Integral $I=\int_0^\infty \frac{\ln(1+x) \operatorname{Li}_2 (-x)}{x^{3/2}} dx$
Hello can you please help me solve this integral $$ \int_0^\infty \frac{\ln(1+x) \operatorname{Li}_2 (-x)}{x^{3/2}} dx=-\frac{2\pi}{3}(\pi^2+24\ln 2). $$ I am trying to work through all logarithmic integrals. Note, the Polylogarithm function is given by $\operatorname{Li}_2(-x)$ and is defined by $$ \operatorname{Li}_2(-x)=\sum_{k=1}^\infty \frac{(-x)^k}{k^2}, \ |-x|<1 $$ and can be extended using analytical continuation for $|-x|>1$. We also know that $$ \frac{d}{dx} \operatorname{Li}_2(-x)=-\frac{\ln(1+x)}{x}. $$Thanks!
Solution 1:
One, rather straightforward way, is the following:
Make the change of variables $x=y^2$ to rewrite the integral as $$\mathcal{I}=2\int_0^{\infty} \frac{\ln(1+y^2)\,\operatorname{Li}_2(-y^2)\,dy}{y^2}. $$
Integrate by parts to kill the logarithm, using that $$\int \frac{\ln(1+y^2)\,dy}{y^2}=2\,\Im \ln(1+iy)-\frac{\ln(1+y^2)}{y}-\pi, \tag{1}$$ and also the expression for the derivative $\operatorname{Li}_2'(-y^2)=-2\frac{\ln(1+y^2)}{y}$. The constant $\pi$ on the right of (1) is needed to ensure non-divergent (actually, vanishing) boundary contribution at $y=\infty$. We thus find $$\mathcal{I}=4\int_0^{\infty}\left(2\,\Im \ln(1+iy)-\frac{\ln(1+y^2)}{y}-\pi\right)\frac{\ln(1+y^2)}{y}dy.\tag{2}$$
The antiderivative of (2) can be expressed in terms of polylogarithms and elementary functions. Substituting the bounds, one finally obtains $$\boxed{\mathcal{I}=-\frac{2\pi}{3}\Bigl(\pi^2+24\ln 2\Bigr)}$$
Solution 2:
I deleted my contour integration approach because it was too long, but I kept the shorter approach that uses Ramanujan's master theorem.
Using the Cauchy product of two power series and partial fraction decomposition, we find that $$ \begin{align} \ln(1+x) \operatorname{Li}_{2}(-x) &= \small \sum_{k=0}^{\infty} (-1)^{k+1} x^{k+2} \sum_{j=0}^{k} \frac{1}{(j+1)^{2}} \frac{1}{k-j+1} \\ &= \small \sum_{k=0}^{\infty} (-1)^{k+1} x^{k+2} \sum_{j=0}^{k} \left( \frac{1}{k+2} \frac{1}{(j+1)^{2}}+ \frac{1}{(k+2)^{2}} \frac{1}{j+1} + \frac{1}{(k+2)^{2}} \frac{1}{k-j+1} \right) \\ &= \small -x^{2}\sum_{k=0}^{\infty} (-x)^{k} \left(\frac{1}{k+2}H_{k+1}^{(2)}+\frac{2}{(k+2)^{2}}H_{k+1} \right) \\&= \small -x^{2}\sum_{k=0}^{\infty} (-x)^{k}\left(\frac{\frac{\pi^{2}}{6} - \psi_{1}(k+2)}{k+2}+\frac{2\left(\gamma+\psi(k+2) \right)}{(k+2)^{2}} \right) , \end{align}$$ where $H_{k}$ is the $k$th harmonic number, $H_{k}^{(2)}$ is the $k$th generalized harmonic number of order $2$, $\psi(x)$ is the digamma function, $\psi_{1}(x)$ is the trigamma function, and $\gamma$ is the Euler-Mascheroni constant.
