Fourier Series and Fourier Transform - II

In our last post, we have introduced the big picture of Fourier Series. In this post, we will continue to explore the Fourier Series.

• Definition of Fourier Series
• Two Examples
• Convergence of Fourier Series
• Dirichlet Kernel
• Orthogonality of the basis
• Some important inequalities
• Application of Rayleigh’s Identity

Definition of Fourier Series

For a periodic function $f(x)$ with period $1$ (you can normalize the period to $1$ by scaling the $x$-axis), we can write it as a Fourier Series:

$f(x) = \sum_{n=-\infty}^{\infty} c_n e^{2\pi i n x}$

where $c_n$ is the Fourier coefficient, which is the projection of $f(x)$ onto the basis $e^{2\pi i n x}$:

$c_n =\langle f(x), e^{2\pi i n x} \rangle = \int_0^1 f(x) e^{-2\pi i n x} dx$

I like using the inner product notation to represent the Fourier coefficient, because it is more intuitive to me. The inner product is the projection of $f(x)$ onto the basis $e^{2\pi i n x}$, which is the same as the Fourier coefficient.

Before we moving on, let’s discuss the properties of the Fourier coefficient $c_n$:

1. $c_n$ is a complex number, which can be written as $c_n = a_n + i b_n$, where $a_n$ and $b_n$ are real numbers.
2. when $f(x)$ is a real function, $c_n = \overline{c_{-n}}$, where $\overline{c_{-n}}$ is the complex conjugate of $c_{-n}$. This is because the basis $e^{2\pi i n x}$ is a complex number, and the inner product of two complex numbers is a complex number. When $f(x)$ is a real function, the inner product of $f(x)$ and $e^{2\pi i n x}$ is a real number, which is the same as the inner product of $f(x)$ and $e^{-2\pi i n x}$.

3. $c_0$ is the average of $f(x)$, which is a real number:$c_0 = \int_0^1 f(x) dx$

4. when $f(x)$ is even, $c_n$ is also even, which means $c_n = c_{-n}$. $\begin{aligned}c_n & = \int_0^1 f(x) e^{-2\pi i n x} dx \\ & = - \int_0^{-1} f(-s) e^{2\pi i n s} ds \quad \text{let } s = -x \\ & = \int_{-1}^0 f(-s) e^{-2\pi i n (-s)} ds \\ & = c_{-s} = c_{-n} \end{aligned}$

5. when $f(x)$ is odd, $c_n$ is odd, which means $c_n = -c_{-n}$.

6. when $f(x)$ is real and even, $c_n$ is real and even, which means $c_n = c_{-n}$.

7. when $f(x)$ is real and odd, $c_n$ is imaginary and odd, which means $c_n = -c_{-n}$.

$c_n = -c_{-n} \quad \text{when } f(x) \text{ is odd} \\ c_n = \overline{c_{-n}} \quad \text{when } f(x) \text{ is real} \\ \longrightarrow - c_{-n} = \overline{c_{-n}} \quad \text{only possible when it is pure imaginary}$

Those properties are very useful when we calculate the Fourier coefficient as they could help us to verify the correctness of our calculation. Since most of signals are real, we can use those properties in practice.

Two Examples

Conside a square wave of period $1$, which is defined as:

$f(x) = \begin{cases} 1 & 0 \leq x < \frac{1}{2} \\ -1 & \frac{1}{2} \leq x < 1 \end{cases}$

The Fourier coefficient of $f(x)$ is:

