Assuming $a\in \mathbb{C}$ has positive real part, I was able to show (using Fubini) that $$\left(\int_{\mathbb{R}} e^{-x^2/(2a)}dx\right)^2 = 2\pi a$$. How can I conclude that $$\int_{\mathbb{R}} e^{-x^2/(2a)}dx = \sqrt{2\pi a}$$, where the square root is the one with positive real part?
I have tried the standard step, i.e, setting $a = \alpha + i\beta$ with $\alpha, \beta \in \mathbb{R}$ and write the real part of the integral as $$\int_{\mathbb{R}} e^{-\alpha x^2/(2|a|^2)}\cos(\beta x^2/(2|a|^2))dx$$ which doesn't seem to clear things up.