<s>[INST] Is $\int_0^1\int_0^1 \frac{Li_3(xy)}{(xy)^2}dxdy$ equal to $3\zeta(2)+\zeta(3)-6$ or to infinity? I am comparing two commercial Computer Algebra Sytems on the integral $\int_0^1\int_0^1 \frac{Li_3(xy)}{(xy)^2}dxdy$.  One says it equals  $3\zeta(2)+\zeta(3)-6=0.1368...$, the other that it does not converge.

which one is correct?

Integrating on the square $[0+\epsilon;1-\epsilon]\times[0+\epsilon;1-\epsilon]$ numerically for smaller and smaller $\epsilon$ gives larger and larger values, which hints at non convergence, but there may well be a cancellation of singularities in the limit.
Any productive comment or answer would be appreciated!
 [/INST] Thanks to metamorphy and Angina Seng, I can answer completely the question:

*

*the integral diverges, since $\int_0^1\int_0^1 \frac{Li_3(xy)}{(xy)^2}dxdy = \int_0^1\int_0^1 \frac{Li_3(xy)-xy}{(xy)^2}dxdy +\int_0^1\int_0^1 \frac{xy}{(xy)^2}dxdy$, where:

a)  the second integral on the rhs diverges
b) the first one on the rhs converges, since $Li_3(xy)=\frac{xy}{1^3} + \frac{(xy)^2}{2^3}+...$ so by canceling the term $xy$ and then simplifying numerator and denominator by $(xy)^2$ the integrand becomes a continuous function on a bounded domain of $\mathbb{R}^2$.


*in fact, we can prove that $\int_0^1\int_0^1 \frac{Li_3(xy)-xy}{(xy)^2}dxdy=3\zeta(2)+\zeta(3)-6$. Indeed, performing explicitely the double integration of the series $\frac{Li_3(xy)-xy}{(xy)^2}$ by $x$ then $y$ we get successively $\int_0^1[\frac{1}{2^3}x+\frac{x^2}{2}\frac{y}{3^3}+\frac{x^3}{3}\frac{y^2}{4^3}+...]_0^1 dy =\int_0^1 (\frac{1}{2^3}+\frac{1}{2}\frac{y}{3^3}+\frac{1}{3}\frac{y^2}{4^3}+...)dy $
$=\frac{1}{2^3}+\frac{1}{2^2}\frac{1}{3^3}+\frac{1}{3^2}\frac{1}{4^3}+... = \frac{1}{2^3}+\sum_{n=2}^{\infty}\frac{1}{n^2(n+1)^3} (*)$.
Now, we have the partial fraction decomposition $\frac{1}{n^2(n+1)^3}=\frac{1}{n^2}+\frac{3}{n+1}+\frac{2}{(n+1)^2}+\frac{1}{(n+1)^3}-\frac{3}{n}$. So the sum from $n=2$ to infinity, taking into account the telescoping cancelations of the terms of $\frac{3}{n+1}$ and $-\frac{3}{n}$, becomes $[\zeta(2)-\frac{1}{1}]+2[\zeta(2)-\frac{1}{1}-\frac{1}{2^2}]+[\zeta(3)-\frac{1}{1}-\frac{1}{2^3}]+[-\frac{3}{2}]$. So when we add the $\frac{1}{2^3}$ from $(*)$ we get:
$\int_0^1\int_0^1 \frac{Li_3(xy)-xy}{(xy)^2}dxdy=3\zeta(2)+\zeta(3)-6$.
And so the first software was somehow computing this (it is not a typo in my code, I've checked by computing it too).
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