$\begin{aligned} c_n & = \int_0^1 f(x) e^{-2\pi i n x} dx \\ & = \int_0^{\frac{1}{2}} e^{-2\pi i n x} dx - \int_{\frac{1}{2}}^1 e^{-2\pi i n x} dx \\ & = \frac{1}{-2\pi i n} e^{-2\pi i n x} \Big|_0^{\frac{1}{2}} - \frac{1}{-2\pi i n} e^{-2\pi i n x} \Big|_{\frac{1}{2}}^1 \\ & = \frac{1}{-2\pi i n}[e^{-\pi i n} - 1] - \frac{1}{-2\pi i n}[e^{-2\pi i n} - e^{-\pi i n}] \\ & = \frac{1}{-2\pi i n}[e^{-\pi i n} - 1 - e^{-2\pi i n} + e^{-\pi i n}] \\ & = \frac{1}{-2\pi i n}[2e^{-\pi i n} -1 - (\cos(-2\pi n) + i \sin(2\pi n))] \\ & = \frac{1}{-2\pi i n}[2e^{-\pi i n} -1 - 1] \\ & = \frac{1}{-2\pi i n}[2e^{-\pi i n} -2] \\ & = \frac{1}{-\pi i n}[e^{-\pi i n} -1] \\ & = \frac{1}{\pi i n}[1- e^{-\pi i n}] \\ \end{aligned}$

Therefore, the fourier series of $f(x)$ is:

$\begin{aligned} f(x) & = \sum_{n=-\infty, n\neq 0}^{\infty} c_n e^{2\pi i n x} \\ & = \sum_{n=-\infty, n \neq 0}^{\infty} \frac{1}{\pi i n}(1- e^{-\pi i n}) e^{2\pi i n x} \end{aligned}$

Notice, $f(x)$ is an odd function, so $c_n$ is imaginary and odd, which means $c_n = -c_{-n}$. Notice that

$\begin{aligned} 1 - e^{-\pi i n} & = 1 - \cos(\pi n) - i \sin(\pi n) \\ & = 1 - (-1)^n - i \sin(\pi n) \\ & = \begin{cases} 0 & n \text{ is even} \\ 2 & n \text{ is odd} \end{cases} \end{aligned}$

So the series can be simplified as:

$\begin{aligned} f(x) & = \sum_{n=-\infty, n\neq 0}^{\infty} c_n e^{2\pi i n x} \\ & = \sum_{n \text{ is odd}} \frac{2}{\pi i n}e^{2\pi i n x} \\ \end{aligned}$

Reflections
We have shown that when the function is real and odd, the fourier coefficients are pure imaginary and odd.

Now, we combine the positive and negative terms together:

$e^{2\pi i n x} - e^{-2\pi i n x} = 2i \sin(2\pi n x)$

let $n = 2k+1$, we have:

$\begin{aligned} f(x) & = \sum_{n \text{ is odd}} \frac{2}{\pi i n}e^{2\pi i n x} \\ & = \sum_{-\infty}^{\infty} \frac{2}{\pi i (2k+1)}e^{2\pi i (2k+1) x} \\ & = \sum_{-\infty}^{\infty} \frac{2}{\pi i (2k+1)}(e^{2\pi i (2k+1) x} - e^{-2\pi i (2k+1) x}) \\ & = \sum_{k=0}^{\infty} \frac{4}{\pi (2k+1)} \sin[2\pi (2k+1) x] \\ & = \frac{4}{\pi} \sum_{k=0}^{\infty} \frac{1}{2k+1} \sin[2\pi (2k+1) x] \end{aligned}$

Here is the visualization of the Fourier Series of the square wave (when $N=100$, you can click the right bottom corner to see the animation):

From the above example,we can see that the fourier series is ‘converging’ to the square wave. The more terms we add, the more similar it is to the square wave. However, we also see discontinuity at the jump points. This is called Gibbs phenomenon. Since both sine and cosine are continuous, the fourier series of a function is also continuous. Therefore the fourier series of a discontinuous function will have discontinuity at the jump points.

