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

Question

Consider $$ y = \int_1^{\infty} \frac{\operatorname{li}(x)^2 (x - 1)}{x^4} dx, $$ where $\operatorname{li}(x)$ is the logarithmic integral. Is there a closed form for y ?

It appears that a good approximation is $ 10 \cdot \operatorname{Ci}\bigl( \frac{56}{19}\bigr)$, where $\operatorname{Ci}(x)$ is the Cosine integral.

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If there is No closed form would it help to allow the function $$ t(x) = \int_1^x \operatorname{li}(t)^2 dt \hspace{10mm} ??$$

Answer 1

Using numerical integration, I found a value of $$1.3707783890401886970603459722050209910157915843390$$ Looking for this number in $ISC$, I found that it is $$10\sum_{n=1}^\infty \frac{1}{4 n^2 \binom{2 n}{n}}$$ which is $\frac{5 }{36}\pi ^2$ !!

Now, $???$

Answer 2

For what it's worth, for $n=0$ and $n=1$ the left-hand side (LHS) and RHS are equal in the below equation, to high numeric precision. I thought I saw a pattern developing for a generalization, but have been unable to ascertain the correct one yet. For $n=0$ and using the dilogarithm at argument 1/2, the already proposed identity results. There is absolutely no rigor in my nascent identity, so I won't present the development.

$$ \int_1^\infty \frac{x-1}{x^{n+4}}\, li(x)^2 \,dx \, \, \, ( \text{for } n=0, 1) \\ =(-1)^{n+1}\Big(\, \frac{\zeta(2)}{2} + \sum_{m=0}^n \frac{(-1)^m}{m+3}\big(\log^{\,2}(m+2)+2\text{Li}_2(1/(m+2))-5\zeta(2)\big)\, \Big)$$

As far as I know, $\text{Li}_2(1/3)$ does not have a 'closed form,' so generalizations in this direction appear unlikely.