TITLE: To compute $\int_0^\infty e^{-(1+ai)^2t^2}dt$ cat I make the substitution $u=(1+ai)t$?, $a\in\Bbb R$
QUESTION [1 upvotes]: To compute $$\int_0^\infty e^{-(1+ai)^2t^2}dt,$$ can I make the substitution $u=(1+ai)t$ ?
Here it works, but can I do that in general or it doesn't work always with complex number ?

REPLY [1 votes]: The prior solutions are altogether too complicated. There is a theorem for the Integral of a Complex Gaussian that shows that
$$\int_{-\infty}^{\infty}e^{-pt^2}\,dt=\sqrt{\frac{\pi}{p}},\quad \forall p\in\mathbb{C};\, \Re\{p\}>0$$
Therefore we can say that
$$\int_0^{\infty}e^{-(1+ia)^2t^2}\,dt=\frac{1}{2}\sqrt{\frac{\pi}{(1+ia)^2}}=\frac{\sqrt{\pi}}{2(1+ia)},\quad\forall\,|a|<1$$
by virtue of the symmetry of the integrand. I have verified this result numerically.
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