Now, let’s see another example - traingle wave - which is defined as:

$f(t) = \frac{1}{2} - |t| = \begin{cases} \frac{1}{2} + t & -\frac{1}{2} \leq t < 0 \\ \frac{1}{2} - t & 0 \leq t < \frac{1}{2} \end{cases}$

The coefficient of $f(t)$ is at $n=0$ is the average of $f(t)$, which is $1/4$. For $n \neq 0$, we have:

$\begin{aligned} c_n & = \int_{-1/2}^{1/2} f(t) e^{-2\pi i n t} dt \\ & = \int_{-1/2}^{1/2} (\frac{1}{2} - |t|) e^{-2\pi i n t} dt \\ & = \frac{1}{2} \int_{-1/2}^{1/2} e^{-2\pi i n t} dt - \int_{-1/2}^{1/2} |t| e^{-2\pi i n t} dt \\ & = - \int_{-1/2}^{1/2} |t| e^{-2\pi i n t} dt; \quad \text{since } \int_{-1/2}^{1/2} e^{-2\pi i n t} dt = 0 \\ & = - \bigg( \int_{-1/2}^{0} - t e^{-2\pi i n t} dt + \int_{0}^{1/2} t e^{-2\pi i n t} dt \bigg) \\ & = \int_{-1/2}^{0} t e^{-2\pi i n t} dt - \int_{0}^{1/2} t e^{-2\pi i n t} dt \\ \end{aligned}$

Now, let $A(n)$ be the first integral and we have:

$\begin{aligned} A(n) & = \int_{-1/2}^{0} t e^{-2\pi i n t} dt \end{aligned}$

It is easy to show that

$\begin{aligned} A(-n) & = \int_{-1/2}^{0} t e^{2\pi i n t} dt \\ & = \int_{1/2}^0 -s e^{-2\pi i n s} - ds \quad \text{let } s = -t \\ & = \int_{1/2}^0 s e^{-2\pi i n s} ds \\ & = - \int_{0}^{1/2} s e^{-2\pi i n s} ds \\ & = - A(n) = - \int_{-1/2}^{0} t e^{-2\pi i n t} dt \\ \end{aligned}$

Therefore, the fourier coefficient can be written as:

$c_n = A(n) + A(-n)$

Now, let’s integrate $A(n)$ by parts:

$\begin{aligned} A(n) & = \int_{-1/2}^{0} t e^{-2\pi i n t} dt \\ & = \frac{1}{-2\pi i n} t e^{-2\pi i n t} \Big|_{-1/2}^0 - \int_{-1/2}^{0} \frac{1}{-2\pi i n} e^{-2\pi i n t} dt \\ & = \frac{1}{-2\pi i n} t e^{-2\pi i n t} \Big|_{-1/2}^0 - \frac{1}{(2\pi i n)^2} e^{-2\pi i n t} \Big|_{-1/2}^0 \\ & = \frac{1}{-2\pi i n} [0 + \frac{1}{2} e^{\pi i n}] - \frac{1}{(2\pi i n)^2} [1 - e^{\pi i n}] \\ & = - \frac{1}{4\pi i n}e^{\pi in} + \frac{1}{4\pi^2n^2} [1 - e^{\pi i n}] \\ & = \frac{\pi i n}{4 \pi^2 n^2} e^{\pi i n} + \frac{1}{4\pi^2n^2} [1 - e^{\pi i n}] \\ & = \frac{1}{4\pi^2n^2} [1 - e^{\pi i n} + \pi i n e^{\pi i n}] \\ & = \frac{1}{4\pi^2n^2} [ 1 + e^{\pi i n} (\pi i n - 1)] \end{aligned}$

Therefore, we could have

$A(-n) = \frac{1}{4\pi^2n^2} [ 1 + e^{-\pi i n} (-\pi i n - 1)]$

The fourier coefficient is:

