Fourier series for the function as defined below

Question

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

I tried following the definition to find the coefficients of the Fourier series for the above function but my answer doesn't match with the one given. Please help; I am a beginner at this.

Since the domain for f(x) is $(-\pi ,\pi)$ but $f(x)=0$ for $\pi /2 \lt x \lt \pi$ make sure your integrals look like $\int_{-\pi}^{\pi/2}$. — herb steinberg, Jul 08 '18 at 21:55

score 1 · Answer 1 · answered Jul 08 '18 at 22:39

You have to solve the following 3 integrals:

$$a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\ dx=\int_{-\pi}^{\pi/2}1\ dx=\frac{3\pi}{2}$$ $$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\ \cos(nx)\ dx=\int_{-\pi}^{\pi/2}\cos(nx)\ dx=\frac{1}{n}(\sin(n\pi/2)-\sin(-n\pi))$$ $$b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\ \sin(nx)\ dx=\int_{-\pi}^{\pi/2}\sin(nx)\ dx=-\frac{1}{n}(\cos(n\pi/2)-\cos(-n\pi))$$

and finally distinguish several cases on $n\in\mathbb{N}$ for actual evaluation of the right hand sides.

--- rk

I did do this but after a lot of trials the answers didn' t match. That's why I asked for a solution. — Pi_die_die, Jul 09 '18 at 03:17

Fourier series for the function as defined below

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