I have the following $2 \pi $ periodic function which is $t$ for $0\leq t <\pi$ and $0$ for $\pi\leq t<2\pi$
I'm asked to find its complex fourier series representation. So I calculate $c_k=\frac{1}{2\pi}\int_{0}^{\pi}te^{-ikt}dt$ which results in $\frac{1}{2\pi} (\frac{\pi(-1)^k}{-ik}+\frac{(-1)^k-1}{k^2})$ and $c_0=\frac{1}{2\pi}\int_{0}^{\pi}tdt=\frac{\pi}{4}$. So our complex fourier series is
$f(t)=\frac{\pi}{4}+\sum_{-\infty, k\neq 0}^{\infty}\frac{1}{2\pi} (\frac{\pi(-1)^k}{-ik}+\frac{(-1)^k-1}{k^2})e^{ikt}=\frac{\pi}{4}+\sum_{-\infty, k\neq 0}^{\infty}\frac{1}{2\pi} (\frac{(-1)^k-1}{k^2})e^{ikt}$
Then I am asked to find the value of the fourier series at $t=\pi$. In their solution, they simply say that the answer is $\frac{\pi}{2}$
But I don't see how they got there.
Plugging in $\pi$, I get:
$f(\pi)=0=\frac{\pi}{4}+\sum_{-\infty, k\neq 0}^{\infty}\frac{-1}{\pi k^2}*(-1)$ for $k$ odd. Now, if I solve, I get $\frac{-\pi^2}{4}$
So I don't see what I did wrong and how they got $\frac{\pi}{2}$.
Thanks for your help !
It's a standard result that the Fourier series evaluated at a jump discontinuity of the (otherwise nice) function is the average of the left and right limits. So, here, that should be the average of $\pi$ from below and $0$ from above, or $\frac{\pi}{2}$. The solution you mention checks out.
So, where did your calculation go wrong?
You've got the right coefficients.
That series for $f(t)$ - the $\frac1k$ terms don't go away just yet. Those correspond to the odd part of $f$ ($\frac t2$ on $(-\pi,\pi)$). We can't drop them until we choose $\pi$ as the point to evaluate at - because there, $e^{ik\pi}=e^{-ik\pi}$ and terms on opposite sides of zero cancel for a principal value sum of zero.
Plugging in $\pi$ - that's a correct expression. And I get $\frac{\pi}{4}+\frac{\pi}{4}=\frac{\pi}{2}$ for it. Your error is somewhere in that last step, for which you didn't show any work.