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$$ \int_{[0,1] \times [0,1]} \frac{dx \, dy}{1-xy}=\frac{\pi^2}{6}, $$ there is a hint for substitution given but I can't seem to get anywhere with it. $$ \begin{cases} x = \frac{u-v}{\sqrt{2}} \\ y= \frac{u+v}{\sqrt{2}} \end{cases} $$ and evaluate by letting $u = \sqrt{2}\sin t$. There should also encounter the term $(1-\sin t)/\cos t$ (Which I couldn't obtain)

I have managed to compute the jacobian of transformation, which is exactly 1, and did the substitution for u and v, obtaining: $$ \int_{[0,\sqrt{2}]\times [-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]} \frac{2}{2-u^2 +v^2} du\, dv $$ doing the subtitution for $u$, we get: $$ \int \frac{2\sqrt{2}\cos t}{2\cos^2 t+v^2}dt\, dv $$ It is then not obvious how to integrate with respect to $t$ now, so change order of integration and integrate with respect to v.

which will be equal to $\int 4\arctan \frac{1}{2\cos t} \, dt$

Now this is a rather complicated integral, how to proceed from here? or is there an alternative method which I overlooked in the previous steps?

Could I have some insight on this problem please??

  • 0
    Welsome to MSE. For formatting math, see [here](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) for a quick tutorial.2012-11-05
  • 0
    Welsome to MSE. For formatting math, see [here](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) for a quick tutorial.2012-11-05
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    A direct method is to expand 1/(1-xy) into the series of general term x^ny^n and, since everything is positive, to integrate this series termwise.2012-11-05

3 Answers 3