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

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 ?

• What have you tried? Commented Apr 14, 2023 at 2:25
• @DavidG.Stork the methods for solving the other two, contour integration , rossers theorem, ramanujan master theorem , laplace transformations , zeta representations and the usual stuff.
– mick
Commented Apr 14, 2023 at 2:27
• I wonder if it is fruitful to split the integral into two divergent integrals , do some kind of summabilty methods and then add them to get the result. But while being formal ofcourse !
– mick
Commented Apr 14, 2023 at 2:33
• It's an interesting question, but when people ask what your attempts are, they want to see those attempts in the question, not in the comments. Everyone would prefer seeing "I tried $x \mapsto 1/x$ so that my integral becomes $\int_0^1 \frac{\operatorname{Li}(1/x)^2 (1/x-1)}{(1/x)^4}\frac{dx}{x^2}$" or something specific like that. Commented Apr 14, 2023 at 3:02
• Ok I have a proof.... but it is waaayy to long ... Will post later when I have time.
– mick
Commented Apr 14, 2023 at 10:52

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 $$0 we have $$\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 $$\{s\in\mathbb{C}:r<|s|0\lor|\Im s|>r)\}$$ (the part of $$|s| to the left of $$\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.

• (+1) Very nice! Commented Apr 20, 2023 at 5:58
• +1 to you and Marco. There exists an elementary proof by Alexander Lemmens, who is a friend of my mentor (and my mentor came with the integral). I wanted to add the pdf here but adding pdf is not possible ?? So I need to type it out completely ?
– mick
Commented Apr 20, 2023 at 10:54

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.