I am calculating the Fourier transform of $\exp(-t^2)$ using contour integration.
I am left with the integral $\int_{-\infty}^\infty \exp(i\omega t)\exp(-t^2)$.
Usually I would now use the residue theorem, but I cannot find the singularities. Can someone help me?
We can write
$$\exp (i\omega t)\exp(-t^2) = \exp\left(-(t-i\omega/2)^2\right)\exp\left((i\omega/2)^2\right).$$
The factor $\exp (-\omega^2/4)$ can be pulled out of the integral, and we are left with
$$\int_{-\infty + i0}^{+\infty+i0} \exp \left(-(z-i\omega/2)^2\right)\,dz.$$
Using the Cauchy integral theorem, we can shift the integration to
$$\int_{-\infty + i\omega/2}^{+\infty+i\omega/2}\exp \left(-(z-i\omega/2)^2\right)\,dz,$$
and parametrising as $z = t + i\omega/2$, that becomes
$$\int_{-\infty}^{\infty} \exp(-t^2)\,dt.$$
The shift of the contour of integration is justified by applying the Cauchy integral theorem to a rectangle with vertices $-R,\, R,\, R+i\omega/2,\, -R+i\omega/2$ and taking the limit $R\to +\infty$.