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

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

  • $\begingroup$ 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, 2015 at 13:39
  • $\begingroup$ @tired it might be. Do you have in mind a particular one? $\endgroup$ Aug 26, 2015 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, 2015 at 18:48

7 Answers 7


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:

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

  • 3
    $\begingroup$ The inner integral of the double integral is easily done via the residue theorem. $\endgroup$
    – Ron Gordon
    Aug 26, 2015 at 14:21
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    $\begingroup$ Good job there (+1). Maybe we can go the same way for the supplementary question. $\endgroup$ Aug 26, 2015 at 14:44

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

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

  • $\begingroup$ Fast and nice by complex analysis. (+1) $\endgroup$ Aug 26, 2015 at 20:28

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

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



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

  • 1
    $\begingroup$ Very creative approach. +1 $\endgroup$ Aug 26, 2015 at 20:16
  • 1
    $\begingroup$ Nice way to go (+1). $\endgroup$ Aug 26, 2015 at 20:24
  • 1
    $\begingroup$ @Chris'sis, thanks a (+1) from you is also a big honor! :) $\endgroup$
    – tired
    Aug 26, 2015 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, 2015 at 23:29
  • 1
    $\begingroup$ @tired haha, indeed, very fast. Also a simple integration by parts solves the problem for $n=1$. :-) $\endgroup$ Aug 26, 2015 at 23:37

Here is a more satisfactory approach to get the same result as RMT (perhaps a little more).

Begin with the original problem. Let $n$ be a positive integer and $$ I_n=\int_0^\infty \text{Li}_n^2(-x^{-2})dx $$ The Mellin transform of $\text{Li}_n(-x)$ is

$$ \begin{align} &\mathcal{M}[\text{Li}_n(-x)]=\int_0^\infty \text{Li}_n(-x)x^{s-1}dx \\ =&\int_0^\infty\frac{x^{s-1}}{n!}\int_0^\infty \frac{-t^{n}}{e^t/x+1}dtdx \\ =&\int_0^\infty\frac{t^{n}}{n!}dt\color{red}{\int_0^\infty \frac{-x^{s}}{e^t+x}dx} \\ =&\frac{\pi}{\sin(\pi s)}\int_0^\infty\frac{t^{n}}{n!}e^{st}dt \\ =&\frac{\pi}{s^n\sin(\pi s)} \end{align} $$

The red integral is well known and can be derived using standard key hole contour.

The classic analytic continuation of $\text{Li}_n(z)$ (by integral) tells us it is always real when $z$ is a negative real number, so after simple substitution $$ I_n=\frac12\int_0^\infty|\text{Li}_n(-x)|^2x^{2c-1}dx $$ where $c=-\dfrac14$.

Then recall the Plancherel's theorem of Mellin transform to get $$ I_n=\frac12\cdot \frac1{2\pi}\int_{-\infty}^\infty\left|\frac{\pi}{(c+it)^n\sin(\pi(c+it))}\right|^2dt=\frac\pi8\int_{-\infty}^\infty\text{sech}(\pi u/2)\frac{16^n}{(1+u^2)^{n}}du $$ where $t=4u$.

Instead of proceed further, take the generating function of both sides, so $$ I(q)=\sum_{n\ge1}I_nq^n=\frac\pi8\int_{-\infty}^\infty\text{sech}(\pi u/2)\left(\sum_{n\ge1}\frac{16^nq^n}{(1+u^2)^{n}}\right)du=2\pi q\int_{-\infty}^\infty\frac{\text{sech}(\pi u/2)}{1+u^2-16q}du $$ the convergence of the series holds if $q$ is sufficiently small, as well as the validity of switching integral and summation.

The integral can be evaluated via direct contour integration, but here I will use the of Fourier transform.

