While reviewing calculus, I noticed that James Stewart's Calculus 3rd ed., chap 14, problems plus (at the end of the chapter), p.965, prob#2 asks the student to show that

[latex]\zeta(2)=\sum_{i=1}^\infty\frac 1{i^2}=\int_0^1 dx\int_0^1 dy\frac1{1-xy}=I_\circ[/latex]

by rotating the integration region by [latex]-\frac {\pi}4[/latex]. Mathworld also evaluates the same integral with the same rotation, http://mathworld.wolfram.com/RiemannZetaFunctionZeta2.html . Mathworld refers to Simmons (1992), which is Simmons, G. F. "Euler's Formula [latex]\sum_{k=1}^\infty\frac 1{k^2}=\frac {\pi^2}6[/latex] by Double Integration." Ch. B. 24 in Calculus Gems: Brief Lives and Memorable Mathematics. New York: McGraw-Hill, 1992, which is in Google Books, but as usual, the needed page is missing.

 

1. I assume both Stewart and mathworld rotate the integration region in an attempt to remove the pole at (x,y)=(1,1). Rotating by

[latex]\begin{pmatrix}u\\v\end{pmatrix}=\begin{pmatrix} \cos\frac {\pi}4&\sin\frac {\pi}4\\-\sin\frac {\pi}4&\cos\frac {\pi}4\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix}[/latex]

gives

[latex]I_\circ=\int_0^{\frac 1{\sqrt2}} du\int_{-u}^u dv\frac 1{1-\frac 12(u^2-v^2)}+\int_{\frac 1{\sqrt2}}^{\sqrt2} du\int_{-\sqrt2+u}^{\sqrt2-u} dv\frac 1{1-\frac 12(u^2-v^2)}[/latex]

which still blows up at (u,v)=([latex]\sqrt2[/latex],0).

 

2. It seems to me that

[latex]I_\circ=\lim_{b\to 1^{-}}\int_0^b dx\lim_{a\to 1^{-}}\int_0^a dy\frac1{1-xy}[/latex]

[latex]I_\circ=\lim_{b\to 1^{-}}\int_0^b dx\lim_{a\to 1^{-}}\int_0^a dy\sum_{i=0}^\infty x^iy^i=\zeta(2)[/latex]

So why rotate if the rotated integrand still blows up?

Please be specific. Please stay as close as possible to the question. Thanks in advance.

outermeasure at http://www.sosmath.com/CBB/viewtopic.php?f=3&t=58677 explained it simply:

The rotation is not to get rid of the pole' date=' but to make it into a form that you can integrate the inner integral to get something you can integrate further for the outer integral. Contrast this with naively integrating the y-integral giving [latex']\displaystyle\int_0^1\frac{1}{1-xy}\,\mathrm{d}y=-\frac{\log(1-x)}{x}[/latex] which does not have a nice (elementary) antiderivative.

 

The pole doesn't pose much problem because the function is actually integrable (i.e. with finite integral), similar to 1/sqrt(|x|) near x=0.

 

Seems obvious now. Rotate to be able to integrate both of the iterated integrals with antiderivatives of elementary functions rather than power series. I should have seen that.