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