It's well known that (any Fourier table) $$ \int_{-\infty}^\infty\text{sech}(\pi u/2)e^{iwu}du=2\text{sech} w\\ \int_{-\infty}^\infty\frac{e^{iwu}}{u^2+b^2}du=\frac{\pi}a e^{-|wb|} $$ Set $b=\sqrt{1-16q}$, then use Parseval's theorem $$ I(q)=2\pi q\cdot \frac1{2\pi}\int_{-\infty}^\infty\frac{2\pi e^{-|w|b}}{b\cosh w}dw=\frac{16\pi q}b\int_0^\infty\frac{e^{(1-b)w}}{e^{2w}+1}dw $$ Finally, use the integral representation of the polygamma function and we get $$ I(q)=\frac{2\pi q}{\sqrt{1-16q}}\left(\psi\left(\frac{3+\sqrt{1-16q}}{4}\right)-\psi\left(\frac{1+\sqrt{1-16q}}{4}\right)\right) $$ $I(q)$ is analytic at $q=0$, so series expansion is valid and values yield $$ \begin{gathered} I_1=4 \pi \log (2)\\ \color{red}{I_2=\frac{4 \pi ^3}{3}+32 \pi \log (2)}\\ I_3=48 \pi \zeta (3)+16 \pi ^3+384 \pi \log (2)\\ I_4=768 \pi \zeta (3)+\frac{112 \pi ^5}{45}+\frac{640 \pi ^3}{3}+5120 \pi \log (2)\\ I_5=11520 \pi \zeta (3)+960 \pi \zeta (5)+\frac{448 \pi ^5}{9}+\frac{8960 \pi ^3}{3}+71680 \pi \log (2)\\ \cdots \end{gathered} $$

Further, this can be generalized to $$ I_n(a)=\int_0^\infty \text{Li}_n^2(-x^{-a})dx\quad a>\frac12 $$ The procedure is mainly the same, except for the convolution becomes $$ I_n(a)=\frac\pi{4a}\int_{-\infty}^\infty\frac{1}{\cosh \left(\pi t/a\right)-\cos \left(\pi/a\right)}\frac{(2a)^{2n}}{(1+u^2)^{n}}du $$ Also take the generating function, then use the Fourier transform (as for its proof, see here for example) $$ \int_{-\infty}^{\infty} \frac{e^{iwx}}{\cosh(ax)-\cos\beta} dx = \frac{2\pi\sinh\left(\frac{w}{a}(\pi-\beta)\right)}{a\sinh\left(\frac{w\pi}{a}\right)\sin\beta} \quad a>0, 0<\beta<2\pi $$ to get $$ I(q;a)=\frac{2 \pi q}{\sin\left(\frac{\pi }{a}\right)\sqrt{1-4 a^2 q}}\left(\psi \left(\frac{2 a+\sqrt{1-4 a^2 q}-1}{2 a}\right)-\psi\left(\frac{1+\sqrt{1-4 a^2 q}}{2 a}\right)\right) $$ For example, $$ I_2(a)=\int_0^\infty \text{Li}_2^2(-x^{-a})dx=2 \pi a \csc \left(\frac{\pi }{a}\right) \left(\psi ^{(1)}\left(\frac{1}{a}\right)-\frac{\pi ^2}6-2 a \psi\left(\frac{1}{a}\right)-2 \gamma a\right)\\ \int_0^\infty \text{Li}_2^2(-x^{-4})dx=\frac{4\pi\sqrt{2}}{3} (48G+24\pi+5 \pi^2+144 \log (2)) $$ where $G$ is the Catalan's constant.


$$I=\int_0^\infty\operatorname{Li}_2^2\left(-\frac1{x^2}\right)\ dx\overset{IBP}{=}-4\int_0^\infty\operatorname{Li}_2\left(-\frac1{x^2}\right)\ln\left(1+\frac1{x^2}\right)\ dx$$

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

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

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


The following is another way to show that $$ \begin{align} \int_{0}^{\infty} \operatorname{Li}_{2}^{2} \left(-x^{-1/a} \right) \, \mathrm dx &= a\int_{0}^{\infty} \frac{\operatorname{Li}_{2}^{2}(-u)}{u^{a+1}} \, \mathrm du \\ &=\frac{2 \pi}{a} \csc (\pi a) \left(\psi_{1}(a)-\frac{\pi ^2}6-\frac{2}{a}\psi(a)-\frac{2 \gamma }{a}\right) \end{align}$$ for $0 < a < 2$, where $\psi(a)$ is the digamma function and $\psi_{1}(a)$ is the trigamma function.

This generalization appears in Po1ynomial's answer. I just replaced $a$ with $1/a$ for convenience.

