Basel Problem using Fourier Series

A very famous mathematical problem known as the “Basel Problem” is solved by Euler in 1734. Basically, it asks for the exact value of \sum_{n=1}^\infty\frac{1}{n^2}.

Three hundred years ago, this was considered a very hard problem and even famous mathematicians of the time like Leibniz, De Moivre, and the Bernoullis could not solve it.

Euler showed (using another method different from ours) that \displaystyle \sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}, bringing him great fame among the mathematical community. It is a beautiful equation; it is surprising that the constant \pi, usually related to circles, appears here.

Squaring the Fourier sine series

Assume that \displaystyle f(x)=\sum_{n=1}^\infty b_n\sin nx.

Then squaring this series formally,
\begin{aligned}  (f(x))^2&=(\sum_{n=1}^\infty b_n\sin nx)^2\\  &=\sum_{n=1}^\infty b_n^2\sin^2 nx+\sum_{n\neq m}b_nb_m\sin nx\sin mx.  \end{aligned}

To see why the above hold, see the following concrete example:
\begin{aligned}  (a_1+a_2+a_3)^2&=(a_1^2+a_2^2+a_3^2)+(a_1a_2+a_1a_3+a_2a_1+a_2a_3+a_3a_1+a_3a_2)\\  &=\sum_{n=1}^3 a_n^2+\sum_{n\neq m}a_na_m.  \end{aligned}

Integrate term by term

We assume that term by term integration is valid.
\displaystyle \frac 1\pi\int_{-\pi}^\pi (f(x))^2\,dx=\frac 1\pi\int_{-\pi}^{\pi}\sum_{n=1}^\infty b_n^2\sin^2{nx}\,dx+\frac{1}{\pi}\int_{-\pi}^\pi\sum_{n\neq m}b_nb_m\sin nx\sin mx\,dx.

Recall that \displaystyle \int_{-\pi}^\pi \sin nx\sin mx\,dx=\begin{cases}0 &\text{if }n\neq m\\  \pi &\text{if }n=m  \end{cases}.

So
\begin{aligned}  \frac 1\pi\int_{-\pi}^{\pi}\sum_{n=1}^\infty b_n^2\sin^2{nx}\,dx&=\frac 1\pi\sum_{n=1}^\infty b_n^2(\int_{-\pi}^\pi\sin^2 nx\,dx)\\  &=\frac 1\pi\sum_{n=1}^\infty b_n^2 (\pi)\\  &=\sum_{n=1}^\infty (b_n)^2.  \end{aligned}

Similarly
\begin{aligned}  \frac{1}{\pi}\int_{-\pi}^\pi\sum_{n\neq m}b_nb_m\sin nx\sin mx\,dx&=\frac 1\pi\sum_{n\neq m}b_nb_m(\int_{-\pi}^{\pi}\sin nx\sin mx\,dx)\\  &=\frac 1\pi\sum_{n\neq m}b_nb_m(0)\\  &=0.  \end{aligned}

So \displaystyle \frac 1\pi\int_{-\pi}^\pi (f(x))^2\,dx=\sum_{n=1}^\infty (b_n)^2. (Parseval’s Identity)

Apply Parseval’s Identity to f(x)=x

By Parseval’s identity,
\displaystyle \frac{1}{\pi}\int_{-\pi}^\pi x^2\,dx=\sum_{n=1}^\infty(\frac{2(-1)^{n+1}}{n})^2.

Simplifying, we get \displaystyle \frac 1\pi\cdot\left[\frac{x^3}{3}\right]_{-\pi}^\pi=\sum_{n=1}^\infty\frac{4}{n^2}.
\begin{aligned}  \frac 1\pi(\frac{2\pi^3}{3})&=\sum_{n=1}^\infty \frac{4}{n^2}\\  \frac{\pi^2}{6}&=\sum_{n=1}^\infty\frac{1}{n^2}.  \end{aligned}