The Basel problem via Fourier series

The Basel problem asks to compute the summation of the reciprocal of the natural numbers squared, and it is known that

    \[\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}.\]

While there are numerous known proofs (our presentation resembles Proof 9 most closely), we give a self-contained one that motivates the usage of Fourier analysis for periodic functions.

Recall that a function f:\mathbb{R}\rightarrow\mathbb{C} is 2\pi-periodic if for every x \in \mathbb{R}, f(x+2\pi)=f(x). From Fourier analysis, every such function can be expressed as a series of cosines and sines. More precisely,

If f is continuous and 2\pi-periodic, then

    \[\left\Vert f - \sum_{n=-\infty}^{\infty} \hat{f}(n) e_n\right\Vert_2 = 0,\]

where e_n(x) = e^{inx}, and \hat{f}(n) is the n-th Fourier coefficient equal to \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)e^{-inx} dx.

We say that \sum_{n=-\infty}^{\infty} \hat{f}(n) e_n is the Fourier series of f. When we define the inner product of two functions to be \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x) \overline{g(x)} dx, then it is easily seen that the “characters” \{e_n\} form an orthonormal system, and the n-th Fourier coefficient can be viewed as the projection of f along the n-th character. The Fourier theorem asserts that f can be approximated in L_2-norm by its Fourier series. In fact, by applying the Weierstrass M-test, we can make the following statement with a stronger convergence guarantee:

If f is continuous and 2\pi-periodic and \sum_{n=-\infty}^{\infty} \left|\hat{f}(n)\right| is convergent, then its Fourier series \sum_{n=-\infty}^{\infty} \hat{f}(n) e_n converges uniformly to f.

Furthermore, the Fourier transform \hat{f} itself encodes many of the properties of the original function f. In particular, we note that the symmetry of f is preserved in its Fourier transform.

If f is even, then for every n \in \mathbb{Z}, \hat{f}(-n) = \hat{f}(n).
If f is odd, then for every n \in \mathbb{Z}, \hat{f}(-n) = -\hat{f}(n).

Now we have the machinery to compute \sum_{n=1}^{\infty} 1/n^2.

(Of Basel) Define f:[-\pi, \pi]\rightarrow \mathbb{R} to be f(x)=|x|. We compute the Fourier coefficient of f. First note that

    \[\hat{f}(0) = \frac{1}{2\pi}\int_{-\pi}^{\pi} |x| dx = \frac{\pi}{2}.\]

For n \neq 0 \in \mathbb{Z}, we have

    \[2\pi\hat{f}(n) = \int_{0}^{-\pi}xe^{-inx} dx + \int_{0}^{\pi}xe^{-inx} dx. \]

Using integration by parts, we have for every a, b \in \mathbb{R},

    \begin{eqnarray*} \int_a^b xe^{-inx}dx & = & \frac{x e^{-inx}}{-in}\Biggr|_a^b - \int_a^b \frac{e^{-inx}}{-in}dx \\ & = & \frac{x e^{-inx}}{-in}\Biggr|_a^b + \frac{e^{-inx}}{n^2}\Biggr|_a^b. \end{eqnarray*}

Then

    \begin{eqnarray*} 2\pi\hat{f}(n) & = & \frac{-\pi(-1)^n}{-in} + \frac{(-1)^n -1 }{n^2} + \frac{\pi(-1)^n}{-in} + \frac{(-1)^n -1 }{n^2} \\ & = & \frac{2((-1)^n - 1)}{n^2}.  \end{eqnarray*}

Thus,

    \[ \hat{f}(n) =  \begin{cases} \frac{-2}{\pi n^2} & \text{if $n$ is odd,} \\ 0 & \text{if $n$ is even and nonzero,} \\ \frac{\pi}{2} & \text{if $n=0$.}  \]

From Euler’s formula, the Fourier series of f can be written as

    \[  \hat{f}(0) + \sum_{n=1}^{\infty} \left[ (\hat{f}(n) + \hat{f}(-n)) \cos(nx) + i(\hat{f}(n) - \hat{f}(-n)) \sin(nx) \right]. \]

Since f is even, by Proposition, \hat{f}(-n)=\hat{f}(n). From our calculation, the Fourier series of f is explicitly

    \[  \frac{\pi}{2} + \sum_{n\geq1, n \text{ odd}} \frac{-4}{\pi n^2} \cos(nx). \]

Note that \sum_{n=1}^{\infty} 1/n^2 is convergent from the Cauchy condensation test. Thus, by the Convergence Theorem, we conclude that

    \[ \frac{\pi}{2} + \sum_{n\geq1, n \text{ odd}} \frac{-4}{\pi n^2} \cos(nx) \longrightarrow |x|, \]

where the convergence is uniform (and in particular, pointwise). Setting x=0, we see that

    \[ \sum_{n\geq1, n \text{ odd}} \frac{1}{n^2} = \frac{\pi^2}{8}. \]

Lastly, observe that

    \begin{eqnarray*} \sum_{n=1}^{\infty} \frac{1}{n^2} & = & \sum_{n\geq1, n \text{ even}} \frac{1}{n^2} + \sum_{n\geq1, n \text{ odd}} \frac{1}{n^2} \\ & = & \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^2} + \frac{\pi^2}{8}, \end{eqnarray*}

finishing the proof.

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