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

Question

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

Some post about the same series (but not the same method): http://math.stackexchange.com/questions/108834/use-fourier-series-for-computing-sum-n-1-infty-frac12n-12 and http://math.stackexchange.com/questions/1454960/how-can-we-show-frac-pi28-1-frac132-frac152-frac172 — Martin Sleziak, Nov 09 '15 at 15:16
A proof of this is given in Aigner, Ziegler: Proofs from the BOOK; p.37 in the third edition, p.55 in the fifth edition. — Martin Sleziak, Nov 09 '15 at 15:21

score 1 · Accepted Answer · answered Nov 10 '15 at 04:34

I collect all hints and write down the answer.

We use the formula for sum of a geometric series as following $$ \frac{1}{1-x^2y^2} = 1 + x^2y^2 +x^4y^4 + \dots. $$
Then we have $$ \int\limits_{R}\frac{1}{1-x^2y^2}\, dxdy = \int\limits_{R}\sum\limits_{n=0}^\infty x^{2n}y^{2n}\, dxdy = \sum\limits_{n=0}^\infty \int\limits_{R}x^{2n}y^{2n}\, dxdy. $$
And $\int\limits_{R}x^{2n}y^{2n}\, dxdy = \int_0^1\int_0^1x^{2n}y^{2n}\, dxdy = 1/(2n+1)^2$.
Finally $$ \int\limits_{R}\frac{1}{1-x^2y^2}\, dxdy = \sum\limits_{n=0}^\infty\frac{1}{(2n+1)^2}. $$

1 Answers1