\int_{-\infty}^\infty e^{-(t+ j\omega/2)^2}dt
Let u= t+ j\omega^2/2. du= dt. The integral becomes
\int_{-\infty}^\infty e^{-u^2}{du}
That's a well known integral. If you haven't seen it before, let
I= \int_{-\infty}^\infty e^{-u^2}{du}
so that, because of the symmetry,
I/2= \int_0^\infty e^{-u^2}{du}
Then it is also true that
I/2= \int_0^\infty e^{-v^2}dv
and, multiplying them
I^2/4= \left(\int_0^\infty e^{-u^2}du\right)\left(\int_0^\infty e^{-v^2}dv\right)= \int_{-\infty}^\infty\int_{-\infty}^\infty e^{-u^2-v^2}dudv
Now change to polar coordinates, r and \theta, with u= r cos(\theta) and v= r sin(\theta). The uv-integral takes each variable from 0 to \infty, the first quadrant, which, in terms of polar coordinates, has r from 0 to \infty and \theta from 0 to \pi/2. Of course, -(u^2+v^2)=-r^2 and dudv= r drd\theta
The integral becomes
\int_{\theta= 0}^{\pi/2}\int_{r=0}^\infty e^{-r^2}rdrd\theta= \frac{\pi}{2}\int_{r=0}^\infty e^{-r^2}rdr
To do that last integral, let x= r^2 so that dx= 2rdr and rdr= (1/2)dx.
Now we have
I^2/4= \frac{\pi}{4}\int_0^\infty e^{-x}dx
That last integral is, of course, 1 so we have I^2/4= \pi/4 and I= \sqrt{\pi}.