$\begin{aligned} c_n & = A(n) + A(-n) \\ & = \frac{1}{4\pi^2n^2} [ 1 + e^{\pi i n} (\pi i n - 1)] + \frac{1}{4\pi^2n^2} [ 1 + e^{-\pi i n} (-\pi i n - 1)] \\ & = \frac{1}{4\pi^2 n^2} [ 2 + e^{\pi i n} (\pi i n - 1) + e^{-\pi i n} (-\pi i n - 1)] \\ & = \frac{1}{4\pi^2 n^2} [ 2 + (\cos(\pi n) + i \sin(\pi n)) (\pi i n - 1) - (\cos(\pi n) - i \sin(\pi n)) (\pi i n + 1)] \\ & = \frac{1}{4\pi^2 n^2} [ 2 + \cos(\pi n)(\pi in - 1) - \cos(\pi n)(\pi in +1)] \\ & = \frac{1}{2\pi^2 n^2} (1 - \cos(\pi n)) \\ & = \begin{cases} 0 & n \text{ is even} \\ \frac{1}{\pi^2 n^2} & n \text{ is odd} \end{cases} \end{aligned}$

Now, let’s write down the fourier series of $f(t)$:

$\begin{aligned} f(t) & = \sum_{n=-\infty}^{\infty} c_n e^{2\pi i n t} \\ & = \sum_{n \text{ is odd}} \frac{1}{\pi^2 n^2} e^{2\pi i n t} \\ & = \sum_{-\infty}^{0} \frac{1}{\pi^2 n^2} e^{2\pi i n t} + \sum_{1}^{\infty} \frac{1}{\pi^2 n^2} e^{2\pi i n t} \\ & = c_{-n} e^{-2\pi i n t} + c_n e^{2\pi i n t} \\ & = c_n (e^{2\pi i n t} + e^{-2\pi i n t}) \\ & = \frac{2}{\pi^2 n^2} \cos (2 \pi n t) \\ & = \frac{1}{4} + \sum_{k=0}^\infty \frac{1}{\pi^2 (2k+1)^2} \cos[2\pi (2k+1) t] \\ \end{aligned}$

For this example, there is no joumping points, so there is no Gibbs phenomenon. The fourier series is converging to the triangle wave. However, since we have infinite terms, the fourier series is not a triangle wave. It is a smooth triangle wave. The fourier series is a smooth approximation of the triangle wave. The more terms we add, the more similar it is to the triangle wave.

This is due to the fact that the fourier series is a linear combination of the basis $e^{2\pi i n t}$. The basis $e^{2\pi i n t}$ is a smooth function, so the fourier series is also a smooth function. Or put it in another way, both sines and cosines are differentiable to any order, so the fourier series is also differentiable to any order.

In summary, a discontinuoity in any order derivative of a periodic function will force an infinite number of terms in the fourier series to approximate the function.

Note also that for the triangle wave the coefficients decrease like $1/n^2$ while for the square wave they decrease like $1/n$. Or, it takes around $N=100$ terms to approximate the square wave, but it only takes around $N=10$ terms to approximate the triangle wave. This has exactly do do wit the fact that the square wave is discontinuous while the triangle wave is continuous but its derivative is discontinuous.

Reflections
I hope those two examples could give you the sense of how the fourier series works and how it converges to the original function in terms of the speed and the smoothness.

Convergence of Fourier Series

Until now, we have assumed that the period is always $1$. Now, let’s assume $f$ is periodic at interval $L$ from $[a, b]$, which means $f(x+L) = f(x)$. We can write the fourier series as:

$c_n = \hat{f}(n) = \frac{1}{L} \int_a^b f(x) e^{-2\pi i n x / L} dx, \quad n \in \mathbb{Z} \tag{1}$

The $N$-th partial sum of the fourier series is:

$S_N(f)(x) = \sum_{n=-N}^{N} \hat{f}(n) e^{2\pi i n x / L} \tag{2}$

Now, we try to answer the following questions:

• Does the fourier series converge to $f(x)$?
• In what sense does $S_N(f)(x)$ converge to $f(x)$ as $N \rightarrow \infty$?

