just as an FYI, this is an exercise from Digital Signal Processing by Proakis & Manolakis 4th Edition.
I need to calculate the spectrum X(F) of a signal given by a graph. That part was easy and I was able to set up the equation. The sign i s a periodic sinusoidal signal of the type:
$$ Asin(\frac{\pi t}{\tau}) $$
The signal is periodic with period $\tau$
So I want to calculate the Fourier series in the following way: $$ C_k = \frac{1}{\tau}\int_{0}^{\tau} Asin(\frac{\pi t}{\tau}) e^{-j2\pi k\frac{t}{\tau} }dt$$ $$ C_k = \frac{1}{j2\tau}\int_{0}^{\tau} [{e^{j\pi k\frac{t}{\tau} } - e^{-j\pi k\frac{t}{\tau} }]e^{-j2\pi k\frac{t}{\tau} }dt}$$ $$ C_k = \frac{A}{j2\tau} \biggr[ \frac{e^{j\pi (1-2k)\frac{t}{\tau}}}{\frac{j \pi}{2}(1-2k)} - \frac{e^{-j\pi (1+2k)\frac{t}{\tau}}}{\frac{-j \pi}{2}(1+2k)}\biggr]_0^{\tau}$$
What I do not understand is how the next passage is derived:
$$ \frac{A}{\pi} \biggr[ \frac{1}{1-2k} + \frac{1}{1+2k} \biggr] $$
The problem goes on doing calculation that I can follow, but i need to know: what has been applied here? My calculations give other values. Here is what I got from the integration:
$$ C_k = \frac{A}{2j\tau} \biggr[ \frac{e^{j\pi (1-2k)}-1}{j \frac{\pi}{\tau}(1-2k} -\frac{e^{-j\pi (1+2k)}-1}{-j \frac{\pi}{\tau}(1+2k)} \biggr] $$
the multiplying the extrnal j and factoring the $\pi$ I get $$ C_k = \frac{A}{2\pi} \biggr[ - \frac{e^{j\pi (1-2k)}-1}{(1-2k)} -\frac{e^{-j\pi (1+2k)}-1}{(1+2k)} \biggr] $$
....am I wrong? If so, can anyone help me understand what I missed?
Edit: Here is a link to the image that has teh full book passage: