I found a lot of videos online showing how to calculate the Fourier transform of
$f(x) = \exp(-ax^2)$, but they all seem to have a mistake or so I think. (I'm not interested in the alternative solution. I'm trying to understand this.)
In these proofs we get by quadratic completion something of the form
$$\int_{-\infty}^\infty \exp\left(-a\left(x + i\frac{y}{2a}\right)^2 \right) \, dx$$
The next argument is to just substitute $z = x + i\frac{y}{2a}$ and get the known integral
$$\int_{-\infty}^\infty \exp(-az^2) \, dz,$$
but can I just blindly say that the boundaries stay the same even though I now integrate from a shifted place in $\mathbb{C}$?
Why should this be the same as
$$\lim_{b\rightarrow \infty}\int_{-b + i\frac{y}{2a}}^{b + i\frac{y}{2a}} \exp(-ax^2) \, dx,$$ which would be my solution?
Are those the same integrals?
$$\int_{-\infty}^\infty \exp\left(-a\left(x + i\frac{y}{2a}\right)^2 \right) \, dx = \lim_{M\,\to\,+\infty} \int_{-M}^M \cdots\cdots $$ Here I will follow up on the suggestion made by Cameron Williams in a comment above. Consider the integral $$ \int_{-M}^M + \int_M^{M+ iy/(2a)} + \int_{M+iy/(2a)}^{-M + iy/(2a)} + \int_{-M+iy/(2a)}^{-M} $$ where each integral is along a straight line segment. If the function being integrated is an entire function, i.e. differentiable everywhere in $\mathbb C,$ which in particular means there are no singularities inside the region bounded by the path along which we are integrating, (and Gaussian functions are indeed entire functions) then this sum is $0.$
Now suppose you can somehow show that the second and fourth terms above approach $0$ as $M\to+\infty.$ Then you can draw a conclusion about the sum of the first and third terms.