$ \int_0^\infty dx\, x^{-3/4} \ln(1+x) \operatorname{Li}_2 \left( -\frac 1 x \right)$

Question

How to show the following identity from Wolfram?

$$ \int_0^\infty dx\, x^{-3/4} \ln(1+x) \operatorname{Li}_2 \left( -\frac 1 x \right) =-2\pi \sqrt 2\left[\frac{5 \pi^2}{3} + 16 \left(3 \ln 2 + G - 4 \right) \right]. $$ Here $G$ is Catalan's constant.

I thought of using $\operatorname{Li}_2(-x) + \operatorname{Li}_2(-1/x) = -\zeta(2) - \frac 1 2 \ln(x)^ 2$, but the result is not much easier.

Answer 1

We start from the integral $$ \int_{0}^{\infty} \frac{x^u }{(1+s x)(1+t x) }dx= \frac{\pi}{\sin \pi u} \cdot \frac{s^u - t^u}{s^u t^u (s-t)} $$ for $s$, $t>0$ and $0 < u < 1$.

Integrating with respect to $t$ we get \begin{eqnarray} \int_{0}^{\infty} \frac{x^u \log ( 1 + \frac{1}{t x})}{(1+s x)}= - \frac{\pi}{\sin \pi u}\cdot \frac{1}{s^{u+1}} \cdot \left(\, B(s/t;1+u, 0) + \log( 1 - s/t\,)\,\right) \end{eqnarray} for $u>0$ and $0 < s < t$, where $B(z;a,b) \colon = \int_0^z v^{a-1}(1-v)^{b-1} d v$ is the incomplete Beta function. Note that we have $B(z;a,0) = z^a \cdot \Phi(z, 1, a)$, where $\Phi(z,s,a) \colon = \sum_{n \ge 0} \frac{z^n}{(n+ a)^s}$ is the Lerch transcendent.

Integrating again with respect to $t$ we get \begin{eqnarray} \int_0^{\infty}\frac{x^u}{1 + s x}\, Li_2(-\frac{1}{t x}) = \frac{\pi}{\sin \pi u \cdot (s t)^{1+u}} \cdot \left( s^{1+u} \Phi(s/t, 2, 1+u) - t^{1+u} Li_2(s/t)\right) \end{eqnarray} again for $u>0$ and $0 < s < t$. In particular, for $u=1/4$ we get $$\int_0^{\infty}\frac{x^{1/4}}{1 + s x}\, Li_2(-\frac{1}{t x}) = \frac{\pi \sqrt{2}}{(s t)^{5/4}} \cdot \left( s^{5/4} \Phi(s/t, 2, 5/4) - t^{5/4} Li_2(s/t)\right)$$ Now integrate with respect to $s$ and get, with $t=1$ and $s = v^4$: \begin{eqnarray} \int_0^{\infty}x^{-3/4}\log(1 + v^4 x)\, Li_2(-\frac{1}{ x}) =-\frac{4 \pi \sqrt{2}}{v}\left( v^5 \phi(v^4,2,5/4) -Li_2(v^4) - 16 v + 8(1-v) \arctan v + 4 (1+v) \log [(1+v)(1+v^2)] +\\+ 4 (1+v)\log(1-v) + 8(1+v) \textrm{arctanh} (v)\right) \end{eqnarray} for $0 < v<1$ and taking $v \to 1$ we get $$\int_0^{\infty}x^{-3/4}\log(1 + x)\, Li_2(-\frac{1}{ x})= \frac{2}{3} \sqrt{2} \pi \,( \, 96 + \pi^2 -144 \log 2 - 6\, \zeta(2, 5/4)\,)= -3.339758...$$

and now use that $\zeta(2, 5/4) = \pi^2 + 8 G - 16$, where $\zeta(z, a)$ is the Hurwitz zeta function, and $G$ is the Catalan's constant.

Answer 2

Using the dilogarithm inversion formula, we can find a closed-form expression for $$\int_{0}^{\infty} \ln(1+x) \operatorname{Li}_{2} \left(-\frac{1}{x} \right) x^{s-1} \, \mathrm dx = -\int_{0}^{\infty} \ln(1+x) \left(\operatorname{Li}_{2} (-x) + \zeta(2)+ \frac{\ln^{2}(x)}{2} \right) x^{s-1} \, \mathrm dx $$ for $-1 < \Re(s) < 1$ in terms of the digamma and trigamma functions.

We will first need to restrict the real part of $s$ to $-1 < \Re(s) <0$ so that we can break up the integral into three convergent integrals.

Since the Mellin transform defines an analytic function where the integral converges absolutely, the final result will hold for $-1 < \Re(s) <1$ due to the identity theorem.

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I used Ramanujan's master theorem in this answer to show that $$\small\int_{0}^{\infty} \frac{\ln(1+x) \operatorname{Li}(-x)}{x^{2}} x^{\alpha -1} \, \mathrm dx = -\pi \csc(\pi \alpha )\left(\frac{\zeta(2) - \psi_{1}(2- \alpha)}{2- \alpha }+\frac{2\left(\gamma+\psi(2- \alpha) \right)}{(2- \alpha)^{2}} \right), \quad 0 < \Re(\alpha) <2.$$

Replacing $\alpha$ with $s+2$, we get $$\small\int_{0}^{\infty} \ln(1+x) \operatorname{Li}(-x) x^{s -1} \, \mathrm dx = -\pi \csc(\pi s)\left(\frac{ \psi_{1}(- s)- \zeta(2)}{s}+\frac{2\left(\gamma+\psi(-s) \right)}{s^{2}} \right), \quad -2 < \Re(s) <0. $$

