$f(x) = 0, -\pi < x< \pi/2$
$f(x) = 1, \pi/2 < x < \pi$
The answer is given as:
$$f(x)=\frac{1}{4} - \frac{1}{\pi} \left ( \frac{cosx}{1} - \frac{cos3x}{3} + \frac{cos5x}{5}... \right )+\frac{1}{\pi} \left (\frac{sinx}{1} - \frac{2sin2x}{2} + \frac{sin3x}{3} -\frac{2sin6x}{6} \right )$$
My attempt:
$$c_n=\frac{1}{2\pi} \int_{-\pi}^{\pi/2} 0dx + \frac{1}{2\pi} \int_{\pi/2}^{\pi} e^{-inx}dx$$ $$=\frac{1}{2\pi} \left [\frac{e^{-in\pi}}{-in} - \frac{e^{-in\pi/2}}{-in} \right ]$$
This evaluates to each of the following, depending on the noted value of $n$:
$=\frac{-1}{2\pi in} (-1+i)$ , $n=1,5,9,13,...$
$=\frac{-1}{\pi in}$, $n=2,6,10,14,...$
$\frac{-1}{2\pi in}(-1-i)$, $n=3,7,11,15,..$
$=0$, $n=4,8,12,16,...$
$c_0=\frac{1}{4}$, of course.
Putting it all together:
$$f(x)=\frac{1}{4}+\frac{1-i}{2\pi i}(\frac{e^{ix}}{1}-\frac{e^-ix}{1}+\frac{e^{5ix}}{5}-\frac{e^{-5ix}}{5}+\ldots)-\frac{1}{\pi i} (\frac{e^{2ix}}{2}-\frac{e^{-2ix}}{2} +\frac{e^{6ix}}{6}-\frac{e^{-6ix}}{6}+\ldots)+$$ $$ \frac{1+i}{2\pi i} (\frac{e^{3ix}}{3}-\frac{e^{-3ix}}{3}+\frac{e^{7ix}}{7}-\frac{e^{-7ix}}{7}+\ldots)$$
$$=\frac{1}{4}+\frac{1-i}{\pi} (\frac{\sin x}{1}+\frac{\sin 5x}{5}+\frac{\sin 9x}{9}+\ldots )-\frac{2}{\pi}(\frac{\sin 2x}{2}+\frac{\sin 6x}{6}+\ldots)$$ $$+\frac{1+i}{\pi}(\frac{\sin 3x}{3}+\frac{\sin7x}{7}+\ldots)$$
Anyone see what I'm doing wrong?