For my thesis project, I am trying to get an analytical expression for a Gaussian low-pass filtered block-impulse or box function. I defined the Gaussian low-pass filter as:
$H(\omega)=\sqrt{ \frac{1}{2\pi\sigma_f}}\cdot e^{-\omega^2/2\sigma_f}$
where $\sigma_f$ is the bandwidth of the filter. The input signal $\theta(t,t_0,T)$ is a box function, at time $t_0$ with width T. In otherwords, $\theta$ is 1 between $t_0-T/2 < t < t_0 + T/2$, and zero everywhere else. My new signal $g(t)$ is therefore:
$g(t) = IFT[G(\omega)] = \int_{-\infty}^\infty{H(\omega')\Theta(\omega')}e^{2\pi i\omega' t}d\omega' $
, where IFT[] is the Inverse Fourier Transform, and $\Theta(\omega)$ is the Fourier Transform of $\theta(t,t_0,T)$. We can get $\Theta(\omega)$ by direct Fourier Transformation
$\begin{split} \Theta(\omega) &=\int_{-\infty}^\infty\theta(t',t_0,T)e^{-2\pi i \omega t'}dt'\\&= \int_{t_0-T}^{t_0+T}e^{-2\pi i\omega t'}dt'\\&=\frac{1}{\pi\omega}e^{-2\pi i\omega t_0}sin(2\pi\omega T)\end{split} $
So the complex integral we end up with is
$g(t) = \sqrt{\frac{1}{2\pi^3\sigma_f}}\int_{-\infty}^\infty e^{-2\pi i\omega(t+t_0)}e^{\omega^2/2\sigma_f}sin(2\pi\omega T)/\omega d\omega$
To use Complex Analysis, we can perform a semi-circular contour integral with an extra semi-circular indent around $\omega = 0$, recognizing the pole at $\omega = 0$. A sketch of the contour is given here. From here I run into problems, as we can write our integrand as $f(\omega) = \frac{\phi(\omega)}{\omega}$, with
$\phi(\omega) = e^{-2\pi i\omega(t+t_0)}e^{\omega^2/2\sigma_f}sin(2\pi\omega T)$.
The residue of $f(\omega)$ at $\omega = 0$ is then shown to be
$Res_{\omega = 0}f(\omega) = \phi(0)$
Which gives in our case $\phi(0) = 0$. According to Cauchy's Integral formula, the real integral from $-\infty$ to $\infty$ is then also 0, given that the other contours go to zero, so then
$g(t) = 0$
Computationally, I can get a non-zero signal for $g(t)$, so I believe I made a mistake. Can someone point me where I went wrong?