To compute $\int_0^\infty e^{-(1+ai)^2t^2}dt$ cat I make the substitution $u=(1+ai)t$?, $a\in\Bbb R$ 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 ?
 A: You can certainly make the substitution, but since $u$ will be complex what you get will be a path integral over a path in the complex plane. 
EDIT: If you want to equate that integral to the corresponding one over the real interval $[0,\infty)$, you'll then want to look at wedge-shaped contours 
such as this:

and see if you can control the integral over the circular arc.
A: 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.
A: First I guess that $a\in \mathbb{R}$ and $|a|<1$ otherwise your integral diverge. For this scope we point out that the function 
$$f(a) = \int_0^\infty e^{-(1+ai)^2t^2}dt,$$ 
is smooth enough on $(-1,1)$ and its  derivative is given as follows
\begin{split}
f'(a) &= &-2i(1+ai) \int_0^\infty t^2 e^{-(1+ai)^2t^2}dt,\\
&=&\frac{i}{1+ai} \int_0^\infty [-2t(1+ai)^2e^{-(1+ai)^2t^2}] tdt,\\
&=&\frac{i}{1+ai} \int_0^\infty \frac{d}{dt}[e^{-(1+ai)^2t^2}] tdt,\qquad\text{(integration by part) }\\
&=&\frac{-i}{1+ai} \underbrace{\left[ te^{-(1+ai)^2t^2}\right]_0^\infty}_{=0} -\frac{i}{1+ai}\int_0^\infty e^{-(1+ai)^2t^2}dt,\\
&=&-\frac{i}{1+ai} f(a)
\end{split}
In fact $|te^{-(1+ai)^2t^2}| = te^{(a^2-1)t^2}\to 0$ since $|a|<1$. The case where $a\in \mathbb{C} $ it suffices to required $ Re [-(1+ai)^2] < 0$.
Whence, f satisfies ODE 
$$f'(a)=-\frac{i}{1+ai} f(a) \qquad \text{for }\qquad  Re [-(1+ai)^2] < 0$$
Which lead to 
$$f(a) = c (1+ai)$$
but $$ -c=f(0)= \int_0^\infty e^{-t^2}dt  = \frac{\sqrt{\pi}}{2}$$
that is 
$$f(a) = \int_0^\infty e^{-(1+ai)^2t^2}dt =-\frac{\sqrt{\pi}}{2}(1+ai) $$
for $ Re [-(1+ai)^2] < 0 $ and this can be analytically extend for $a\in \mathbb{C}$ using complex analysis tools.
