Prove that $\int_0^{\infty} \frac{{\rm Li}_2(-x)}{1+x^2}\, {\rm d}x = - \frac{7 \pi^3}{96}$

Question

Here is something interesting I came up while playing with dilogs.

Prove that

$$\int_0^{\infty} \frac{{\rm Li}_2(-x)}{1+x^2}\, {\rm d}x = - \frac{7 \pi^3}{96}$$

maybe we can derive a general form. For instance

$$\int_0^{\infty} \frac{{\rm Li}_2(-x)}{1+x^\alpha} \, {\rm d}x , \; \; \alpha \in \mathbb{N} \setminus \{1\}$$

Even more interesting is that if we replace $\text{Li}_2(-x)$ with $\text{Li}_2(x)$, the result is $\frac{5\pi^3}{96} - i\pi G$

$G$ being Catalan's Constant.

It would not be too much of a leap to imagine the results with $-x$ and $x$ are somehow connected to each other.

Answer 1

Let $x\mapsto \frac{1}{x}$, then $$ I=\int_0^{\infty} \frac{\operatorname{Li_2}(-x)}{1+x^2} d x =\int_0^{\infty} \frac{\operatorname{Li_2}\left(-\frac{1}{x}\right)}{1+\frac{1}{x^2}} \frac{d x}{x^2}=\int_0^{\infty} \frac{\operatorname{Li_2}\left(-\frac{1}{x}\right)}{1+x^2} d x $$ Adding them together yields $$ \begin{aligned} 2 I & =\int_0^{\infty} \frac{\operatorname{Li_2}(-x)+ \operatorname{Li_2}\left(-\frac{1}{x}\right)}{1+x^2} d x \\ & =\int_0^{\infty} \frac{-\frac{\pi^2}{6}-\frac{1}{2} \ln ^2 x}{1+x^2} d x \cdots(*)\\ & =-\frac{\pi^2}{6}\left[\tan ^{-1} x\right]_0^{\infty}-\frac{1}{2} \int_0^{\infty} \frac{\ln ^2 x}{1+x^2} d x \\ & =-\frac{\pi^3}{12}-\frac{1}{2}\left(\frac{\pi^3}{8}\right) \cdots(**)\\&=-\frac{7 \pi^3}{48} \end{aligned} $$ We can now conclude that $$I= -\frac{7 \pi^3}{96} $$ where $(*)$ using the identity: $\operatorname{Li_2}(-x)+ \operatorname{Li_2}\left(-\frac{1}{x}\right) =-\frac{\pi^2}{6}-\frac{1}{2} \ln ^2 x $ and $(**)$ using the result:$\int_0^{\infty} \dfrac{\ln ^2 x}{1+x^2} d x=\dfrac{\pi^3}{8}$.

Answer 2

$$ \begin{eqnarray} I&=&\int_0^{\infty} \frac{\operatorname{Li_2}(-x)}{1+x^2} d x =\int_0^{1} \frac{\operatorname{Li_2}\left(-x\right)}{1+{x^2}} dx+\int_1^{\infty} \frac{\operatorname{Li_2}\left(-x\right)}{1+x^2} \overset{x\to1/x}{d x}\\ &=&\int_0^{1} \frac{\operatorname{Li_2}\left(-x\right)}{1+{x^2}} dx+\int_0^{1} \frac{\operatorname{Li_2}\left(-\frac1x\right)}{1+x^2} {d x}\\ &=&\int_0^{1} \frac{\operatorname{Li_2}\left(-x\right)+\operatorname{Li_2}\left(-\frac1x\right)}{1+{x^2}} dx\\ &\overset{IBP}=&\bigg[\operatorname{Li_2}\left(-x\right)+\operatorname{Li_2}\left(-\frac1x\right)\bigg]\arctan(x)\bigg|_0^1+\int_0^1\frac1x\arctan(x)\ln xdx\tag{*}\\ &=&-\frac{\pi^3}{24}-\frac{\pi^3}{32}\\ &=&-\frac{7\pi^3}{96}. \end{eqnarray} $$ The last integral in (*) can be evaluated by series expansion of $\arctan(x)$.

Answer 3

Let $$ f(z) = \frac{\operatorname{Li}_{s}(-z) \ln(-z)}{1+z^{2}}, \quad s >0,$$ where $\operatorname{Li}_{s}(z)$ is the principal branch of the polylogarithm of order $s$, and $\ln(z)$ is the prinicpal branch of the logarithm.

$\operatorname{Li}_{s}(-z)$ has a branch cut on $(- \infty, -1]$.

On the upper side of the branch cut, the imaginary part of $\operatorname{Li}_{s}(-z)$ is $- \frac{\pi i}{\Gamma(s)} \ln^{s-1}(-z)$, $z \le -1$.

