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inkliing
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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
\zeta(2)=\sum_{i=1}^\infty\frac 1{i^2}=\int_0^1 dx\int_0^1 dy\frac1{1-xy}=I_\circ
by rotating the integration region by -\frac {\pi}4. 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 \sum_{k=1}^\infty\frac 1{k^2}=\frac {\pi^2}6 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
\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}
gives
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)}
which still blows up at (u,v)=(\sqrt2,0).
2. It seems to me that
\begin{align*}I_\circ&=\int_0^1 dx\int_0^1 dy\frac1{1-xy}\\&=\lim_{b\to 1^{-}}\int_0^b dx\lim_{a\to 1^{-}}\int_0^a dy\frac1{1-xy}\\<br /> &=\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)\end{align*}
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.
\zeta(2)=\sum_{i=1}^\infty\frac 1{i^2}=\int_0^1 dx\int_0^1 dy\frac1{1-xy}=I_\circ
by rotating the integration region by -\frac {\pi}4. 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 \sum_{k=1}^\infty\frac 1{k^2}=\frac {\pi^2}6 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
\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}
gives
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)}
which still blows up at (u,v)=(\sqrt2,0).
2. It seems to me that
\begin{align*}I_\circ&=\int_0^1 dx\int_0^1 dy\frac1{1-xy}\\&=\lim_{b\to 1^{-}}\int_0^b dx\lim_{a\to 1^{-}}\int_0^a dy\frac1{1-xy}\\<br /> &=\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)\end{align*}
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.
