$$f(x)=\begin{cases}0&-5\leq x<0\\3&0\leq x<5\end{cases}\\\text{with } f(x+10)=f(x)$$
is the function I need to make a Fourier series on, I know where to start, but I get confused along the way and I don't know in general how to "finish up" a Fourier series.
My own solution so far so you can follow my thoughts, used variables etc: page 1 page 2
Hope my handwriting is readable and thank you in advance!
Since the quality of the images aren't good enough, I'll just write it out on here: $\tilde{f}_k = \int_{-5}^{5} f(x)e^{-ikx}dx$
$\tilde{f}_k = \int_{-5}^{0} 0e^{-ikx}dx + \int_{0}^{5} 3e^{-ikx}dx$
Integrating this, I got:
$\tilde{f}_k = \frac{1}{ik}(3-3e^{-5ik})$
for k = 0 I got $\tilde{f_{0}} = \frac{3}{2}$
After this it gets fuzzy for me for what exactly I need to do, I filled in the formula but the steps after this become weird and unclear. It's kinda different in each example I've seen.
The solution to this problem should be:
$f(x) = \frac{3}{2} + \frac{6}{\pi}\sum_{n=0}^{+\infty}\frac{1}{2n+1}\sin(\frac{(2n+1)\pi x}{5})$