And on the lower side of the branch cut, the imaginary part of $\operatorname{Li}_{s}(-z)$ is $\frac{\pi i}{\Gamma(s)} \ln^{s-1}(-z)$, $z \le - 1$.

(See here and this question.)

Integrating $f(z)$ around the same double keyhole contour that I used in this answer to evaluate the Mellin transform of $\operatorname{Li}_{s}(-x) $, we get

$$\begin{align} - 2 \pi i \int_{0}^{\infty} \frac{\operatorname{Li}_{s}(-x)}{1+x^{2}} \, \mathrm dx - \frac{2 \pi i }{\Gamma(s)} \int_{1}^{\infty} \frac{\ln^{s}(x)}{1+x^{2}} \, \mathrm dx&= 2 \pi i \left(\operatorname*{Res}_{z=i} f(z) + \operatorname*{Res}_{z=-i} f(z) \right) \\ &= 2 \pi i \left(\frac{\operatorname{Li}_{s}(-i) (-\frac{ i \pi}{2})}{2i} + \frac{\operatorname{Li}_{s}(i) (\frac{i \pi}{2})}{-2i}\right) \\ &\overset{(1)}{=} 2 \pi i \left(-\frac{\pi \operatorname{Li}_{s}(-1) }{2^{s+1}} \right) \\ &\overset{(2)}= 2 \pi i \left(\frac{\pi \, \eta(s) }{2^{s+1}} \right). \end{align}$$

Therefore, $$ \begin{align} \int_{0}^{\infty} \frac{\operatorname{Li}_{s}(-x)}{1+x^{2}} \, \mathrm dx &= - \frac{1}{\Gamma(s)} \int_{1}^{\infty} \frac{\ln^{s}(x)}{1+x^{2}} \, \mathrm dx - \frac{\pi \, \eta(s)}{2^{s+1}} \\&= - \frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{u^{s}}{1+e^{2u}} \, e^{u} \, \mathrm du - \frac{\pi \, \eta(s)}{2^{s+1}} \\&= -\frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{u^{s}}{e^{-2u}+1} \, e^{-u} \, \mathrm du - \frac{\pi \, \eta(s)}{2^{s+1}} \\&\overset{(3)}= -\frac{1}{\Gamma(s)} \, \Gamma(s+1) \beta(s+1) - \frac{\pi \, \eta(s)}{2^{s+1}} \\&= -s \beta(s+1) - \frac{\pi \, \eta(s)}{2^{s+1}} . \end{align}$$

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$(1)$ Polylogarithm square relationship

$(2)$ https://en.wikipedia.org/wiki/Polylogarithm#Relationship_to_other_functions

$(3)$ https://en.wikipedia.org/wiki/Dirichlet_beta_function#Definition

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For $s=1$, we have $$\int_{0}^{\infty} \frac{\operatorname{Li}_{1}(-x)}{1+x^{2}} \, \mathrm dx = - \int_{0}^{\infty} \frac{\ln(1+x)}{1+x^{2}} \, \mathrm dx = -\beta(2) - \frac{\pi \eta(1)}{2^{2}} = -G - \frac{\pi \ln(2)}{4}. $$

And for $s=2$, we have $$\int_{0}^{\infty} \frac{\operatorname{Li}_{2}(-x)}{1+x^{2}} \, \mathrm dx = -2 \beta(3) - \frac{\pi \eta(2)}{2^{3}} = - \frac{\pi^{3}}{16} - \frac{\pi^{3}}{96} = - \frac{7 \pi^{3}}{96}.$$

Answer 4

I suppose we could also use the integral representation of polylogarithms $\displaystyle \operatorname{Li}_{\nu}(x) = \frac{x}{\Gamma(\nu)}\int_0^{\infty} \frac{t^{\nu -1}}{e^t - x}\,dt$ (where, $\Re(\nu) > 0$) and rewrite our integral as:

\begin{align}\int_0^{\infty} \frac{\operatorname{Li}_2(-x)}{1+x^2}\,dx &= \int_0^{\infty} \int_0^{\infty} \frac{-xt}{(e^t+x)(x^2+1)}\,dt\,dx\\&= -\int_0^{\infty} \int_0^{\infty} \frac{t}{(1+e^{2t})(1+x^2)} + \frac{te^{t}}{1+e^{2t}}\left(\frac{x}{1+x^2} - \frac{1}{x+e^t}\right)\,dx\,dt\\&= -\frac{\pi}{2}\int_0^{\infty}\frac{t}{1+e^{2t}}\,dt - \int_0^{\infty}\frac{t^2e^{t}}{1+e^{2t}}\,dt\\&= -\frac{\pi}{8}\int_0^{\infty}\frac{t}{1+e^{t}}\,dt - \frac{1}{8}\int_0^{\infty}\frac{t^2e^{t/2}}{1+e^{t}}\,dt\\&= \frac{\pi}{8}\operatorname{Li}_2(-1) - 2\sum\limits_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}\\&= -\frac{\pi^3}{96} - \frac{\pi^3}{16} = -\frac{7\pi^3}{96}\end{align}

where, we have used the partial fraction decomposition $$\frac{x}{(x+a)(1+x^2)} = \frac{1}{(1+a^2)(1+x^2)} + \frac{a}{1+a^2}\left(\frac{x}{1+x^2} - \frac{1}{x+a}\right)$$ in the second step.

