A reproduction of a proof by Leonard Euler, which I dug out of the back of a calculus textbook with the help of Brian Burrell. This was in MATH 233 at the University of Massachusetts, Amherst, Fall 2015.
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\title{
The Sum of the Reciprocals of the Squares }
\author{Samuel Castillo}
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\lhead{Samuel Castillo}
\rhead{March 2, 2016}
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\centering{\textbf{\large THE SUM OF THE RECIPROCALS OF THE SQUARES \\
\large A Proof by Leonhard Euler}}
\begin{flushleft}
As if one needed further evidence for the genius of Leonhard Euler, here is one of his solutions to the summation of a famous series. The sum of the reciprocals of the squares of the natural numbers was a question first posed in 1644 by Pietro Mengoli, and left unsolved until Leonhard Euler 1734 [1].
The original method that Euler used was not what follows, but an expansion of the series of the sine and cosine functions. What makes this particular method appealing is a reliance on multivariate calculus techniques [2]. It was well-known at the time that the series $\sum_{n=1}^{\infty}\frac{1}{n^p}$ diverges for $p<1$ to some finite value; finding that specific value, however, is a far greater challenge.
The object of this paper is to find, and prove, the exact value that this series converges to.
\end{flushleft}
$$\lim_{n\to\infty}(1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \ldots + \frac{1}{n^2}) = \sum_{n=1}^{\infty} \frac{1}{n^2}$$
Euler accomplished this by showing first that this series is equal to the following integrated region, and then finding the exact value of the definite integral.
$$\int\limits_{0}^{1} \int\limits_{0}^{1} \frac{1}{1-xy} dy dx$$
This double integral over the region $D = \left\{ {x,y | 0\leq x \leq1, 0\leq y \leq 1 }\right\} $, which is represented graphically below, at first appears to have nothing in common with this series; however, the integrand can be rewritten. In particular, $\frac{1}{1-xy}$ is of the form $\frac{1}{1-p}$ for $|p|<1$ can be expanded as an infinite series.
$$S_n = 1 + p + p^2 + p^3 + \ldots + p^n$$
$$p*S_n = p + p^2 + p^3 + \ldots + p^{n+1}$$
$$s_n - p*s_n = 1 - p^{n+1} = s_n(1-p) => s_n = \frac{1-p^{n+1}}{1-p}$$
$$\lim_{n\to\infty} s_n = \lim_{n\to\infty} \frac{1-p^{n+1}}{1-p} = \frac{1}{1-p}$$
$$ p = xy \Rightarrow \frac{1}{1-xy} = xy + x^2y^2 + x^3y^3 + \ldots $$
$$\int\limits_{0}^{1} \int\limits_{0}^{1} \frac{1}{1-xy} dy dx = \int\limits_{0}^{1} \int\limits_{0}^{1} ( xy + x^2y^2 + x^3y^3 + \ldots ) dy dx = \sum_{n=1}^{\infty} \frac{1}{n^2}$$
So we know that this integral is equal to the sum of the reciprocals of the squares. Euler performed a transformation of variables here to find the exact value of the double integral.
Let $ x = \frac{u+v}{\sqrt{2}}$, and let $y = \frac{u-v}{\sqrt{2}} $.
Because the determinant of the Jacobian matrix is 1, or equivalently because rotation is a linear transformation, $dy dx = du dv$.
Intuitively this makes sense as the area of the transformed region is $1 * 1 = 1$, as the region D is rotated counterclockwise by 90 degrees, now changing the limits of integration accordingly to the bounds on the U-V axis.
\newline
Here is an image of the unit square representing the region $D$.
\newline
\includegraphics[width=2 in,scale=0.06]{default_region.PNG}
Now here is the rotated region of integration.
\includegraphics[width=2 in,scale=0.06]{rotated_region.PNG}
Here we substitute in $(u,v)$ for $(x,y)$ and evaluate.
$$\int\limits_{0}^{1} \int\limits_{0}^{1} \frac{1}{1-xy} dy dx = \int\limits_{0}^{\frac{\sqrt{2}}{2}} \int\limits_{-u}^{u} \frac{2}{(2-u^2)+v^2} dv du +
\int\limits_{\frac{\sqrt{2}}{2}}^{\sqrt{2}} \int\limits_{u - \sqrt{2}}^{-u+\sqrt{2}} \frac{2}{(2-u^2)+v^2} dv du =\textbf{\textit{ A + B}}$$
\section{Evaluating the inside integral of \textbf{\textit{A}}:}
Let $\alpha = \sqrt{2 - u^2}$,
$$\Rightarrow \int\limits_{-u}^{u} \frac{2}{(2-u^2)+v^2} dv = 2 \int\limits_{-u}^{u} \frac{1}{\alpha^2 + v^2} dv = \frac{2}{\alpha}\arctan{\frac{v}{\alpha}}\Big|_{-u}^{u} = \frac{4}{\alpha} \arctan{\frac{u}{\alpha}} $$
$$= \frac{4}{\sqrt{2 - u^2}} \arctan{\frac{u}{\sqrt{2 - u^2}}} = \int\limits_{0}^{\frac{\sqrt{2}}{2}} \frac{4}{\sqrt{2 - u^2}} \arctan{\frac{u}{\sqrt{2 - u^2}}} du$$
This can be evaluated with the substitution $u = \sqrt{2}\sin{\theta}, du = \sqrt{2}\cos{\theta} d\theta$
$$ \Rightarrow \int\limits_{a}^{b} \frac{4}{\sqrt{2 - {\sqrt{2}\sin{\theta}}^2}} \arctan{\frac{\sqrt{2}\sin{\theta}}{\sqrt{2 - {\sqrt{2}\sin{\theta}}^2}}} * \sqrt{2}\cos{\theta} d\theta = 2*{\theta}^2\Big|_{a}^{b}$$ where $u = \sqrt{2}\sin{\theta} \Rightarrow \theta = \arcsin{\frac{u}{\sqrt{2}}}$
This means that the total of \textbf{\textit{A}} is just $$2*{\arcsin{(\frac{u}{\sqrt{2}}})}^2\Big|_{0}^{\frac{\sqrt{2}}{2}} = \frac{{\pi}^2}{18}$$
\section{Evaluating the Integral \textit{\textbf{B}}:}
$$\int\limits_{\frac{\sqrt{2}}{2}}^{\sqrt{2}} \int\limits_{u - \sqrt{2}}^{-u+\sqrt{2}} \frac{2}{(2-u^2)+v^2} dv du$$
From our earlier result, it was shown that the inside integral stays the same, and only the limits of integration change.
