Question

Find the half range Fourier sine series of the function

Answer 1

The full range is $-\pi$ (length $2L=2\pi$), so the half range is $L=\pi$. The half range sine series would then be given by

$f(x)=\displaystyle\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L=\sum_{n\ge1}b_n\sin nx$

where

$b_n=\displaystyle\frac2L\int_0^Lf(x)\sin\dfrac{n\pi x}L\,\mathrm dx=\frac2\pi\int_0^\pi(\pi-x)\sin nx\,\mathrm dx$

Essentially, this is the same as finding the Fourier series for the function

$\begin{cases}g(x)=\begin{cases}\pi-x&\text{for }0$

Integrating by parts yields

$b_n=\dfrac2\pi\left(\dfrac\pi n-\dfrac{\sin n\pi}{n^2}\right)=\dfrac2n$

So the half range sine series for this function is simply

$f(x)=\displaystyle\sum_{n\ge1}\frac{2\sin nx}n$