Hi: I'm reading a text "Fourier Transforms for Pedestrians" and it's a nice text but it skips steps that I sometimes don't understand.
The current example that I don't follow is one where the lorentzian is convoluted with the cosine. The convolution is denoted as
$$H(\omega) = \int_{-\infty}^{\infty} \frac{\sigma}{\pi} \frac{1}{\omega^2 + \sigma^2} \cos(\omega - \omega_{0})t \,d\omega .$$
It is assumed that $h(t) = f(t) g(t)$. So, therefore, using the fact that the convolution $ H(\omega) = \frac{1}{2 \pi}F(\omega) \ast G(\omega)$, the terms can be picked out of the integral so that
$F(\omega) = \frac{\sigma}{\pi} \frac{1}{\omega^2 + \sigma^2} $
and
$$G(\omega_{0}) = 2 \pi \cos(\omega_{0}).$$
The question is to find $h(t)$.
I can show that
$$g(t) = 2 \pi \left(\frac{(\delta(t_{0} - t)}{2} + \frac{\delta(t_{0} + t)}{2}\right).$$
but I can't get $f(t)$ which means that I can't get $h(t)$. The answer is that
$f(t) = \exp(-\sigma t_{0})$. Thanks for your help in advance.
I think this one is easier evaluated directly. Write
$$h(t) = \frac{\sigma}{\pi} \operatorname{Re}{\left [e^{-i \omega_0 t} \int_{-\infty}^{\infty} d\omega \frac{e^{i \omega t}}{\omega^2+\sigma^2} \right ]} $$
Note that we are now just dealing with the direct, FT of the Lorentzian, which is simply $(\pi/\sigma) e^{-\sigma |t|}$. Thus,
$$h(t) = e^{-\sigma |t|} \cos{\omega_0 t}$$