Using the principal branch of the dilogarithm and the branch of $z^{-a-1}$ where $0 \le \arg(z) < 2 \pi$, let's integrate the function $$f(z) = \frac{\operatorname{Li}_{2}^{2}(-z)}{z^{a+1}} $$ around the following double keyhole contour:

enter image description here

From the dilogarithm "inversion" formula it follows that on the upper side of the branch cut on $[-\infty, -1]$, $$\operatorname{Li}_{2}(-z) = - \operatorname{Li}_{2} \left(-\frac{1}{z} \right) + \frac{\pi^{2}}{3} - \frac{1}{2} \ln^{2}(-z) - i \pi \ln(-z), \quad z <-1.$$

While on the lower side of the branch on $(-\infty, -1]$, $$\operatorname{Li}_{2}(-z) = - \operatorname{Li}_{2} \left(-\frac{1}{z} \right) + \frac{\pi^{2}}{3} - \frac{1}{2} \ln^{2}(-z) + i \pi \ln(-z), \quad z <-1.$$

The integral vanishes on the two big semicircles as their radii go to to infinity since $f(z)$ is asymptotic to $\frac{1}{4}\frac{\ln(z)^{2}}{z^{a+1}} $ as $|z| \to \infty$ and $a >0$. (See here.)

The integral also vanishes on the small circle about the origin as its radius goes to zero since $\lim_{z \to 0} f(z) = 0$ if $a$ is less than $2$. And the integral vanishes on the small semicircle about $z=-1$ regardless of the value of $a$.

So we have $$\small \left(1-e^{-2 \pi i(a+1) } \right)\int_{0}^{\infty} \frac{\operatorname{Li}_{2}^{2}(-x)}{x^{a+1}} \, \mathrm dx + \int_{1}^{\infty} \frac{1}{(xe^{\pi i})^{a+1} }\left(4 \pi i \operatorname{Li}_{2} \left(\frac{1}{x} \right) \ln(x) + 2 \pi i \ln^{3}(x) - \frac{4 \pi^{3} i }{3} \ln(x) \right) \, \mathrm dx =0.$$

Making the substitution $t= \frac{1}{x}$ in the second integral and rearranging, we have

$$\int_{0}^{\infty} \frac{\operatorname{Li}_{2}^{2}(-x)}{x^{a+1}} \, \mathrm dx= \pi \csc(\pi a) \int_{0}^{1} t^{a-1} \left(-2 \operatorname{Li}_{2}(t) \ln(t) - \ln^{3}(t) + \frac{2\pi^{2}}{3} \ln(t) \right) \, \mathrm dt, $$

where $$\int_{0}^{1} t^{a-1} \ln^{3}(t) \, \mathrm dt = \frac{\mathrm d^{3}}{\mathrm d a^{3}} \frac{1}{a} = - \frac{6}{a^{4}}, $$

$$\int_{0}^{1} t^{a-1} \ln (t) \, \mathrm dt = \frac{\mathrm d}{\mathrm d a} \frac{1}{a} = - \frac{1}{a^{2}}, $$

and $$ \begin{align} \int_{0}^{1} t^{a-1} \operatorname{Li}_{2}(t)\ln(t) \, \mathrm dt &= \int_{0}^{1} t^{a-1} \ln(t) \sum_{n=1}^{\infty} \frac{t^{n}}{n^{2}} \, \mathrm dt \\ &= - \sum_{n=1}^{\infty} \frac{1}{n^{2}(a+n)^{2}} \\ &= \frac{2}{a^{3}} \sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{a+n} \right) - \frac{1}{a^{2}} \sum_{n=1}^{\infty} \frac{1}{n^{2}} - \frac{1}{a^{2}} \sum_{n=1}^{\infty} \frac{1}{(a+n)^{2}} \\ &= \frac{2}{a^{3}} \left(\psi(a+1) + \gamma \right) - \frac{\pi^{2}}{6a^{2}} - \frac{1}{a^{2}} \left(\psi_{1}(a) -\frac{1}{a^{2}} \right). \end{align} $$