Roughly speaking, there are three senses of convergence:

1. Pointwise Convergence: $S_N(f)(x)$ converges to $f(x)$ for every $x$.
2. Uniform Convergence: $S_N(f)(x)$ converges to $f(x)$ uniformly. In words, when $N$ is large, the partial sum $S_N(f)(x)$ is close to $f(x)$ for every $x$ over the entire interval $[a, b]$.
3. Mean Square Convergence: $S_N(f)(x)$ converges to $f(x)$ in the mean square sense. In words, the average of the square of the difference between $S_N(f)(x)$ and $f(x)$ converges to $0$ as $N \rightarrow \infty$, meaning:

$\lim_{N \rightarrow \infty} \int_a^b |S_N(f)(x) - f(x)|^2 dx = 0 \tag{3}$

We will not prove the convergence of the fourier series here. We refer the readers to the two examples we have shown above. The square wave is discontinuous, so the fourier series converges to the square wave in the mean square sense. The triangle wave is continuous, so the fourier series converges to the triangle wave uniformly. Generally speaking, uniform convergence is the strongest form of convergence. Pointwise convergence is the weakest form of convergence. Mean square convergence is in between, which is also very subtle to study.

Dirichlet Kernel

After introducing the partial sum, it is natural to ask how good is the partial sum $S_N(f)(x)$ in approximating $f(x)$. The answer is given by the Dirichlet kernel. Now, let’s examine the partial sum $S_N(f)(x)$ (to simplify the notation, we assume $L=1$):

$\begin{aligned} S_N(f)(x) & = \sum_{n=-N}^{N} \hat{f}(n) e^{2\pi i n x / L} \\ & = \sum_{n=-N}^{N} \frac{1}{L} \int_a^b f(t) e^{-2\pi i n t / L} dt \ e^{2\pi i n x / L} \\ & = \int_a^b f(t) \sum_{n=-N}^{N} e^{-2\pi i n (t-x) } dt \\ & = \int_a^b f(t) D_N(t-x) dt \\ \end{aligned}$

where $D_N(x)$ is the Dirichlet kernel:

$D_N(x) = \sum_{n=-N}^{N} e^{-2\pi i n x} = \frac{\sin[(N+\frac{1}{2})2\pi x]}{\sin(\pi x)} \tag{4}$

We will not discuss the derivation of the Dirichelt kernel here. We will learn more about the Dirichlet kernel in the future when we talk about the convolution.

Orthogonality of the basis

In the previous post, we have show that

$e_n(t) = e^{2\pi i n t}$

is an orthogonal basis. From this, we could derive Pythagoras’s Theorem for the inner product:

$\langle f, g \rangle = \int_a^b f(x) \overline{g(x)} dx$

For our basis, we have:

$\begin{aligned} \langle e_n, e_m \rangle & = \int_a^b e^{2\pi i n t} \overline{e^{2\pi i m t}} dt \\ & = \begin{cases} 1 & n = m \\ 0 & n \neq m \end{cases} \end{aligned}$

The Pythagoras’s Theorem for the inner product is:

$\bigg | \bigg | \sum_{n=-N}^{N} e_n \bigg | \bigg |^2 = \sum_{n=-N}^{N} |e_n|^2 \tag{5}$

Here is the proof:

$\begin{aligned} \bigg | \bigg | \sum_{n=-N}^{N} e_n \bigg | \bigg |^2 & = \bigg \langle \sum_{n=-N}^{N} e_n, \sum_{n=-N}^{N} e_n \bigg \rangle \\ & = \sum_{n=-N}^N \sum_{m=-N}^N \langle e_n, e_m \rangle \quad \text{by linearity} \\ & = \sum_{n=-N}^N \sum_{m=-N}^N \begin{cases} <e_n, e_m> & n = m \\ 0 & n \neq m \end{cases} \\ & = \sum_{n=-N}^N |e_n|^2 \end{aligned}$

Some important inequalities

Before we finish this post, let’s introduce some important inequalities that are useful when we study the fourier series:

