I'm trying to show
$$\int_{0}^{\infty}e^{-ax^2}dx=\frac{1}{2}\sqrt\frac{\pi}{a}$$
where $a$ is a complex number.
I made the substitution $x=y\sqrt{\bar a}$, where $\bar a $ is $a$ conjugate.
$$\Rightarrow \int_{0}^{\infty}e^{-ax^2}dx=\sqrt{\bar a}\int_{0}^{\infty}e^{-\lvert a \rvert^2 y^2}dy \tag 1$$ $$=\sqrt{\bar a}I$$ where the standard Integral $I$ (evaluated by contour integration) is $\frac{1}{2}\sqrt\frac{\pi}{\lvert a \rvert ^2}$ $$\Rightarrow\int_{0}^{\infty}e^{-ax^2}dx=\frac{\sqrt{\bar a}}{2}\sqrt\frac{\pi}{\lvert a \rvert ^2}=\frac{1}{2}\sqrt\frac {\pi}{a}$$
But I never used the fact that $Re(a)>0$ which is necessary for the convergence. Can anyone point out where is the mistake in above? I suspect it is in the integration limits when changing the variables.
Does anyone know how to evaluate the integral using contour?