inkliing Posted October 16, 2012 Report Posted October 16, 2012 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 [math]\zeta(2)=\sum_{i=1}^\infty\frac 1{i^2}=\int_0^1 dx\int_0^1 dy\frac1{1-xy}=I_\circ[/math]by rotating the integration region by [math]-\frac {\pi}4[/math]. 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 [math]\sum_{k=1}^\infty\frac 1{k^2}=\frac {\pi^2}6[/math] 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[math]\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}[/math]gives [math]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)}[/math]which still blows up at (u,v)=([math]\sqrt2[/math],0). 2. It seems to me that [math]I_\circ=\int_0^1 dx\int_0^1 dy\frac1{1-xy}[/math][math]I_\circ=\lim_{b\to 1^{-}}\int_0^b dx\lim_{a\to 1^{-}}\int_0^a dy\frac1{1-xy}[/math][math]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)[/math]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. Quote
Aethelwulf Posted October 16, 2012 Report Posted October 16, 2012 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 [math]\zeta(2)=\sum_{i=1}^\infty\frac 1{i^2}=\int_0^1 dx\int_0^1 dy\frac1{1-xy}=I_\circ[/math]by rotating the integration region by [math]-\frac {\pi}4[/math]. 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 [math]\sum_{k=1}^\infty\frac 1{k^2}=\frac {\pi^2}6[/math] 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[math]\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}[/math]gives [math]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)}[/math]which still blows up at (u,v)=([math]\sqrt2[/math],0). 2. It seems to me that [math]I_\circ=\int_0^1 dx\int_0^1 dy\frac1{1-xy}[/math][math]I_\circ=\lim_{b\to 1^{-}}\int_0^b dx\lim_{a\to 1^{-}}\int_0^a dy\frac1{1-xy}[/math][math]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)[/math]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. I'll see if I can answer this for you... may take a bit of time... Quote
inkliing Posted October 17, 2012 Author Report Posted October 17, 2012 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 [math']\displaystyle\int_0^1\frac{1}{1-xy}\,\mathrm{d}y=-\frac{\log(1-x)}{x}[/math] 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. :) Quote
Aethelwulf Posted October 17, 2012 Report Posted October 17, 2012 outermeasure at http://www.sosmath.com/CBB/viewtopic.php?f=3&t=58677 explained it simply: 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. :) It's been that long since I last looked at the zeta function... I was certainly taking my time with it but I am very pleased you seem happy enough to have found the reason you were looking for. Quote
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