• Bessel’s inequality: $\sum_{n=-\infty}^{\infty} |\hat{f}(n)|^2 \leq \frac{1}{L} \int_a^b |f(x)|^2 dx \tag{6}$
• Rayleigh’s Identity (a.k.a. Parseval’s theorem): $\frac{1}{L} \int_a^b |f(x)|^2 dx = \sum_{n=-\infty}^{\infty} |\hat{f}(n)|^2 \tag{7}$
• Cauchy-Schwarz inequality: $\bigg| \int_a^b f(x) \overline{g(x)} dx \bigg| \leq \sqrt{\int_a^b |f(x)|^2 dx} \sqrt{\int_a^b |g(x)|^2 dx} \tag{8}$

The norm of a function $f(x)$ is defined as:

$||f(x)|| = \sqrt{\langle f(x), f(x) \rangle} = \sqrt{\int_a^b |f(x)|^2 dx} \tag{9}$

If you forget how we calculate the absolute value of a complex number, here is a quick review:

$|z| = \sqrt{z \overline{z}} = \sqrt{a^2 + b^2} \tag{10}$

where $z = a + bi$ and $\overline{z} = a - bi$. Therefore,

$\langle f(x), f(x) \rangle = \int_a^b f(x) \overline{f(x)} dx = \int_a^b |f(x)|^2 dx = ||f(x)||^2$

With the defintion of norm, let’s prove the Bessel’s inequality. For the complex inner product, we have:

$\begin{aligned} ||f + g||^2 & = \langle f + g, f + g \rangle \\ \end{aligned}$

$\begin{aligned} 0 \leq \bigg | \bigg | f(x) - \sum_{n=-N}^{N} \langle f(x), e^{2\pi i n x} \rangle e^{2\pi i n x} \bigg | \bigg |^2 & = \langle f(x) - \sum_{n=-N}^{N} \langle f(x), e^{2\pi i n x} \rangle e^{2\pi i n x}, f(x) - \sum_{n=-N}^{N} \langle f(x), e^{2\pi i n x} \rangle e^{2\pi i n x} \rangle \\ & = ||f(x)||^2 - \sum_{n=-N}^{N} |\langle f(x), e^{2\pi i n x} \rangle|^2 \\ & = ||f(x)||^2 - \sum_{n=-N}^{N} |\hat{f}(n)|^2 \end{aligned}$

This proves the Bessel’s inequality in equation (6). The complete proof of Bessel’s inequality can be found here.

Now, let’s derive the Rayleigh’s Identity. We will assume $L=1$ for simplicity:

$\begin{aligned} \langle f, f \rangle & = \int_0^1 f(x) \overline{f(x)} dx \\ & = \int_0^1 |f(x)|^2 dx \\ & = \bigg \langle \sum_{n=-\infty}^{\infty} \hat{f}(n) e^{2\pi i n x}, \sum_{m=-\infty}^{\infty} \hat{f}(m) e^{2\pi i m x} \bigg \rangle \\ & = \bigg \langle \sum_{n=-\infty}^{\infty} \langle f, e_n \rangle e_n, \sum_{m=-\infty}^{\infty} \langle f, e_m \rangle e_m \bigg \rangle \\ & = \sum_{n, m} \langle f, e_n \rangle \overline{ \langle f, e_m \rangle} \langle e_n, e_m \rangle \quad \text{using linearity} \\ & = \sum_{n, m} \langle f, e_n \rangle \overline{ \langle f, e_m \rangle} \delta_{n, m} \quad \text{using orthogonality} \\ & = \sum_{n} |\langle f, e_n \rangle|^2 \\ & = \sum_{n} |\hat{f}(n)|^2 \end{aligned}$

This proves the Rayleigh’s Identity in equation (7). This means the energy of the function $f(x)$ is the sum of the energy of the fourier coefficients.

We will not prove the Cauchy-Schwarz inequality here as this one is so well-known. The proof can be found anywhere on the internet.

