How to find the Fourier's coefficient $a_n$ of the Fourier's series of $\sin(x)$ on $(0,\pi]$, $0$on $(-\pi,0]$

Question

Considering $g(x)$, periodical with a period of $2\pi$ defined by

\begin{equation*} g(x)= \begin{cases} 0 & \text{for $x \in (-\pi;0]$} \\ \sin(x) & \text{for $x \in [0;\pi)$} \end{cases} \end{equation*}

I want to find the Fourier's series $S(g)$ of $g$.

\begin{align*} a_n&=\frac{1}{\pi}\int\limits_0^\pi \sin(x)\cos(x)d(x)\\ &=[sin^2(x)]-\int\limits_0^\pi \cos^2(x)d(x)\\ &=-\frac{1}{2}\int\limits_0^\pi (\cos(2x)+1)d(x)\\ &= \sin(2x)+\frac{x}{2}\\ \end{align*}

I've the feeling, seeing the solution of \begin{equation*} g(x)= \begin{cases} 0 & \text{for $x \in (-\pi;0]$} \\ x\sin(x) & \text{for $x \in [0;\pi)$} \end{cases} \end{equation*}

That I'm wrong...

Answer 1

For

\begin{equation*} g(x)= \begin{cases} 0 & \text{$x \in [-\pi,0)$} \\ \sin(x) & \text{$x \in [0,\pi)$} \end{cases} \end{equation*}

Write

$$g(x) = \frac12 a_0 + \sum_{n=1}^{\infty} \left (a_n \cos{n x} + b_n \sin{n x} \right ) $$

$$a_0 = \frac1{\pi} \int_0^{\pi} dx \, \sin{x} = \frac{2}{\pi}$$ $$a_1 = \frac1{\pi} \int_0^{\pi} dx \, \sin{x} \cos{x} = 0$$ $$b_1 = \frac1{\pi} \int_0^{\pi} dx \, \sin^2{x} = \frac12$$

For $n \gt 1$:

$$\begin{align}a_n &= \frac1{\pi} \int_0^{\pi} dx \, \sin{x} \cos{n x}\\ &= \frac1{2 \pi} \int_0^{\pi} dx \, \left [\sin{(n+1) x} - \sin{(n-1) x} \right ]\\ &= \frac1{2 \pi} \left [\frac{1-(-1)^{n+1}}{n+1} - \frac{1-(-1)^{n-1}}{n-1} \right ] \\ &= -\frac{1+(-1)^n}{\pi (n^2-1)} \end{align}$$

$$\begin{align}b_n &= \frac1{\pi} \int_0^{\pi} dx \, \sin{x} \sin{n x}\\ &= \frac1{2 \pi} \int_0^{\pi} dx \, \left [\cos{(n-1) x} - \cos{(n+1) x} \right ]\\ &= 0 \end{align}$$

Thus,

$$g(x) = \frac1{\pi} + \frac12 \sin{x} - \frac{2}{\pi} \sum_{n=1}^{\infty} \frac{\cos{2 n x}}{4 n^2-1}$$