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Is it possible to use change of variable in order to solve the following problem?

Calculate $\int_0^1\int_0^1 \frac{1}{1-x^2y^2}dxdy$ using the following transformation $x=\frac{\sin u}{\cos v},$ and $y= \frac{\sin v}{\cos u}$.

My attempt is to pick the triangle region $R$ in $u$$v$ plane, where $R$ is the trianle with three vertexes $(0,0), (\frac{\pi}{2},0), (0,\frac{\pi}{2}).$ Then $|J|=1-x^2y^2$ and $\int_0^1\int_0^1 \frac{1}{1-x^2y^2}dxdy=\int\int_R 1 dudv=\frac{\pi^2}{8}.$ Is it right? Note that integtand $\frac{1}{1-x^2y^2}$ is singular at $(1,1).$

I would be thankful for any comments.

04170706
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  • You can check out Jonathan Sondow's articles about this type of integral by typing his name in google. He's done some research about it quite seriously. – DeepSea Jun 09 '17 at 05:37
  • Thank you for the answer. I will check Jonathan Sondow's articles. – 04170706 Jun 09 '17 at 06:37
  • This is essentially Apostol's proof of $\zeta(2)=\frac{\pi^2}{6}$: https://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-limits-k-1-infty-frac1k2 – Jack D'Aurizio Jun 09 '17 at 09:59
  • You may easily notice that $$\iint_{(0,1)^2}\frac{dx,dy}{1-x^2y^2} =\iint_{(0,1)^2}(1+x^2y^2+x^4y^4+\ldots),dx,dy\stackrel{\text{Fubini}}{=}1+\frac{1}{3^2}+\frac{1}{5^2}+\ldots = \frac{3}{4}\zeta(2).$$ – Jack D'Aurizio Jun 09 '17 at 10:01

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