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Integral of a Complex Gaussian



Theorem:

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



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,

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

as claimed.



Subsections
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``Spectral Audio Signal Processing'', by Julius O. Smith III, W3K Publishing, 2011, ISBN 978-0-9745607-3-1.
Copyright © 2016-07-18 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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