About

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)f'(x).
  • Integration: If f(x)f(x) is piecewise continuous, the integral of its Fourier series from aa to xx converges to ∫axf(t) dt\int_a^x f(t)\,dt.

Exercises

Exercise 1
Question

Given

x=4π(sin⁡πx2−12sin⁡2πx2+13sin⁡3πx2−⋯ ),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)=x2f(x)=x^2 with period 2π2\pi and evaluate

∑n=1∞(−1)n−1n2.\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}.
Show solution ↓
Solution

The given series is a sine series for f(x)=xf(x)=x on (−π,π)(-\pi,\pi).
Integrate both sides term by term from 00 to xx:

∫0xt dt=4π∫0x(sin⁡πt2−12sin⁡2πt2+13sin⁡3πt2−⋯ )dt.\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:

∫0xt dt=x22.\int_0^x t\,dt = \frac{x^2}{2}.

Integrating the right-hand side:

x22=4π[−2πcos⁡πx2+222πcos⁡2πx2−232πcos⁡3πx2+⋯ ]+C.\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=0x=0 to determine CC, we obtain the Fourier series of x2x^2:

x2=π23+4∑n=1∞(−1)nn2cos⁡nπx2.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

∑n=1∞(−1)n−1n2,\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2},

set x=0x=0:

0=π23−4∑n=1∞(−1)n−1n2.0=\frac{\pi^2}{3} -4\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}.

Thus,

∑n=1∞(−1)n−1n2=π212.\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2} =\frac{\pi^2}{12}.
Exercise 2
Question

Given

x=2∑n=1∞(−1)n−1sin⁡nxn,x = 2\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\sin nx}{n},

find the Fourier series of

f(x)=x2andf(x)=π2x−x33f(x)=x^2 \quad \text{and} \quad f(x)=\frac{\pi^2x-x^3}{3}

with period 2π2\pi.

Show solution ↓
Solution

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

Step 1: Find the series for x2x^2

Integrate term by term from 00 to xx:

∫0xt dt=2∑n=1∞(−1)n−1n∫0xsin⁡(nt) dt.\int_0^x t\,dt = 2\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} \int_0^x \sin(nt)\,dt.

This gives

x22=2∑n=1∞(−1)n−1n2(1−cos⁡nx).\frac{x^2}{2} = 2\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2} (1-\cos nx).

Hence,

x2=π23+4∑n=1∞(−1)nn2cos⁡nx.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 x2x^2 term by term:

2x=−4∑n=1∞(−1)nnsin⁡nx.2x = -4\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\sin nx.

Integrating once more,

π2x−x33=4∑n=1∞(−1)n−1n3sin⁡nx.\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.