over $[0,1]$ contributes nothing since it is purely real.
$(3)$: Change the path of integration to the unit semicircular arc in the upper half plane. We pick up the residue at $z=1$ in the process.
$(4)$: The first integral in the previous line is purely real; Expand $\displaystyle\frac{1}{\ln{z}(z-1)^2}$ as a Laurent series.


Approach 2:

Here is another approach that avoids complex analysis. \begin{align} 2\int^1_0\frac{{\rm d}x}{\pi^2+\ln^2\left(\frac{1-x}{1+x}\right)} &=-\frac{4}{\pi}\mathrm{Im}\int^1_0\frac{{\rm d}x}{\ln(-x)(1+x)^2}\\ &=\frac{4}{\pi}\int^\infty_0\int^1_0\frac{x^s\sin(\pi s)}{(1+x)^2}\ {\rm d}x\ {\rm d}s\\ &=\frac{4}{\pi}\sum^\infty_{k=0}(-1)^k\int^1_0\int^1_0\frac{x^{s+k}\sin(\pi s)}{(1+x)^2}\ {\rm d}x\ {\rm d}s\\ &=\frac{4}{\pi}\int^1_0\sin(\pi s)\int^1_0\frac{x^{s}}{(1+x)^3}\ {\rm d}x\ {\rm d}s\\ &=\frac{4}{\pi}\int^1_0\sin(\pi s)\left[-\frac{1}{8}-\frac{s}{4}+\frac{s(s-1)}{4}\left(\psi_0\left(\frac{s}{2}\right)-\psi_0\left(\frac{s-1}{2}\right)\right)\right]\ {\rm d}s\\ &=\frac{1}{\pi}\int^1_0\sin(\pi s)s(s-1)\left[\psi_0\left(\frac{s}{2}\right)-\psi_0\left(\frac{s+1}{2}\right)\right]\ {\rm d}s\\ &=\frac{2}{\pi}\int^1_0\ln\frac{\Gamma\left(\frac{s}{2}\right)}{\Gamma\left(\frac{s+1}{2}\right)}\left(\frac{{\rm d}}{{\rm d}s}\sin(\pi s)s(1-s)\right)\ {\rm d}s \end{align} Using the fact $$\ln\frac{\Gamma\left(\frac{s}{2}\right)}{\Gamma\left(\frac{s+1}{2}\right)}=\sum_{n\in2\mathbb{N}_0+1}\left[\frac{1}{n}\cos(n\pi s)+2\frac{\gamma+\ln(2n\pi)}{n\pi}\sin(n\pi s)\right]\ \ , \ \ 0<s\le1$$ in tandem with the results \begin{align} \int^1_0\sin((2n+1)\pi s)\left(\frac{{\rm d}}{{\rm d}s}\sin(\pi s)s(1-s)\right)\ {\rm d}s &=0\\ \int^1_0\cos((2n+1)\pi s)\left(\frac{{\rm d}}{{\rm d}s}\sin(\pi s)s(1-s)\right)\ {\rm d}s &= \begin{cases} \frac{\pi}{12}+\frac{1}{4\pi},& \text{if $n=0$}\\ \frac{2n+1}{4\pi}\left(\frac{1}{(n+1)^2}-\frac{1}{n^2}\right),& \text{if $n\in\mathbb{N}$} \end{cases} \end{align} yields \begin{align} 2\int^1_0\frac{{\rm d}x}{\pi^2+\ln^2\left(\frac{1-x}{1+x}\right)} &=\frac{2}{\pi}\left[\frac{\pi}{12}+\frac{1}{4\pi}+\frac{1}{4\pi}\sum^\infty_{n=1}\left(\frac{1}{(n+1)^2}-\frac{1}{n^2}\right)\right]=\frac{1}{6} \end{align}


Supplementary Problem:

