prove that :
$$\int_{0}^{1}\Big(\frac{\operatorname{li}(x)}{x}\Big)^2dx< \frac{\pi^2}{6}$$
Where $\operatorname{li}(x)$ denotes the logarithm integral .
The inequality is very very sharp .I have tried integration by parts and power series but we need a lot of terms to get the result.So i was thinking to another methods.Since we have an integral representation of $\zeta(2)$ why not make a comparison of two functions?
It's the kind of result which is misleading (is there a question for that I think yes )
So if you have a tricky way you are very welcome .
Any helps is greatly appreciated .
Thanks a lot for your contributions.
Update :
I work with WA wich gives a difference between this two quantities
So I'm really lost to prove this :
$$\int_{0}^{1}\Big(\frac{\operatorname{li}(x)}{x}\Big)^2dx= \frac{\pi^2}{6}$$
My work :
The integral is equivalent to :
$$\int_{0}^{\infty}\frac{\operatorname{li}^2\Big(\frac{x}{x+1}\Big)}{x^2}dx$$
The next idea is to apply the RMT but I can't find a power series to the function :$f(x)=\operatorname{li}^2\Big(\frac{x}{x+1}\Big)$
My question:
can someone achieve my work ?
Best Answer
I also don't know how to find the series required to apply Ramanujan's master theorem, but here's the simpler method from my old answer:
We have $$ \mathrm{e}^{t} - 1 = \sum \limits_{k=1}^\infty \frac{t^k}{k!} \, , \, t \in \mathbb{R} \, , ~ \stackrel{t = \ln(x)}{\Leftrightarrow} ~ x - 1 = \sum \limits_{k=1}^\infty \frac{\ln^k(x)}{k!} \, , \, x > 0 \, .$$ Therefore, we can integrate by parts twice and use $\operatorname{li}' = \frac{1}{\ln}$ to obtain \begin{align} \int \limits_0^1 \frac{\operatorname{li}^2(x)}{x^2} \, \mathrm{d} x &= \left[\operatorname{li}^2(x) \left(1 - \frac{1}{x}\right) \right]_{x=0}^{x=1} - \int \limits_0^1 2 \operatorname{li}(x) \operatorname{li}'(x) \left(1 - \frac{1}{x}\right) \, \mathrm{d} x \\ &= 0 + 2 \int \limits_0^1 \frac{-\operatorname{li}(x)}{x} \frac{x-1}{\ln(x)} \mathrm{d} x = 2 \sum \limits_{k=1}^\infty \frac{1}{k!} \int \limits_0^1 -\operatorname{li}(x) \frac{\ln^{k-1}(x)}{x} \, \mathrm{d} x \\ &= 2 \sum \limits_{k=1}^\infty \frac{1}{k!} \left\{\left[-\operatorname{li}(x) \frac{\ln^k(x)}{k}\right]_{x=0}^{x=1} - \int \limits_0^1 -\operatorname{li}'(x) \frac{\ln^k(x)}{k} \mathrm{d} x \right\} \\ &= 2 \sum \limits_{k=1}^\infty \frac{1}{k!} \left\{0 + \frac{1}{k} \int \limits_0^1 \ln^{k-1}(x)\mathrm{d} x \right\} \stackrel{x = \mathrm{e}^{-u}}{=} 2 \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{k! k} \int \limits_0^\infty u^{k-1} \mathrm{e}^{-u} \, \mathrm{d} u \\ &= 2 \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{k! k} (k-1)! = 2 \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{k^2} = 2 \operatorname{\eta}(2) = 2 \frac{\pi^2}{12} = \frac{\pi^2}{6} \, , \end{align} where $\eta$ is the Dirichlet eta function. Interchanging summation and integration can be justified using the dominated convergence theorem.