$i$ 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!
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.
Yes! In fact: $\int_{-\infty}^\infty e^{-(x+b)^2/c^2}dx= c\sqrt{\pi}$. In your case $b=-iut$ and $c=\sqrt{2t}$ hence the result follows. A justification is in order: when you make the change of variables $x \to x-iut$ you just integrate on a contour (an axis) parallel to the $x$-axis and the result holds even with the presence of complex numbers in the exponential.