Substituting $x\mapsto\tanh{x}$ then $2x\mapsto\ln{x}$, \begin{align} \int^1_{-1}\frac{\ln(1-x)}{\pi^2+\ln^2\left(\frac{1-x}{1+x}\right)}\ {\rm d}x &=\int^1_{-1}\frac{\ln{2}}{\pi^2+\ln^2\left(\frac{1-x}{1+x}\right)}\ {\rm d}x+\int^\infty_{-\infty}\frac{\ln\left(\frac{1-\tanh{x}}{2}\right)}{(\pi^2+4x^2)\cosh^2{x}}\ {\rm d}x\\ &=\frac{\ln{2}}{6}-\int^\infty_{0}\frac{2\ln(1+x)}{(\pi^2+\ln^2{x})(1+x)^2}\ {\rm d}x\\ &=\frac{\ln{2}}{6}-\int^1_0\frac{1}{2\pi i}\int_Cf(z,x)\ {\rm d}z\ {\rm d}x \end{align} where $f(z,x)=\dfrac{2z}{(1+xz)(\ln{z}-\pi i)(1+z)^2}$ and $C$ is the keyhole contour deformed along the branch cut $[0,\infty]$. We note that $$\frac{1}{(\ln{z}-\pi i)(1+z)^2}\sim_{-1}-\frac{1}{(z+1)^3}+\frac{1}{2(z+1)^2}+\frac{1}{12(z+1)}+\mathcal{O}(1)$$ and by the residue theorem, the contour integral is \begin{align} \frac{1}{2\pi i}\int_Cf(z,x)\ {\rm d}z =&\ \operatorname*{Res}_{z=-1/x}f(z,x)+\operatorname*{Res}_{z=-1}f(z,x)\\ =&\ \frac{2\left(-\frac{1}{x}\right)}{x\left(\ln\left(-\frac{1}{x}\right)-\pi i\right)\left(1-\frac{1}{x}\right)^2}-\frac{1}{2}\left.\left(\frac{{\rm d}^2}{{\rm d}z^2}+\frac{{\rm d}}{{\rm d}z}\right)\frac{2z}{1+xz}\right|_{z=-1}-\frac{1}{6(1-x)}\\ =&\ \frac{2x}{(1-x)^3}+\frac{1}{(1-x)^2}-\frac{1}{6(1-x)}+\frac{2}{(1-x)^2\ln{x}} \end{align} Next, let $$I(s)=\int^1_0\left(\frac{2x}{(1-x)^3}+\frac{1}{(1-x)^2}-\frac{1}{6(1-x)}+\frac{2}{(1-x)^2\ln{x}}\right)x^s\ {\rm d}x$$ such that the integral we seek is $\displaystyle\frac{\ln{2}}{6}-I(0)$. Differentiating twice and integrating by parts, \begin{align} I''(s) &=1+\int^1_0\left(\frac{s^2x^{s-1}\ln^2{x}}{1-x}+\frac{2sx^{s-1}\ln{x}}{1-x}-\frac{x^s\ln^2{x}}{6(1-x)}\right)\ {\rm d}x\\ &=1-s^2\psi_2(s)-2s\psi_1(s)+\frac{1}{6}\psi_2(s+1) \end{align} and we may integrate back to obtain \begin{align} I'(s) &=\frac{1}{2}+s-s^2\psi_1(s)+\frac{1}{6}\psi_1(s+1)\\ I(s) &=L-s\ln(2\pi)-\frac{s}{2}+\frac{3s^2}{2}+\frac{1}{6}\psi_0(s+1)-s^2\psi_0(s)+2\ln{G(s+1)} \end{align} where \begin{align} L &=\lim_{s\to\infty}\left(s\ln(2\pi)+\frac{s}{2}-\frac{3s^2}{2}-\frac{1}{6}\psi_0(s+1)+s^2\psi_0(s)-2\ln{G(s+1)}\right)\\ &=\lim_{s\to\infty}\left(s\ln(2\pi)+\frac{s}{2}-\frac{3s^2}{2}-\frac{1}{6}\ln{s}+s^2\ln{s}-\frac{s}{2}-\frac{1}{12}-s^2\ln{s}+\frac{3s^2}{2}-s\ln(2\pi)+\frac{1}{6}\ln{s}-2\zeta'(-1)+\mathcal{O}\left(\frac{1}{s}\right)\right)\\ &=-2\zeta'(-1)-\frac{1}{12} \end{align} Letting $s=0$ and noting that $\psi_0(1)=-\gamma$, we arrive at $$\int^1_{-1}\frac{\ln(1-x)}{\pi^2+\ln^2\left(\frac{1-x}{1+x}\right)}\ {\rm d}x=2\zeta'(-1)+\frac{\gamma}{6}+\frac{\ln{2}}{6}+\frac{1}{12}$$