Application of Rayleigh’s Identity

Now, let’s use the Rayleigh’s Identity to prove the following identity:

$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \tag{11}$

Euler proved this identity in 1735. This is also a special case of zeta function, which is defined as:

$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} \tag{12}$

All those equations are also realted to the Basel problem, which is fun to read.

To derive the identity, we first define a function $f(x)$ as:

$f(x) = x \quad \text{on } (-\pi, \pi) \tag{13}$

The period of $f(x)$ is $2\pi$. Therefore, we can write the fourier coefficient of $f(x)$ as:

$\begin{aligned} \hat{f}(n) & = \frac{1}{2\pi} \int_{-\pi}^{\pi} x e^{-2\pi i n x/L} dx \\ & = \frac{1}{2\pi} \int_{-\pi}^{\pi} x e^{- i n x} dx \\ & = \frac{1}{2\pi} \bigg( \frac{1}{-i n} x e^{ -i n x} \Big|_{-\pi}^{\pi} - \int_{-\pi}^{\pi} \frac{1}{-i n} e^{ -i n x} dx \bigg) \\ & = \frac{1}{2\pi} \bigg( \frac{1}{-i n} x e^{ -i n x} \Big|_{-\pi}^{\pi} - \frac{1}{(i n)^2} e^{ -i n x} \Big|_{-\pi}^{\pi} \bigg) \\ & = \frac{1}{2\pi} \bigg( \frac{1}{i n} (\pi e^{ -i n \pi} - (-\pi) e^{i n \pi}) - \frac{1}{(i n)^2} (e^{ -i n \pi} - e^{- i n \pi}) \bigg) \\ & = \frac{1}{2\pi} \bigg( \frac{1}{-i n} \pi (e^{in\pi} + e^{-in\pi}) - \frac{1}{(i n)^2} \frac{\sin n \pi}{2i} \bigg) \\ & = \frac{1}{2\pi} \frac{1}{-i n} \pi \cdot 2 \cos n \pi \\ & = \frac{\cos n \pi}{-in} \\ & = \frac{(-1)^{n+1}}{in} \quad n \neq 0 \end{aligned}$

Therefore we have the fourier series of $f(x)$:

$\begin{aligned} f(x) & = \sum_{n=-\infty}^{\infty} \hat{f}(n) e^{2\pi i n x/L} \\ & = \sum_{n=-\infty, n\neq 0}^{\infty} \frac{(-1)^{n+1}}{in} e^{ i n x} \end{aligned}$

From the above equation, we can see that $f(x)$ is an odd function, so the fourier coefficient is imaginary and odd. Based on Rayleigh’s Identity, we have:

$\begin{aligned} \frac{1}{2\pi} \int_{-\pi}^\pi |f(x)|^2 dx & = \sum_{n=-\infty}^{\infty} |\hat{f}(n)|^2 \\ & = \sum_{-\infty}^{-1} |\hat{f}(n)|^2 + \sum_{1}^{\infty} |\hat{f}(n)|^2 \\ & = \sum_{-\infty}^{-1} \bigg | \frac{(-1)^{n+1}}{in} \bigg |^2 + \sum_{n=1}^{\infty} \bigg | \frac{(-1)^{n+1}}{in} \bigg |^2 \\ & = 2 \sum_{n=1}^{\infty} \frac{1}{n^2} \\ & = \frac{1}{2\pi} \int_{-\pi}^\pi x^2 dx \\ & = \frac{1}{2\pi} \frac{x^3}{3} \Big|_{-\pi}^{\pi} \\ & = \frac{\pi^2}{3} \\ \end{aligned}$

Therefore, we have:

$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$

Reflections
I have mentioned that learning Fourer Series will help us to understand probability theory. Here is an example. The above example is related to the Basel problem and Zeta function, which are related to Gamma function. When you study probability theory, you will see Gamma function a lot, for example, the Gamma distribution or the Gamma distribution as the conjugate prior of the Poisson distribution, etc.