Differentiation and Integration of Fourier Series

Theory

Fourier series can be differentiated or integrated term by term.

Differentiation: The derivative of a Fourier series converges to the average of the left and right limits of $f'(x)$ .
Integration: If $f(x)$ is piecewise continuous, the integral of its Fourier series from $a$ to $x$ converges to $\int_a^x f(t)\,dt$ .

Exercise 1

Given

x = \frac{4}{\pi}\left( \sin\frac{\pi x}{2} -\frac{1}{2}\sin\frac{2\pi x}{2} +\frac{1}{3}\sin\frac{3\pi x}{2} -\cdots \right),

find the Fourier series of $f(x)=x^2$ with period $2\pi$ and evaluate

\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}.

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Solution

The given series is a sine series for $f(x)=x$ on $(-\pi,\pi)$ .
Integrate both sides term by term from $0$ to $x$ :

\int_0^x t\,dt = \frac{4}{\pi}\int_0^x \left( \sin\frac{\pi t}{2} -\frac{1}{2}\sin\frac{2\pi t}{2} +\frac{1}{3}\sin\frac{3\pi t}{2} -\cdots \right)dt.

The left-hand side gives:

\int_0^x t\,dt = \frac{x^2}{2}.

Integrating the right-hand side:

\frac{x^2}{2} = \frac{4}{\pi} \left[ -\frac{2}{\pi}\cos\frac{\pi x}{2} +\frac{2}{2^2\pi}\cos\frac{2\pi x}{2} -\frac{2}{3^2\pi}\cos\frac{3\pi x}{2} +\cdots \right] + C.

Using $x=0$ to determine $C$ , we obtain the Fourier series of $x^2$ :

x^2=\frac{\pi^2}{3} +4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2} \cos\frac{n\pi x}{2}.

To find the sum

\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2},

set $x=0$ :

0=\frac{\pi^2}{3} -4\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}.

Thus,

\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2} =\frac{\pi^2}{12}.

Exercise 2

Given

x = 2\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\sin nx}{n},

find the Fourier series of

f(x)=x^2 \quad \text{and} \quad f(x)=\frac{\pi^2x-x^3}{3}

with period $2\pi$ .

Show solution ↓

Solution

The given series is the Fourier sine series of $f(x)=x$ on $(-\pi,\pi)$ .

Step 1: Find the series for $x^2$

Integrate term by term from $0$ to $x$ :

\int_0^x t\,dt = 2\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} \int_0^x \sin(nt)\,dt.

This gives

\frac{x^2}{2} = 2\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2} (1-\cos nx).

Hence,

x^2 = \frac{\pi^2}{3} +4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos nx.

**Step 2: Find the series for $\frac{\pi^2x-x^3**{3}$ }

Differentiate the series for $x^2$ term by term:

2x = -4\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\sin nx.

Integrating once more,

\frac{\pi^2x-x^3}{3} = 4\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^3}\sin nx.

This is the required Fourier series.