TITLE: A tricky double integral
QUESTION [5 upvotes]: What is
$$\int_0^1 \int_0^1 \frac{ dx \; dy}{1+xy+x^2y^2} ? $$
Can you do one of the integrals and turn it into a single integral?
I get lost in a sea of inverse tangents.

REPLY [6 votes]: You can rewrite the integral as
$$\int_0^1 \int_0^1 {1 - xy \over 1 - x^3y^3}\,dx\,dy$$
Expanding ${\displaystyle{1 \over 1 - x^3y^3}}$ as a geometric series $\sum_n x^{3n}y^{3n}$
this becomes
$$\int_0^1 \int_0^1 \sum_{n=0}^{\infty} x^{3n}y^{3n} - \sum_{n=0}^{\infty} x^{3n+1}y^{3n+1}\,dx\,dy$$
Integrating this termwise this becomes
$$\sum_{n = 0}^{\infty} {1 \over (3n + 1)^2 } - \sum_{n = 0}^{\infty} {1 \over (3n + 2)^2 } $$
I don't think these sums have a closed form in terms of well-known functions; in effect these are the polygamma functions that Julian Aguirre put in his answer.