$$ \frac{4}{\sqrt{2 - u^2}} \arctan{\frac{v}{\sqrt{2 - u^2}}}\Big|_{u - \sqrt{2}}^{-u + \sqrt{2}} =
\int\limits_{\frac{\sqrt{2}}{2}}^{\sqrt{2}} \frac{1}{\sqrt{2-u^2}} [\arctan{(\frac{u - \sqrt{2}}{\sqrt{2 - u^2}})} - \arctan{(\frac{-u + \sqrt{2}}{\sqrt{2 - u^2}}})]du$$
Just as last time let $ u = \sqrt{2}\sin{\theta}, du = \sqrt{2}\cos{\theta} $ and the integral simplifies to
$$2\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{2}}\arctan{(\frac{-\sqrt{2}\sin{\theta} + \sqrt{2}}{\sqrt{2\cos^2{\theta}}}) - \arctan{(\frac{\sqrt{2}\sin{\theta} - \sqrt{2}}{\sqrt{2\cos^2{\theta}}}}})d\theta$$
$$ = 2\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{2}}\arctan{(\frac{-\sin{\theta} + 1}{\cos{\theta}})d\theta -2\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{2}} \arctan{(\frac{\sin{\theta} - 1}{cos{\theta}}}})d\theta $$
$$= 4\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{2}}\arctan{(\frac{-\sin{\theta} + 1}{\cos{\theta}}})d\theta = 4\theta \arctan{(\frac{-\sin{\theta} + 1}{\cos{\theta}}})d\theta - 4 \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{2}} \theta \frac{1}{1+\Delta^2}\frac{d\Delta}{d\theta}d\theta$$
\begin{flushleft}For simplicity, \end{flushleft}
\begin{center}
\begin{itemize}
\item[] $\Delta = \frac{1-sin{\theta}}{\cos{\theta}}$
\item[] $\frac{d\Delta}{d\theta} = \frac{\sin{\theta} -1}{\cos^2{\theta}}$
\item[] $\Delta^2 = \frac{1 - 2\sin{\theta} + \sin^2{\theta}}{\cos^2{\theta}}$
\item[] $1 + \Delta^2 = \frac{1 - 2\sin{\theta} + \sin^2{\theta} + \cos^2{\theta}}{\cos^2{\theta}} = \frac{2 - 2\sin{\theta}}{\cos^2{\theta}}$
\item[] $\frac{1}{1+\Delta^2} = \frac{1}{2}\frac{\cos^2{\theta}}{1 - 1\sin{\theta}}$
\end{itemize}
\end{center}
By substituting these back in, it can be seen that the integral miraculously simplifies.
$$4\theta \arctan{(\frac{-\sin{\theta} + 1}{\cos{\theta}}})d\theta - 4 \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{2}} \theta \frac{1}{1+\Delta^2}\frac{d\Delta}{d\theta}d\theta$$
$$\theta \arctan{(\frac{-\sin{\theta} + 1}{\cos{\theta}}})d\theta - 4 \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{2}} \theta \frac{1}{2}\frac{\cos^2{\theta}}{(1 - \sin{\theta})} \frac{(-1)(1 - \sin{\theta})}{\cos^2{\theta}} d\theta$$
$$ = \theta \arctan{(\frac{-\sin{\theta} + 1}{\cos{\theta}}})d\theta - 2\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{2}} \theta d\theta = 4 \frac{\pi^2}{36} = \frac{\pi^2}{12} = \textbf{\textit{B}}$$
Adding \textbf{\textit{B}} to \textbf{\textit{A}} and equating it to the original geometric series, we find that the summation of the infinite series of the reciprocals of the squares of all positive integers,
$$4 \frac{\pi^2}{36} + \frac{\pi^2}{18} = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \ldots = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \space \space \blacksquare$$
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\begin{thebibliography}{9}
\bibitem{lamport94}
Ayoub, Raymond,
\emph{Euler and the zeta function},
Amer. Math. Monthly 81: 1067–86
1974.
\bibitem{lamport94}
Stewart, James,
\emph{Calculus: Early Transcendentals},
Cengage Learning, Boston,
7th edition,
2013.
\end{thebibliography}
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