Ramanujan's master theorem also shows that $$\int_{0}^{\infty} \frac{\ln(1+x)x^{\alpha-1}}{x} \, \mathrm dx = \Gamma(\alpha)\frac{\Gamma(1-\alpha)}{1-\alpha} = \frac{\pi \csc(\pi \alpha)}{1-\alpha}, \quad 0<\Re(\alpha) <1 . $$

Replacing $\alpha$ with $s+1$, we get $$\int_{0}^{\infty} \ln(1+x) x^{s-1} \, \mathrm dx = \frac{\pi \csc(\pi s)}{s}, \quad -1 < \Re(s) < 0. $$

And differentiating under the integral sign twice, we get $$\begin{align} &\int_{0}^{\infty} \ln(1+x) \ln^{2}(x) x^{s-1} \, \mathrm dx \\ &= \frac{\mathrm d^{2}}{\mathrm d s^{2}} \frac{\pi \csc(\pi s)}{s} \\ &= \pi \left(\csc (\pi s)\right)^{''} \frac{1}{s} + 2\pi \left(\csc (\pi s)\right)' \left(\frac{1}{s} \right)^{'} + \pi \csc(\pi s) \left(\frac{1}{s} \right)^{''} \\ &= \pi^{2} \left(- \csc(\pi s) \cot(\pi s) \right)^{'} \frac{1}{s} + \frac{2\pi^{2}\csc(\pi s) \cot(\pi s)}{s^{2}} + \frac{2\pi \csc(\pi s)}{s^{3}} \\ &= \frac{\pi^{3}\csc(\pi s)^{3}+ \pi^{3}\csc(\pi s) \cot^{2}(\pi s) }{s}+ \frac{2\pi^{2}\csc(\pi s) \cot(\pi s)}{s^{2}} + \frac{2\pi \csc(\pi s)}{s^{3}} \\ &= \pi \csc(\pi s) \left(\frac{\pi^{2}\left( \csc^{2}(\pi s)+ \cot^{2}(\pi s)\right)}{s} + \frac{2\pi \cot(\pi s)}{s^{2}} + \frac{2}{s^{3}}\right). \end{align}$$

Therefore, $$ \begin{align} I(s) &= \int_{0}^{\infty} \ln(1+x) \operatorname{Li}_{2} \left(-\frac{1}{x} \right) x^{s-1} \, \mathrm dx \\ &= \small \pi \csc(\pi s)\left(\frac{ \psi_{1}(- s)- 2 \zeta(2)}{s}+\frac{2\left(\gamma+\psi(-s) \right)}{s^{2}} -\frac{\pi^{2}\left( \csc^{2}(\pi s)+ \cot^{2}(\pi s)\right)}{2s} - \frac{\pi \cot(\pi s)}{s^{2}} - \frac{1}{s^{3}}\right)\end{align}$$ for $ -1 <\Re(s) < 1$.

The singularity at $s=0$ is removable.

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Using the recurrence relation and the reflection formula of the digamma function, we have $$\psi \left(- \frac{1}{4} \right) = \psi \left(\frac{3}{4} \right) -\frac{1}{\left(-\frac{1}{4}\right)}= \psi\left(\frac{1}{4} \right) + \pi \cot \left(\frac{\pi}{4} \right)+4 = \psi \left(\frac{1}{4} \right) + \pi + 4, $$ where, by Gauss' digamma theorem , $$\psi \left(\frac{1}{4} \right) = - \gamma - 3\ln(2) - \frac{\pi}{2}. $$

And using the recurrence relation and reflection formula of the trigamma function, we have $$\psi_{1} \left(-\frac{1}{4} \right) = \psi_{1} \left(\frac{3}{4} \right) + \frac{1}{\left(-\frac{1}{4} \right)^{2}} = \pi^{2} \csc^{2} \left(\frac{\pi}{4} \right)- \psi_{1} \left(\frac{1}{4} \right)+16= 2 \pi^{2} - \psi_{1} \left(\frac{1}{4} \right)+16,$$

where $$\begin{align} \psi_{1} \left(\frac{1}{4} \right) &= 16\sum_{n=0}^{\infty} \frac{1}{(4n+1)^{2}} \\ &= 8 \left( \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} + \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2}}\right)\\ &= 8 \left((1-2^{-2})\zeta(2)+G \right) \\ &= \pi^{2} + 8 G. \end{align}$$

Therefore, $$ \begin{align} I(1/4) &= \int_{0}^{\infty} \ln(1+x) \operatorname{Li}_{2} \left(- \frac{1}{x} \right) x^{-3/4} \, \mathrm dx \\ &= \sqrt{2} \pi \left(4 \left(\pi^{2}-8G+16- \frac{\pi^{2}}{3} \right)+ 32 \left(\frac{\pi}{2}-3 \ln (2) +4 \right) - 2 \pi^{2} (3) -16 \pi -64 \right) \\ &= \sqrt{2} \pi \left(\frac{8 \pi^{2}}{3}-32 G +64 +16 \pi-96 \ln(2) +128 - 6 \pi^{2} -16 \pi - 64\right) \\ &= \sqrt{2} \pi \left(- \frac{10\pi^{2}}{3}-32G +128 -96 \ln(2) \right) \\ &= - 2 \sqrt{2} \pi \left(\frac{5\pi^{2}}{3}+ 16 \left(3 \ln(2)+G-4 \right) \right), \end{align}$$ confirming the value given on The Wolfram Functions Site.