Show that $ \int_1^{\infty} \frac{\operatorname{li}(x)^2 (x - 1)}{x^4} dx = \frac{5 \pi^2}{36} $

Question

We know

$$\int_{0}^{1}\Big(\frac{\operatorname{li}(x)}{x}\Big)^2dx= \frac{\pi^2}{6}$$

and

$$\int_0^1 4 \space\operatorname{li}(x)^3 \space (x-1) \space x^{-3} dx = \zeta(3) $$

see the proofs :

prove that $\int_{0}^{1}\Big(\frac{\operatorname{li}(x)}{x}\Big)^2dx= \frac{\pi^2}{6}$

Show that $\int_0^1 4 \space\operatorname{li}(x)^3 \space (x-1) \space x^{-3} dx = \zeta(3) $

But we also have this similar one :

$$ \int_1^{\infty} \frac{\operatorname{li}(x)^2 (x - 1)}{x^4} dx = \frac{5 \pi^2}{36} $$

That one is much harder to explain. Notice it is probably harder because the powers of $\operatorname{li}(x)^2$ and $x^4$ do not match and the integral is going from $1$ to $\infty$ !

How to prove this equality ?

Answer 1

I'm just finishing the answer by Marco Cantarini, with $$\newcommand{\li}{\operatorname{li}}\newcommand{\dilog}{\operatorname{Li}_2}\int_1^\infty\li(x)x^{s-1}\,dx=\frac{\log(-1-s)}{s}$$ (for $\Re s<-1$) touched in the past (which calls for the Mellin transform of $\li(x)^2$ via multiplication/convolution). It turns out that the latter has a closed form in terms of the dilogarithm function (which suggests doing integrals with $\li(x)^3$ or $\li(x)^4$ the same way).

As a result, for $a\in\mathbb{C}$ and $0<c<\Re a$ we have $$\int_1^\infty\frac{\li(x)^2}{x^{3+a}}\,dx=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\log s}{1+s}\frac{\log(a-s)}{1+a-s}\,ds.$$

The integral above is transformed into a principal-value integral along the negative real axis: $$\int_1^\infty\frac{\li(x)^2}{x^{3+a}}\,dx=\text{p.v.}\int_0^\infty\frac{\log(a+x)}{1+a+x}\frac{dx}{x-1}.$$ This is done via the Cauchy integral theorem applied to the boundary of $$\{s\in\mathbb{C}:r<|s|<R\land\Re s<c\land(\Re s>0\lor|\Im s|>r)\}$$ (the part of $|s|<R$ to the left of $\Re s=c$, with a notch around the negative real axis), and taking $R\to\infty$ and $r\to 0$. The latter should be done with care, since $s=-1$ is a singularity on the branch cut of the integrand, but its contributions (at the edges of the cut) cancel each other.

A straightforward (but tedious) way to get a closed form of the last integral (in terms of the dilogarithm) is to write $\text{p.v.}\int_0^\infty=\lim\limits_{R\to\infty}\lim\limits_{r\to 0}\left(\int_0^{1-r}+\int_{1+r}^R\right)$ and evaluate the integrals individually. After simplification using dilogarithm identities, if we put $a=s-1$, we get $$f(s):=\int_1^\infty\frac{\li(x)^2}{x^{2+s}}\,dx=\frac1{1+s}\left[\frac{\pi^2}6+\log(s)^2+2\dilog\left(\frac1s\right)\right]$$ for $\Re s\geqslant 1$. The values of $\dilog(1)$ and $\dilog(1/2)$ yield $f(1)=\pi^2/4$ and $f(2)=\pi^2/9$.

Finally, the value of the given integral is $f(1)-f(2)=5\pi^2/36$ as expected.

Answer 2

A partial answer. We start from the Mellin transform $$\int_{1}^{+\infty}x^{s-1}\text{li}(x)dx=\frac{-\log\left(-s-1\right)}{s},\,\text{Re}(s)<-1.$$ Then, from Plancherel's theorem, we get $$\int_{1}^{+\infty}\frac{\text{li}(x)^{2}}{x^{3}}dx=\frac{1}{2\pi}\int_{\mathbb{R}}\frac{\left|\log\left(-it\right)\right|^{2}}{1+t^{2}}dt$$ where we are assuming that $\log(x)$ is the principal value $\text{Log}(z)$. Hence, the problem boils down to the evaluation of $$\frac{1}{\pi}\int_{0}^{+\infty}\frac{\log\left(t\right)^{2}}{1+t^{2}}dt=\frac{\pi^{2}}{8},$$ which can be easily deduced from the Mellin transform $$\int_{0}^{+\infty}\frac{t^{s}}{1+t^{2}}dt=\frac{\pi}{2}\sec\left(\frac{\pi s}{2}\right),\,\text{Re}(s)>-1$$ so $$\int_{1}^{+\infty}\frac{\text{li}(x)^{2}}{x^{3}}dx=\frac{\pi^{2}}{4}.$$ In a similar manner, we have $$\int_{1}^{+\infty}\frac{\text{li}(x)^{2}}{x^{4}}dx=\frac{1}{2\pi}\int_{\mathbb{R}}\frac{\left|\log\left(\frac{1}{2}-it\right)\right|^{2}}{\frac{9}{4}+t^{2}}dt$$ but this time the evaluation of the integral is less obvious. Still working on it.