Output of LTI (in time and frequency $\omega$ domain) : when input goes through LPF I would like to raise a mathematical question :
Let's say we are been given :
$$x(t) = \begin{cases} \cos(\pi t) &amp; |t| \leq 0.5 \\
                                       0 &amp; \textrm{otherwise} \end{cases}$$
If $x(t)$ goes through an LPF : $H(\omega) = \begin{cases} \pi - 0.5|\omega| &amp; |\omega| \leq 1.5\pi\\
0 &amp; \textrm{otherwise} \end{cases}$
Compute and sketch :

Fourier coefficients of $x(t)$, as well as $X(\omega)$
$y(t)$ and $Y(\omega)$ where $y$ is the output of the LTI. Is y(t) periodic? If so compute it's period and its trigonmetric fourier anlysis coefficients
I did the math and I've got this:
$a_n = \frac{2}{\pi}\frac{\cos(\frac{n\pi}{2})}{1-n^2}$ which can not be defined for $n=-1$ or $ n=1$
update:* We have to compute $a_1$ and $a_{-1}$ in seperate: If we do the computations we gain : $a_1= a_{-1}=0$
I then did :
$$c_m = \frac{1}{2}(a_m + jbm) \rightarrow c_m = \frac{a_m}{2}$$
$$x(t) = \sum_{m=-\infty}^{\infty}c_me^{jm \omega_o t} \rightarrow 
X(\omega) =2\pi \sum_{m=-\infty}^{\infty}c_m\delta(\omega - m\omega_o)$$

What I tried:
Now we know for an LTI : $y(t) = x(t) * h(t) \rightarrow Y(\omega) = X(\omega)H(\omega)$
but since $H(\omega)$ exists only in $\delta=[-1.5\pi, 1.5\pi]$ and $\omega_o = \pi$ then we only need $X(\omega)$ in $\delta$ .
But , $X(\omega)$ is $0$  for $|m|=1 \rightarrow |\omega| =\pi$ and therefore $X(\omega) \neq 0 \rightarrow m=0$ in $\delta$ so :
$X(\omega) = 2\pi a_0 \delta(\omega)$ in $\delta$
we also know $\int_{-a}^{a}f(x)\delta(x)dx=f(0)$  so:
\begin{align}
y(t) &amp;= \frac{1}{2\pi}\int_{-\infty}^{\infty}Y(\omega)e^{j\omega t}d\omega \rightarrow \int_{-1.5\pi}^{1.5\pi}\frac{1}{2\pi}Y(\omega)e^{j\omega t}d\omega \\
&amp;\rightarrow \int_{-1.5\pi}^{1.5\pi}\frac{1}{2\pi}(\pi^2a_0 -0.5a_0|\omega|)e^{j\omega t}\delta(\omega)d\omega = \frac{1}{2\pi}\pi^2 a_0 = \frac{1}{2\pi}\pi^2 \frac{2}{\pi} = 1
\end{align}
I assume that it should have been periodic but it's not. So where did my solution go wrong?
Since this appears to be homework, I can give you a few generic pointers

LTI systems maintain periodicity: if the input is periodic so is the output and vice versa.
If a signal is finite in length it can't be periodic
If a signal has finite bandwidth, it must be infinite in time.
You can model the input as $x(t)$ as cosine multiplied with a rectangular function.
Multiplication in the time domain is convolution in the frequency domain.