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This paper http://math.ucsb.edu/~cmart07/Evaluating%20Integrals.pdf hints at a way to compute the sum $$ \sum_{n=1}^\infty \frac{1}{n^2} $$ by expanding it into the double integral $$\int_0^1 \int_0^1 \frac{\mathrm{d}x \, \mathrm{d}y}{1-xy}.$$ Now for solving this integral, the paper suggests rotating the area $[0,1]^2$ by $45^\circ$ for then to rewrite it in polar coordinates.

I have made a sketch of the area below, but I am having problems rewriting the integral in polar coordinates.

The area to be evaluated

Dissregarding the function I was thinking the limits would have to be

$$\int_{-\pi/4}^{\pi/4} \int_0^{\sqrt{\cos^4\theta + (\sqrt{2}-\cos^2\theta)^2}}r \mathrm{d}r\mathrm{d}\theta$$ but the upper limit is wrong for the radius. Hmm... I was basically finding the distance from the function $f(x) = \sqrt{2}-x$ to origo, then converting this to polar..

Any help computing the sum using the double integral transform would be very appreceated. I already know several methods for computing the bessel identity, however this one stumped me.

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Nice question (+1) –  Chris's sis Dec 31 '12 at 13:14
    
The proof appears in Aigner & Zeigler. I'll copy it out. –  Alex J Best Dec 31 '12 at 13:15
    
The upper limits of the original integral are $\infty$ or $1$? –  leonbloy Dec 31 '12 at 13:16
    
Fixed it! Sorry! –  N3buchadnezzar Dec 31 '12 at 13:18
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2 Answers

up vote 3 down vote accepted

I think if you use Change of Variables in Double Integrals by using Jacobian such that $$u=y+x,v=y-x\;\; u\Big|_{0}^{\sqrt{2}},v\Big|_{0}^{\sqrt{2}}$$ your integral would be easy. See http://www.math24.net/change-of-variables-in-double-integrals.html for more.

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\Big| makes the evaluation bar look "prettier" ;-) –  amWhy Feb 26 '13 at 14:36
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The start: The idea is that once the area is transformed, it is symmetric about the axis. We substitute $u=\frac{y+x}{2}$ and $v=\frac{y-x}{2}$. This transforms our function as follows $\frac{1}{1-xy} = \frac{1}{1-u^2+x^2}$ we also need to add a factor of 2.

From "Proofs from The Book", Aigner & Ziegler.

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