calculus - Prove that $int_R frac{1}{1-x^2y^2} ,dx ,dy = sum_{n=0}^{infty} frac{1}{(2n+1)^2}$

The region $R$ is the unit square with corners at $(0,0), (1,0), (0,1)$ and $(1,1)$.

The idea is to consider the geometric series.

Any help would be appreciated. Thank you

Answer

I collect all hints and write down the answer.

1. We use the formula for sum of a geometric series as following
   $$\frac{1}{1-x^2y^2} = 1 + x^2y^2 +x^4y^4 + \dots.$$


2. Then we have
   $$\int\limits_{R}\frac{1}{1-x^2y^2}\, dxdy = \int\limits_{R}\sum\limits_{n=0}^\infty x^{2n}y^{2n}\, dxdy = \sum\limits_{n=0}^\infty \int\limits_{R}x^{2n}y^{2n}\, dxdy.$$


3. And $\int\limits_{R}x^{2n}y^{2n}\, dxdy = \int_0^1\int_0^1x^{2n}y^{2n}\, dxdy = 1/(2n+1)^2$.


4. Finally
   $$\int\limits_{R}\frac{1}{1-x^2y^2}\, dxdy = \sum\limits_{n=0}^\infty\frac{1}{(2n+1)^2}.$$