Fourier series saw function coeficient expression

Question

so im trying to apply the fourier series to a saw function and following this website and if you look at eq 2, i dont understand the last step, i tried solving it myself but the second term disapears over there... i got:$$c_n=\frac{A(\pi2ni+1)}{4\pi^2n^2}$$ is it because only the complex part is used? and why isnt $c_0=0$?

Answer 1

You have $$ c_n = \frac1T \int_{0}^{T} f(t) e^{-i\frac{2\pi nt}{T}}dt \\ f(t) = \frac ATt $$

Then

$$ c_0 = \frac1T \int_{0}^{T} f(t) dt = \frac1T [\frac{At^2}{2T} ]_{0}^{T} = \frac1T(\frac{AT}2) = \frac A2 $$

For $c_n$ you should get arrive at

$$ -Ae^{i2\pi n}\frac{-2i \pi n + e^{i 2 \pi n } -1 }{4\pi^2 n^2} = \frac{iA}{2\pi n}. $$