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Do you see any fast way of calculating this one? $$\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^2 \, dx$$

Numerically, it's about

$$\approx 111.024457130115028409990464833072173251135063166330638343951498907293$$

or in a predicted closed form

$$\frac{4 }{3}\pi ^3+32 \pi \log (2).$$

Ideas, suggestions, opinions are welcome, and the solutions are optionals.

Supplementary question for the integrals lovers: calculate in closed form

$$\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^3 \, dx.$$

As a note, it would be remarkable to be able to find a solution for the generalization below

$$\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^n \, dx.$$

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  • $\begingroup$ i would guess that using the functional identies for $Li_2$ this may be reduced to something already calculated on this site. en.wikipedia.org/wiki/Spence%27s_function $\endgroup$ – tired Aug 26 '15 at 13:39
  • $\begingroup$ @tired it might be. Do you have in mind a particular one? $\endgroup$ – user 1357113 Aug 26 '15 at 13:53
  • $\begingroup$ i wrote something up using another starting point, because i didn't find a particular meaningful functional identity $\endgroup$ – tired Aug 26 '15 at 18:48
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Ok i want to tackle the problem from a different starting point:

Recall that the $\text{Li}_2(x)$ has the following integral representation which is closely related to the Debye functions, well known in condensed matter physics:

$$ \text{Li}_2(z)=\frac{z}{\Gamma(2)}\int_{0}^{\infty}\frac{t}{e^t-z}dt \quad \quad(1) $$

This leads us to the following representation of our Integral $$ I=\int_{0}^{\infty}\left(\text{Li}_2\left(\frac{1}{x^2}\right)\right)^2dx=\\ \int_{0}^{\infty}\mathrm{d}t_1t_1\int_{0}^{\infty}\mathrm{d}t_2t_2\int_{0}^{\infty}dx\frac{1}{x^2 e^{t_1}+1}\frac{1}{x^2 e^{t_2}+1} $$

The inner integral is quite a straightforward exercise in residue calculus and yields (using parity)

$$ I=8\pi\int_{0}^{\infty}\mathrm{d}t_1t_1\int_{0}^{\infty}\mathrm{d}t_2t_2\frac{1}{e^{t_1}+e^{t_2}} $$

Using (1) again to perform for example the $t_1$ integral, we get

$$ I=8\pi\int_{0}^{\infty}\mathrm{d}t_2t_2 e^{-t_2} \text{Li}_2\left(-e^{t_2}\right) $$

Defining $p(t)=t e^{-t} $ and $ q(t)=\text{Li}_2\left(-e^{t}\right)$ (relabeling $t=t_2$) we perform two integration by parts ending up with ( $p^{(-n)}(x)$ denotes the $n$ primitive with respect to $x$)

$$ \frac{1}{8\pi}I=\underbrace{[p^{(-1)}(t)q(t)-p^{(-2)}(t)q'(t)]_0^{\infty}}_{=\frac{\pi ^2}{12}+2\log (2)}+\underbrace{\int_0^{\infty}\overbrace{\frac{t+2}{e^t+1}}^{p^{(-2)}(t)q"(t)}dt}_{J} $$

Here we have used the special value $\text{Li}_2\left(-1\right)=-\frac{\pi^2}{12}\quad (2)$ (See also the appendix for a proof of this fact). The $J$ is also straightforwardly calculated in the appendix and yields

$$ J=\frac{\pi ^2}{12}+2\log (2) $$

Ande therefore

$$ I=\frac{4\pi^3}{3}+32\pi \log(2) $$

Q.E.D

Appendix

Calculation of $J$

$$ J=2\underbrace{\int_{0}^{\infty}\frac{1}{1+e^t}}_{J_1}+\underbrace{\int_{0}^{\infty}\frac{t}{1+e^t}}_{J_2} $$

$J_1$ can be done by done by letting $e^t=z$

$$ J_1=\int_{1}^{\infty}\frac{1}{z(z+1)}=[\log (z)-\log (z+1)]_0^{\infty}=\log(2) $$

$J_2$ may be done by different methods, but most straightforwardly i think by using geometric series and integrating term wise

$$ J_2 =\sum_{n=1}^{\infty}(-1)^n\int_{0}^{\infty}e^{-(n+1)x}=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}=\eta(2)=\frac{\pi^2}{12} \quad (3) $$

here we used a special value of the Dirichlet $\eta$-function

Note that by a change of variables $e^{-z}=q$ the last integral in (3) we also have $\sum_{n=1}^{\infty}\frac{(-1)^{(n+1)}}{n^2}=-\text{Li}_2(-1)$ which proves (2).

