TITLE: Is $\frac{1}{\sqrt{2\pi t}} \int_{\mathbb{R}} e^{-\frac{(x-iut)^2}{2t}} \, dx=1$?
QUESTION [5 upvotes]: $i$ is such that $i^2=-1$. I am not familiar with complex integral. Is
$$
\frac{1}{\sqrt{2\pi t}} \int_{\mathbb{R}}  e^{-\frac{(x-iut)^2}{2t}} \, dx=1
$$
as if computing the probability of a normal density function despite the mean is imaginary. .
Thanks!

REPLY [6 votes]: Alternate proof.  Note that for complex $z$,
$$
\frac{1}{\sqrt{2\pi t}} \int_{\mathbb{R}}  e^{-(x-z)^2/(2t)} \, dx
$$
exists, and is an entire function of $z$.  But it is the constant $1$ for $z$ on the real line, so it is therefore the constant $1$ for $z$ in the whole complex plane.