So, I suppose a natural way to proceed with the calculation of $\displaystyle \int_0^{\infty} \frac{\operatorname{Li}_2(-x)}{1+x^\alpha}\,dx$ (where, $\alpha \in \mathbb{N}\setminus\{1\}$) might be to use partial fraction decomposition of $\displaystyle \frac{x}{(1+x^\alpha)(x+a)}$ (but I doubt the calculation would be pretty .. :P)

For example, when $\alpha = 2n \ge 4$ (even integers) we have a partial fraction decomposition: $$\frac{x}{(1+x^{2n})(x+a)} = \frac{1}{(1+a^{2n})}\sum\limits_{j=1}^{2n-2} (-1)^ja^{j+1} \frac{x^{2n-1-j}}{1+x^{2n}} + \frac{a}{(1+a^{2n})}\left(\frac{x^{2n-1}}{1+x^{2n}} - \frac{1}{x+a}\right)$$ and using the fact that $\displaystyle \int_0^\infty\frac{x^m}{1+x^n}\,dx = \frac{\pi}{n}\csc\left(\frac{\pi (m+1)}{n}\right)$ we have: $$\int_0^{\infty} \frac{x}{(1+x^{2n})(x+a)}\,dx = \frac{\pi}{2n(1+a^{2n})}\sum\limits_{j=1}^{2n-2} (-1)^ja^{j+1} \csc\left(\frac{\pi j}{2n}\right) + \frac{a\log a}{(1+a^{2n})}$$

Thus, \begin{align*}\int_0^{\infty} \frac{\operatorname{Li}_2(-x)}{1+x^{2n}}\,dx &= \frac{\pi}{2n}\sum\limits_{j=1}^{2n-2} (-1)^{j-1}\csc\left(\frac{\pi j}{2n}\right) \int_0^{\infty} \frac{te^{(j+1)t} }{1+e^{2nt}}\,dt - \int_0^{\infty}\frac{t^2e^{t}}{(1+e^{2nt})}\,dt\\&=\frac{\pi}{(2n)^3} \sum\limits_{j=1}^{2n-2} (-1)^{j-1}\csc\left(\frac{\pi j}{2n}\right) \int_0^{\infty} \frac{te^{\frac{(j+1)t}{2n}} }{1+e^{t}}\,dt - \frac{1}{(2n)^3}\int_0^{\infty}\frac{t^2e^{\frac{t}{2n}}}{(1+e^{t})}\,dt\end{align*} which may be expressed in terms of Trigamma function and Hurwitz-zeta function with,

$\displaystyle \int_0^{\infty} \frac{te^{at}}{1+e^{t}}\,dt = \frac{1}{4}\left(\psi^{(1)}\left(\frac{1}{2} - \frac{a}{2}\right) - \psi^{(1)}\left(1 - \frac{a}{2}\right)\right)$

and $\displaystyle \int_0^{\infty} \frac{t^2e^{at}}{1+e^{t}} = \frac{1}{4}\left(\sum\limits_{n=0}^{\infty} \frac{1}{\left(n + \frac{1}{2} - \frac{a}{2}\right)^{3}} - \sum\limits_{n=0}^{\infty} \frac{1}{\left(n + 1 - \frac{a}{2}\right)^{3}}\right) = \frac{1}{4}\left(\zeta\left(3,\frac{1}{2}-\frac{a}{2}\right) - \zeta\left(3,1-\frac{a}{2}\right)\right)$ where, $\Re (a) < 1$.