Using $J=2J_1+J_2$ the stated result follows

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    $\begingroup$ Very creative approach. +1 $\endgroup$ – Random Variable Aug 26 '15 at 20:16
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    $\begingroup$ Nice way to go (+1). $\endgroup$ – user 1357113 Aug 26 '15 at 20:24
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    $\begingroup$ @Chris'sis, thanks a (+1) from you is also a big honor! :) $\endgroup$ – tired Aug 26 '15 at 20:43
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    $\begingroup$ @Chris'ssistheartist this method is incredibly fast in solving the integral in question for the case $n=1$ just in case u wanna give it a try $\endgroup$ – tired Aug 26 '15 at 23:29
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    $\begingroup$ @tired haha, indeed, very fast. Also a simple integration by parts solves the problem for $n=1$. :-) $\endgroup$ – user 1357113 Aug 26 '15 at 23:37
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With a substitution and a step of integration by parts, the problems boils down to computing: $$ I = -4\int_{0}^{+\infty}\frac{\log(1+x^2)\,\text{Li}_2(-x^2)}{x^2}\,dx. \tag{1}$$ Integrating by parts again, the problem boils down to computing: $$ I_1 = \int_{0}^{+\infty}\frac{\log(1+x^2)^2}{x^2}\,dx,\qquad I_2=\int_{0}^{+\infty}\arctan\left(\frac{1}{x}\right)\frac{\log(1+x^2)}{x}\,dx.\tag{2} $$ The first integral is straightforward: $$ I_1 = \frac{1}{2}\left.\frac{d^2}{d\alpha^2}\int_{0}^{+\infty}\frac{(1+x)^{\alpha}-1}{x^{3/2}}\,dx\,\right|_{\alpha=0} \tag{3}$$ since the innermost integral can be evaluated in terms of the beta function.

That leads to $I_1=4\pi\log 2$. Now we just need to compute: $$ J = \int_{1}^{+\infty}\int_{0}^{+\infty}\frac{\log(1+x^2)}{1+t^2 x^2}\,dx\,dt \tag{4}$$ to recover the value of $I_2=\frac{\pi^3}{8}$. That proves the conjecture:

$$ I = \frac{4\pi^3}{3}+32\pi\log 2.\tag{5}$$

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    $\begingroup$ The inner integral of the double integral is easily done via the residue theorem. $\endgroup$ – Ron Gordon Aug 26 '15 at 14:21
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    $\begingroup$ Good job there (+1). Maybe we can go the same way for the supplementary question. $\endgroup$ – user 1357113 Aug 26 '15 at 14:44
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Another way to evaluate the integral $$I_{2} = \int_{0}^{\infty} \frac{\arctan (\frac{1}{x}) \log(1+x^{2})}{x} \, dx$$ in Jack D'Aurizio's answer (which according to Wolfram Alpha evaluates to $\frac{\pi^{3}}{12}$) is to consider the complex function $$f(z) = \frac{\arctan \left(\frac{1}{z} \right) \log(1-iz)}{z} = \frac{\text{arccot}(z) \log(1-iz)}{z}.$$

Using the principal branch of the logarithm, $\text{arccot}(z)$ has a branch cut on $[-i, i]$ and $\log(1-iz)$ has a branch cut on $(-i\infty, -i]$.