Therefore, $$ \begin{align} \int_{0}^{\infty} \operatorname{Li}_{2}^{2} \left(-x^{-1/a}\right) \, \mathrm dx &= a\int_{0}^{\infty} \frac{\operatorname{Li}_{2}^{2}(-u)}{u^{a+1}} \, \mathrm du \\ &= \small \pi a \csc(\pi a) \left(-\frac{4}{a^{3}} \left(\psi(a) + \frac{1}{a} + \gamma \right) + \frac{2\pi^{2}}{6a^{2}} + \frac{2}{a^{2}} \left(\psi_{1}(a) -\frac{1}{a^{2}} \right) + \frac{6}{a^{4}} - \frac{2 \pi^{2}}{3a^{2}} \right) \\ &= \frac{2 \pi}{a} \csc(\pi a) \left(\psi_{1}(a) - \frac{\pi^{2}}{6}- \frac{2}{a} \psi(a) - \frac{2\gamma}{a} \right) . \end{align}$$

For the case $a=1$, we need to take the limit as $a \to 1$.

The integral $$ \int_{0}^{\infty} \operatorname{Li}_{2}^{\color{red}{3}} \left(-x^{-1/a} \right) \, \mathrm dx$$ can be expressed in terms of the double series $$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m^{2}n^{2} (m+n+a)^{2}}. $$ I don't know if that has a closed-form expression


A potentially baffling solution is to rewrite in terms of Meijer's $G$ function, use the formula for integrating a pair of $G$s, then evaluate.

\begin{align*} \int_0^\infty \, \mathrm{Li}_2^2(-1/x^2) \,\mathrm{d}x &= \int_0^\infty \left( \frac{-1}{x^2} \,{}_3F_2 \left( \begin{matrix} 1,1,1 \\ 2,2 \end{matrix} \middle| -1/x^2 \right) \right)^2 \,\mathrm{d}x \\ &= \int_0^\infty \frac{1}{x^4} \left( \frac{\Gamma(2)^2}{\Gamma(1)^3} G_{3\ 3}^{3\ 1} \left(x^2 \middle| \begin{matrix}1 & 2 & 2 \\ 1 & 1 & 1 \end{matrix} \right) \right)^2 \,\mathrm{d}x \\ &= \frac{1}{2} \int_0^\infty \frac{2x}{x^5} \left( G_{3\ 3}^{3\ 1} \left(x^2 \middle| \begin{matrix}1 & 2 & 2 \\ 1 & 1 & 1 \end{matrix} \right) \right)^2 \,\mathrm{d}x \\ &= \frac{1}{2} \int_0^\infty \frac{1}{u} G_{3\ 3}^{3\ 1} \left(u \middle| \begin{matrix}1 & 2 & 2 \\ 1 & 1 & 1 \end{matrix} \right) \frac{1}{u^{3/2}} G_{3\ 3}^{3\ 1} \left(u \middle| \begin{matrix}1 & 2 & 2 \\ 1 & 1 & 1 \end{matrix} \right) \,\mathrm{d}u \\ &= \frac{1}{2} \int_0^\infty G_{3\ 3}^{3\ 1} \left(u \middle| \begin{matrix}0 & 1 & 1 \\ 0 & 0 & 0 \end{matrix} \right) G_{3\ 3}^{3\ 1} \left(u \middle| \begin{matrix}-1/2 & 1/2 & 1/2 \\ -1/2 & -1/2 & -1/2 \end{matrix} \right) \,\mathrm{d}u \\ &= \frac{1}{2} G_{6\ 6}^{4\ 4} \left( 1 \middle| \, \begin{matrix}0 & 1/2 & 1/2 & 1/2 & 1 & 1 \\ 0 & 0 & 0 & 1/2 & -1/2 & -1/2 \end{matrix} \right) \\ &= \frac{4}{3} \pi (\pi^2 + 24 \ln(2)) \text{.} \end{align*}

I poked at using $G^{1\ 3}_{3\ 3}$ for one ${}_3F_2$ and $G^{3\ 1}_{3\ 3}$ for the other, but this didn't seem to help much.

  • $\begingroup$ In the $G_{6\ 6}^{4\ 4}$ there should be a way to reduce the first two $1/2$s in the top row with the $-1/2$s in the bottom row and the first two $0$s in the bottom row with the first two $1$s in the top row, reducing the order of $G$ and maybe picking up a rational algebraic component in the integrand, but again, this didn't seem to help. $\endgroup$ Mar 1 at 3:58

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