I wish to evaluate $y(t) = \mathscr{F}^{-1} [ \cot{a \omega} \times \mathscr{F} \{ U(t) \sin{\omega_0 t} \} ] $, where $\mathscr{F}$ represents the Fourier transform, and U(t) represents the Heaviside step function.
I seek to use contour integration to remove the singularities introduced by $\cot{a \omega}$ and express the result as an infinite sum of residues. I have attempted this myself and believe there must be a mistake because the answer doesn't seem to make sense. Can anybody help me?
My attempt at the problem:
1) Substitute the Fourier transform of the right-hand sine into the problem.
$$ y(t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \cot{a \omega} \Big[ \frac{\pi}{2j} [ \delta(\omega - \omega_0) - \delta(\omega + \omega_0) ] + \frac{\omega_0}{\omega_0^2 - \omega^2} \Big] e^{j \omega t} \, d \omega $$
2) Use the translation property of the delta function.
$$ y(t) = - \frac{j}{2} \cot{a \omega_0} \cos{\omega_0 t} + \frac{1}{2 \pi} \int_{-\infty}^{\infty} \frac{\omega_0 \cot{a \omega} }{ \omega_0^2 - \omega^2} e^{j \omega t} \, d \omega$$
3) Evaluate the infinite integral $I = \int_{-\infty}^{\infty} \frac{\omega_0 \cot{a \omega} }{ \omega_0^2 - \omega^2} e^{j \omega t} \, d \omega$ by indenting the contour to avoid the set of poles lying on the real axis, and assuming the integral of the path around the large arc is 0 (Jordan's Lemma). Thus
$$ I = - j \pi \sum \text{res}, $$
where the residues at the infinite set of first order poles $\omega = \pm n\pi /a $, ($n = 0,1,2...$), and the two first order poles $\omega = \pm \omega_0$ are given by
$$ \mathrm{res}_\omega = \frac{ \omega_0 \cos{a\omega} e^{j \omega t} }{ (\omega_0^2 - \omega^2) a \cos{a \omega} - 2 \omega \sin{a \omega} },$$
and so
$$ \begin{align} I & = - j \pi \Big( - \frac{1}{2} \cot{a \omega_0} e^{j \omega_0 t} - \frac{1}{2} \cot{a \omega_0} e^{-j \omega_0 t} \Big) - j \pi \Big( \sum_{n=-\infty}^{\infty} \frac{\omega_0 \cos{a \omega_n} e^{j \omega_n t}}{ (\omega_0^2 - \omega_n^2) a \cos{a \omega_n} } \Big) \\ & = j \pi \cot{a \omega_0} \cos{\omega_0 t} + j \pi \sum_{n=0}^{\infty} \epsilon_n \frac{\cos{w_n t} }{a (\omega_n^2 - \omega_0^2) }\end{align}$$
4) Substitute the result into the expression for $y(t)$...
$$ y(t) = 0 + \frac{j}{2a} \sum_{n=0}^{\infty} \epsilon_n \frac{\cos{w_n t} }{ (\omega_n^2 - \omega_0^2) } $$
I believe that this is incorrect since I would anticipate that
$$ y(t) \rightarrow \cot{a \omega_0} \sin{\omega_0 t} \quad \text{as} \quad t \rightarrow \infty. $$
This fact may make things a little easier:
$$ \mathscr{F}^{-1} (\hat{f} \hat{g}) = f * g $$
where $\hat f = \mathscr{F}( f)$ and $f*g$ is convolution.