Fourier series of $f(x) = 1$ on the interval $\pi/2 < |x| < \pi$

Question

I am trying to calculate the Fourier series of $f(x) = 1$ on the interval $\pi/2 < |x| < \pi$ and $f(x) = 0$ otherwise.

$f(x) = 1$ is an even function. Therefore, $b_n = 0$.

I am troubling how to proceed with the interval due to the absolute value in this case. Any help would be grateful.

Answer 1

We have

$f(x)=0$ for $x \in [-\pi/2, \pi/2]$

and

$f(x)=1$ for $x \in ( -\pi,\pi) \setminus [-\pi/2, \pi/2]$

Can you proceed ?

Answer 2

When in doubt, plot it out. Your function looks like this:

It's $0$ in the interval $(-\pi/2,\pi/2)$ and $1$ in the intervals $(-\pi, -\pi/2)$ and $(\pi/2,\pi)$. From this we can determine $$ \int_{-\pi}^\pi f(x)\cos(k x)dx = \int_{-\pi}^{-\pi/2}\cos(kx)dx + \int_{\pi/2}^\pi \cos(k x)dx. $$ You can use standard integration techniques to evaluate this integral and find the Fourier coefficients. If you do it all right, you should get $a_0 = 1/2$, $a_{2n} = 0$, and $a_{2n-1} = (-1)^n 2/[(2n-1)\pi]$.