Applying Ramanujan's master theorem, we get $$ \begin{align} \int_{0}^{\infty} \frac{\ln(1+x)\operatorname{Li}_{2}(-x) }{x^{2}} \, x^{\alpha-1} \, \mathrm dx &= -\Gamma(\alpha) \left(\frac{\frac{\pi^{2}}{6} - \psi_{1}(2- \alpha)}{2- \alpha}+\frac{2\left(\gamma+\psi(2- \alpha) \right)}{(2- \alpha)^{2}} \right) \Gamma(1- \alpha) \\ &= -\frac{\pi}{\sin (\alpha \pi)}\left(\frac{\frac{\pi^{2}}{6} - \psi_{1}(2- \alpha)}{2- \alpha }+\frac{2\left(\gamma+\psi(2- \alpha) \right)}{(2- \alpha)^{2}} \right), \end{align}$$
which holds for $0 < \alpha < 2$.
(For $\alpha =1$, the right side of the equation should be treated as a limit.)
Letting $\alpha= \frac{3}{2}$, we get $$ \begin{align} \int_{0}^{\infty} \frac{\ln(1+x)\operatorname{Li}_{2}(-x) }{x^{3/2}} \, \mathrm dx &= \pi \left(\frac{\frac{\pi^{2}}{6} - \psi_{1} \left(\frac{1}{2} \right)}{\frac{1}{2}} +\frac{2 \left(\gamma +\psi \left(\frac{1}{2} \right) \right)}{\frac{1}{4}}\right) \\ &= \pi \left(\frac{\frac{\pi^{2}}{6}- \frac{\pi^{2}}{2}}{\frac{1}{2}} - \frac{4 \ln 2}{\frac{1}{4}} \right) \\ &= \pi \left(-\frac{2 \pi^{2}}{3} - 16 \ln 2 \right) \\ &= -\frac{2 \pi}{3} \left(\pi^{2} + 24 \ln 2 \right). \end{align}$$
Similarly, Ramanujan's master theorem shows that $$ \small \int_{0}^{\infty} \frac{\ln(1+x) \operatorname{Li}_{{\color{red}{3}}}(-x)}{x^{2}} \, x^{\alpha-1} \, \mathrm dx = - \frac{\pi}{\sin (\alpha \pi)} \left(\frac{\zeta(3)+\frac{\psi_{2}(2- \alpha)}{2}}{(2- \alpha)} + \frac{\frac{\pi^{2}}{6} - \psi_{1}(2- \alpha)}{(2-\alpha)^{2}} + \frac{2\left(\gamma +\psi(2 - \alpha)\right)}{(2-\alpha)^{3}} \right).$$
Letting $\alpha= \frac{3}{2}$, we get $$ \begin{align} \int_{0}^{\infty} \frac{\ln(1+x) \operatorname{Li}_{3}(-x)}{x^{3/2}} \, \mathrm dx &= \pi \left(-\frac{6 \zeta(3)}{\frac{1}{2}} - \frac{\frac{\pi^{2}}{3}}{\frac{1}{4}} - \frac{4 \ln 2}{\frac{1}{8}} \right) \\ &= - \frac{4 \pi}{3} \left(9 \zeta(3) + \pi^{2} + 24 \ln 2 \right). \end{align}$$
Appendix:
$\psi \left(\frac{1}{2} \right) = - \gamma + \sum_{n=0}^{\infty} \left(\frac{1}{n+1} - \frac{2}{2n+1}\right) = - \gamma + 2 \lim_{N \to \infty} \left(H_{N} - H_{2N} \right) = - \gamma - 2 \ln 2 $
$\psi_{1} \left(\frac{1}{2} \right) = 4 \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} = 4 (1-2^{-2})\zeta(2) = \frac{\pi^{2}}{2}$
$\psi_{2} \left(\frac{1}{2} \right) = -16 \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{3}} = -16 (1-2^{-3})\zeta(3) = -14 \zeta(3)$