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    $\begingroup$ Nice way to go (+1) $\endgroup$ – user 1357113 Jul 31 '15 at 14:42
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    $\begingroup$ How would one prove the formula for $~\ln\dfrac{\Gamma(t)}{\Gamma\Big(t+\tfrac12\Big)}~$ ? I am not able to evaluate $~\displaystyle\sum~\ln n\cdot\frac{\sin nx}n~$ in any meaningful manner, except to note that it is the derivative with regard to a of $~\displaystyle\sum\frac{\sin nx}{n^a}$ $\endgroup$ – Lucian Aug 1 '15 at 8:09
  • $\begingroup$ @Lucian It follows from the Fourier expansion of $\ln\Gamma(x)$, which was derived here. math.stackexchange.com/questions/1008732/… $\endgroup$ – M.N.C.E. Aug 1 '15 at 8:11
  • $\begingroup$ It looks like you're done with the second one too. Well-done! :-) $\endgroup$ – user 1357113 Aug 2 '15 at 8:40
  • $\begingroup$ @Chris'ssistheartist Thanks. $\endgroup$ – M.N.C.E. Aug 2 '15 at 11:42
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Sub $x=\tanh{u}$, $dx = \operatorname{sech^2}{u} \, du$. Then the integral is

$$\int_{-\infty}^{\infty} du \, \frac{\operatorname{sech^2}{u}}{\pi^2+4 u^2} $$

Now, use Parseval. The Fourier transforms of the pieces of the integrand are

$$\int_{-\infty}^{\infty} du \, \frac{e^{i u k}}{\pi^2+4 u^2} = \frac14 \frac{\pi}{\pi/2} e^{-\pi |k|/2} $$

$$\int_{-\infty}^{\infty} du \, \operatorname{sech^2}{u} \, e^{i u k} = \pi k \operatorname{csch}{\left (\frac{\pi k}{2} \right )}$$

so by Parseval...

$$\int_{-\infty}^{\infty} du \, \frac{\operatorname{sech^2}{u}}{\pi^2+4 u^2} = \frac12 \frac{\pi}{2 \pi} \int_{-\infty}^{\infty} dk \, k \operatorname{csch}{\left (\frac{\pi k}{2} \right )} e^{-\pi |k|/2} = \int_0^{\infty} dk \frac{k \, e^{-\pi k/2}}{e^{\pi k/2}-e^{-\pi k/2}}$$

Expand the denominator:

$$\int_{-\infty}^{\infty} du \, \frac{\operatorname{sech^2}{u}}{\pi^2+4 u^2} = \sum_{m=0}^{\infty} \int_0^{\infty} dk \, k \, e^{-(1+m) \pi k} = \frac{1}{\pi^2} \sum_{m=0}^{\infty} \frac1{\left (1+m \right )^2} = \frac1{6}$$

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  • $\begingroup$ (+1) for the powerful tool that is good to know and that leads to an elegant solution. However, the interesting thing is getting the answer by real analysis exclusively, not using not even complex numbers. I'm curious if in the math literature there is any tool like that mentioned. $\endgroup$ – user 1357113 Jul 30 '15 at 7:52
  • $\begingroup$ By the way, I don't think I saw anyone on this site showing the exceptional power of Parseval tool more than you. Perhaps it's one of your favourite tools. :-) $\endgroup$ – user 1357113 Jul 30 '15 at 8:01
  • $\begingroup$ @Chris'ssistheartist: I'm surprised I don't see it more. I use it with respect to Fourier transforms because I know a lot of FTs off the top of my head. Parseval is not limited to FTs and so it is an incredibly versatile tool for evaluating integrals and sums. $\endgroup$ – Ron Gordon Jul 30 '15 at 8:25
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$\newcommand{\sech}{\operatorname{sech}}\newcommand{\arctanh}{\operatorname{arctanh}}\newcommand{\Res}{\operatorname*{Res}}$ $\Res\limits_{z=\frac\pi2i}\left(\frac{\sech^2(z)}{\pi^2+4z^2}\right)=-i\frac{3+\pi^2}{12\pi^3}$ and for $k\ge1$, $\Res\limits_{z=\frac{(2k+1)\pi}2i}\left(\frac{\sech^2(z)}{\pi^2+4z^2}\right)=\frac{8z}{\left(\pi^2+4z^2\right)^2}$. Therefore, summing the residues in the upper half-plane, we get $$ \begin{align} \int_{-1}^1\frac1{\pi^2+(2\arctanh(x))^2}\,\mathrm{d}x &=\int_{-\infty}^\infty\frac1{\pi^2+4u^2}\,\mathrm{d}\tanh(u)\\ &=\int_{-\infty}^\infty\frac{\sech^2(u)}{\pi^2+4u^2}\,\mathrm{d}u\\ &=2\pi i\left(-i\frac{3+\pi^2}{12\pi^3}+\frac{i}{4\pi^3}\sum_{k=1}^\infty\frac{2k+1}{\left(k^2+k\right)^2}\right)\\ &=2\pi i\left(-i\frac{3+\pi^2}{12\pi^3}+\frac{i}{4\pi^3}\sum_{k=1}^\infty\left(\frac1{k^2}-\frac1{(k+1)^2}\right)\right)\\[4pt] &=\frac16 \end{align} $$ once we collapse the telescoping series.

