Could it possibly have a nice closed form? $\int _0^1\int _0^1\frac{x y}{(x+1) (y+1) \log (x y)}\ dx \ dy$
Solution 1:
Just a note. We can decompose the integrand as the following. $$ \frac{x y}{(x+1) (y+1) \log (x y)} = \frac{x}{(x+1)\log (x y)} - \frac{x}{(x+1) (y+1) \log (x y)} $$ Then we notice that $$ \int _0^1\int _0^1 \frac{x}{(x+1)\log (x y)} \ dx \ dy = \int_0^1 \frac{\operatorname{li}(x)}{x+1} \ dx $$ where $\operatorname{li}$ is the logarithmic integral. This integral doesn't have closed-form as I know.
The rest of the work is to find a closed-form of the following integral: $$\int _0^1\int _0^1\frac{x}{(x+1) (y+1) \log (x y)}\ dx \ dy.$$