$2^{nd}$ attempt

This is a consequence of the nice behaviour of $dx/(1+x^2)$ under $x \mapsto 1/x$ and the dilog identity $$\text{Li}_2(-x ) +\text{Li}_2( -1/x) = -\frac{\pi^2}{6} - \frac12 \ln^2 x.$$

This gives $$\int_0^\infty \frac{\text{Li}_2(-x)}{1+x^2}dx = \int_0^\infty \frac{\text{Li}_2(-1/x)}{1+x^2}dx \\ = \frac12 \int_0^\infty \frac1{1+x^2} \left( -\frac{\pi^2}{6}-\frac12 \ln^2 x \right)dx \\= -\frac12 \frac{\pi^2}{6} \frac{\pi}{2} - \frac12 \int_0^1 \frac{\ln^2 x}{1+x^2} dx \\= - \frac{\pi^3}{24}- \frac12 \sum_{n=0}^{\infty} \frac{2!\,(-1)^n}{(2n+1)^3} \\= - \frac{7 \pi^3}{96}.$$ $\displaystyle \frac{x^{s/2-1}}{1+x^s} dx$ behaves nicely under $x \mapsto 1/x$, and that gives: $$\large \int_0^{\infty} \frac{x^{s/2-1}}{1+x^s} \text{Li}_2(-x) dx=- \frac{\pi^3}{4} \left( \frac1{3 s}+ \frac1{ s^3}\right).$$

Answer 5

Prove that $$\boxed{\int_0^{\infty} \frac{\operatorname{Li_2}(x)}{1+x^2} d x=\frac{5 \pi^3}{96}-i \pi G}$$ Proof:

$$ \begin{aligned} \int_0^{\infty} \frac{\operatorname{Li_2}(x)}{1+x^2} d x = & \int_0^1 \frac{\operatorname{Li_2}(x)}{1+x^2} d x+\int_1^{\infty} \frac{\operatorname{Li_2}(x)}{1+x^2} d x \\ = & \int_0^1 \frac{\operatorname{Li_2}(x)}{1+x^2} d x+\int_0^1 \frac{\operatorname{Li_2}\left(\frac{1}{x}\right)}{1+x^2} d x, \textrm{ via } x\mapsto\frac{1}{x} \\ = & \int_0^1 \frac{\operatorname{Li_2}(x)+ \operatorname{Li_2}\left(\frac{1}{x}\right)}{1+x^2} d x \end{aligned} $$ Plugging the identity

$$ \begin{aligned} \operatorname{Li}_2(x)+\operatorname{Li}_2\left(\frac{1}{x}\right)=i \pi \ln x-\frac{1}{2} \ln ^2 x+\frac{\pi^2}{3} \end{aligned} $$

into the last sum in the numerator of the integrand yields $$ \begin{aligned} I & =i \pi \int_0^1 \frac{\ln x}{1+x^2} d x-\frac{1}{2} \int_0^1 \frac{\ln ^2 x}{1+x^2} d x+\frac{\pi^2}{3} \int_0^1 \frac{1}{1+x^2} d x \\ & =-i \pi G-\frac{1}{2} \cdot \frac{\pi^3}{16}+\frac{\pi^3}{12} \\ & =\frac{5 \pi^3}{96}-i \pi G \end{aligned} $$ Wish you enjoy the solution!

Answer 6

\begin{align}J&=\int_0^{\infty} \frac{{\rm Li}_2(-x)}{1+x^2}\, {\rm d}x\\ &=-\int_0^\infty \frac{1}{1+x^2}\left(\underbrace{\int_0^x \frac{\ln(1+t)}{t}dt}_{\text{IBP}}\right)\\ &=-\underbrace{\int_0^\infty \frac{\ln(1+x)\ln x}{1+x^2}dx}_{=A}+\int_0^\infty\frac{1}{1+x^2}\left(\int_0^x\frac{\ln t}{1+t}dt\right)dx\\ &=-A+\int_0^\infty\left(\int_0^1\frac{x\ln(tx)}{(1+x^2)(1+tx)}dt\right)dx\\ &=-A+\int_0^\infty\left(\int_0^1\frac{x\ln x}{(1+x^2)(1+tx)}dt\right)dx+\int_0^1\left(\int_0^\infty\frac{x\ln t}{(1+x^2)(1+tx)}dx\right)dt\\ &=-A+\int_0^\infty \left[\frac{\ln(1+tx)\ln x}{1+x^2}\right]_{t=0}^{t=1}dx+\int_0^1 \left[\frac{\ln\left(\frac{\sqrt{1+x^2}}{1+tx}\right)\ln t}{1+t^2}+\frac{t\arctan x\ln t}{1+t^2}\right]_{x=0}^{x=\infty}\!\!\!dt\\ &=-A+A-\int_0^1\frac{\ln^2 t}{1+t^2}dt+\frac{\pi}{2}\underbrace{\int_0^1\frac{t\ln t}{1+t^2}dt}_{u=t^2}\\ &=-\underbrace{\int_0^1\frac{\ln^2 t}{1+t^2}dt}_{=\frac{\pi^3}{16}}+\frac{\pi}{8}\underbrace{\int_0^1\frac{\ln u}{1+u}du}_{=-\frac{\pi^2}{12}}=\boxed{-\frac{7\pi^3}{96}} \end{align}