Integral of a Complex Gaussian

Theorem:

$$\zbox{\int_{-\infty}^\infty e^{-p t^2}dt = \sqrt{\frac{\pi}{p}}, \quad \forall p\in \mathbb{C}: \; \mbox{re}\left\{p\right\}>0}$$

Proof: Let $I(p)$ denote the integral. Then

$$\begin{eqnarray*} I^2(p) &=& \left(\int_{-\infty}^\infty e^{-p x^2}dx\right) \left(\int_{-\infty}^\infty e^{-p y^2}dy\right)\\ &=& \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-p (x^2+y^2)}dx\,dy\\ &=& \int_0^{2\pi}\int_0^\infty e^{-p r^2}r\,dr\,d\theta\\ &=& 2\pi\int_0^\infty e^{-p r^2}r\,dr\\ &=& \left. 2\pi\frac{1}{-2p} e^{-p r^2}\right\vert _0^\infty = \frac{\pi}{-p} (0 - 1) = \frac{\pi}{p} \end{eqnarray*}$$

where we needed re$\left\{p\right\}>0$ to have $e^{-p r^2}\to 0$ as $r\to\infty$ . Thus,

$$I(p) = \sqrt{\frac{\pi}{p}}$$

as claimed.