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  • $\begingroup$ That was a fast one! (+1) $\endgroup$ – user 1357113 Aug 1 '15 at 14:40
  • $\begingroup$ @Chris'ssistheartist: Yeah. However, the computation of the residues was long, but mechanical. $\endgroup$ – robjohn Aug 2 '15 at 9:35
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After making the substitution $u = \text{arctanh}(x)$, we could use the Laplace transform $$\int_{0}^{\infty} \cos(ax) \, e^{-bx} \, dx = \frac{b}{a^{2}+b^{2}} \, , \, \text{Re} (b) >0 $$

and then switch the order of integration.

Specifically,

$$ \begin{align} \int_{-\infty}^{\infty} \frac{\text{sech}^{2}(u)}{4u^{2} + \pi^{2}} \, du &= \frac{1}{\pi} \int_{-\infty}^{\infty} \text{sech}^{2}(u) \int_{0}^{\infty} \cos(2ut)\, e^{- \pi t} \, dt \, du \\ &= \frac{1}{\pi} \int_{0}^{\infty} e^{- \pi t} \int_{-\infty}^{\infty} \text{sech}^{2} (u) \cos(2tu) \, du \, \ dt. \end{align}$$

The inside integral is basically the second Fourier transform that Ron Gordon uses in his answer.

A relatively quick way to evaluate it is to integrate the complex function $f(z) = \text{sech}^{2}(z) \, e^{2itz}$ around a rectangular contour in the upper half-plane of height $\pi$ and use the fact that $\text{sech}^{2} (z)$ is $i \pi$-periodic.

Doing so we get

$$\begin{align} \int_{-\infty}^{\infty} \text{sech}^{2}(u) \, e^{2itu} \, du - \int_{-\infty}^{\infty} \text{sech}^2(u) \, e^{2it(u+ i \pi)} \, du &= 2 \pi i \, \text{Res}[f(z), i \pi /2] \\ &= 2 \pi i \, \text{Res} [f(z+ i \pi/2), 0] \\ &= 2 \pi i \, \text{Res} \left[-\text{csch}^{2}(z) e^{2i tz} e^{-\pi t} ,0 \right] \\ &= 2 \pi i \left(-2ite^{- \pi t} \right) \end{align} $$

since $\text{csch}^{2}(z) = \frac{1}{z^{2}} + \mathcal{O}(1).$

Then combining the two integrals on the right, we get

$$ \int_{-\infty}^{\infty} \text{sech}^{2}(x) \, e^{2itu} \, du = \int_{-\infty}^{\infty} \text{sech}^{2}(x) \cos(2tu) \, du = 2 \pi t \, \frac{2 \, e^{- \pi t}}{1-e^{- 2 \pi t}} = 2 \pi t \, \text{csch} (\pi t). $$

(I couldn't' think of an approach that avoided complex analysis entirely.)

Therefore,

$$ \int_{-\infty}^{\infty} \frac{\text{sech}^{2}(u)}{\pi^{2}+4u^{2}} \, du = 2 \int_{0}^{\infty} t \, \text{csch} (\pi t) \, e^{- \pi t}\, dt. $$

After making the substitution $w = 2t$, we end up with the same integral that results from using Parseval's theorem. You can refer to Ron Gordon's answer to complete the evaluation.

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  • $\begingroup$ Cute approach (+1) $\endgroup$ – user 1357113 Jul 31 '15 at 14:42

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