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Integral of a Complex Gaussian


$\displaystyle \zbox {\int_{-\infty}^\infty e^{-p t^2}dt = \sqrt{\frac{\pi}{p}}, \quad \forall p\in {\bf C}: \; \mbox{re}\left\{p\right\}>0}$ (D.7)

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

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}

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

$\displaystyle I(p) = \sqrt{\frac{\pi}{p}}$ (D.8)

as claimed.

Area Under a Real Gaussian

Corollary: Setting $ p=1/(2\sigma^2)$ in the previous theorem, where $ \sigma>0$ is real, we have

$\displaystyle \int_{-\infty}^\infty e^{-t^2/2\sigma^2}dt = \sqrt{2\pi\sigma^2}, \quad \sigma>0$ (D.9)

Therefore, we may normalize the Gaussian to unit area by defining

$\displaystyle f(t) \isdef \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{t^2}{2\sigma^2}}.$ (D.10)


$\displaystyle f(t)>0\;\forall t$   and$\displaystyle \quad \int_{-\infty}^\infty f(t)\,dt = 1,$ (D.11)

it satisfies the requirements of a probability density function.

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Gaussian Integral with Complex Offset
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Infinite Flatness at Infinity