So integrating around a semicircle in the upper half-plane deformed around the branch cut and using the fact $\text{arccot}(z) \sim \frac{1}{z}$ when $z$ is large in magnitude, we get

$$\int_{-\infty}^{\infty} \frac{\arctan(\frac{1}{x}) \log(1-ix)}{x} \, dx + \int_{0}^{1} \frac{\frac{i}{2} (2 \pi i) \log(1+t)}{it} \, i \, dt =0.$$

Then equating the real parts on both sides of the equation, we get

$$\int_{0}^{\infty} \frac{\arctan(\frac{1}{x}) \log(1+x^{2})}{x} \, dx = \pi \int_{0}^{1} \frac{\log(1+t)}{t} \, dt = - \pi \, \text{Li}_{2}(-1) = \frac{\pi^{3}}{12}.$$

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  • $\begingroup$ Fast and nice by complex analysis. (+1) $\endgroup$ – user 1357113 Aug 26 '15 at 20:28
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$$I=\int_0^\infty\operatorname{Li}_2^2\left(-\frac1{x^2}\right)\ dx\overset{IBP}{=}-4\int_0^\infty\operatorname{Li}_2\left(-\frac1{x^2}\right)\ln\left(1+\frac1{x^2}\right)\ dx$$

By using $$\int_0^1\frac{x\ln^n(u)}{1-xu}\ du=(-1)^n n!\operatorname{Li}_{n+1}(x)$$ setting $n=1$ and replacing $x$ with $-\frac1{x^2}$ we can write

$$\operatorname{Li}_2\left(-\frac1{x^2}\right)=\int_0^1\frac{\ln u}{u+x^2}\ du$$

Then \begin{align} I&=-4\int_0^1\ln u\left(\int_0^\infty\frac{\ln\left(1+ \frac1{x^2}\right)}{u+x^2}\ dx\right)\ du\\ &=-4\int_0^1\ln u\left(\frac{\pi}{\sqrt{u}}\ln\left(\frac{1+\sqrt{u}}{\sqrt{u}}\right)\right)\ du, \quad \color{red}{\sqrt{u}=x}\\ &=-16\pi\int_0^1\ln x\left(\ln(1+x)-\ln x\right)\ dx\\ &=-16\pi\left(2-\frac{\pi^2}{12}-2\ln2-2\right)\\ &=\boxed{\frac43\pi^3+32\pi\ln2} \end{align}


Proof of $\ \displaystyle\int_0^\infty\frac{\ln\left(1+ \frac1{x^2}\right)}{u+x^2}\ dx=\frac{\pi}{\sqrt{u}}\ln\left(\frac{1+\sqrt{u}}{\sqrt{u}}\right)$

Let $$I=\int_0^\infty\frac{\ln\left(1+\frac1{x^2}\right)}{u+x^2}\ dx\overset{x\mapsto\ 1/x}{=}\int_0^\infty\frac{\ln(1+x^2)}{1+ux^2}\ dx$$

and $$I(a)=\int_0^\infty\frac{\ln(1+ax^2)}{1+ux^2}\ dx, \quad I(0)=0,\quad I(1)=I$$

\begin{align} I'(a)&=\int_0^\infty\frac{x^2}{(1+ux^2)(1+ax^2)}\ dx\\ &=\frac1{a-u}\int_0^\infty\left(\frac1{1+ux^2}-\frac1{1+ax^2}\right)\ dx\\ &=\frac1{a-u}*\frac{\pi}{2}\left(\frac1{\sqrt{u}}-\frac1{\sqrt{a}}\right)\\ &=\frac{\pi}{2}\frac{1}{\sqrt{u}a+u\sqrt{a}} \end{align}

Then $$I=\int_0^1 I'(a)\ da=\frac{\pi}{2}\int_0^1\frac{da}{\sqrt{u}a+u\sqrt{a}}\\=\frac{\pi}{2}\left(\frac{2}{\sqrt{u}}\ln\left(\frac{1+\sqrt{u}}{\sqrt{u}}\right)\right)=\frac{\pi}{\sqrt{u}}\ln\left(\frac{1+\sqrt{u}}{\sqrt{u}}\right)$$

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