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How do I find out the Cauchy Principal value of $\int_{-\infty}^{\infty}e^{-ax^2}\cos(2abx) \,dx\,\,\,\,\,\,\,\,a,b>0$ using complex integration? The answer is $\sqrt{\frac{\pi}{a}}e^{-ab^2}$, and apparently this integral is called the Poisson integral.

My attempt: I am guessing, I compute the integral $\int_{\gamma}e^{-az^2}e^{i2abz} $, where $\gamma$ is the semicircular contour, of infinite radius. However this gives the answer zero, as there are no singularities inside the contour, so the total integral is 0. Also the integral along the semicircular arc is 0, as $|e^{-aR^2 (\cos{\theta}+i\sin{\theta})}|=e^{-aR^2\cos{\theta}}$ which tends to 0, as $R \to \infty$. Hence Jordan's Lemma implies that the integral is 0.

Where am I going wrong, any help will be appreciated.

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3 Answers

up vote 4 down vote accepted

Consider $f(z) = e^{-ax^2}$, and let $C$ be the rectangular contour with vertices at $-R,R,R+ib$ and $-R+ib$.
As $R \to \infty$, the two sides parallel to the imaginary axis disappear, then we have, by Cauchy's theorem:

$$ 0 = \oint_C f(z) \, dz = \int_{-\infty}^\infty f(x)\, dx + \int_{\infty}^{-\infty} e^{-a(x+ib)^2} \implies\\ \int_{-\infty}^\infty e^{-ax^2 -2 a i x b +a b^2}\, dx = \int_{-\infty}^\infty f(x)\, dx = \sqrt{\frac{\pi}{a}} \implies \\ \int_{-\infty}^\infty e^{-ax^2 -2 a i x b}\, dx = \sqrt{\frac{\pi}{a}}e^{-ab^2} $$

Taking real parts of both sides, the result follows.

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First of all, you are not seeking the Cauchy principal value. As you say: where are the poles? It makes no sense to speak of this here. You only refer to the PV when you have to deform the integration contour about a pole.

Next, the integral about your contour does not vanish. Your integrand looks like

$$e^{-a R^2 \cos{2 \theta}} e^{-2 a b R \sin{\theta}}$$

Keep in mind that, in the upper half plane, $\theta \in [0,\pi]$. Thus, the first factor diverges along sections of the contour as $R \to \infty$. This integration contour will not help.

In fact, Cauchy's integral theorem is of no help for this integral as far as I can see. @DonAntonio summarizes the standard way of attacking this integral. This theorem, however, is very useful for integrals like

$$\int_{-\infty}^{\infty} dx \: e^{i a x^2}$$

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Thanks Ron. +1. –  ramanujan_dirac Apr 28 '13 at 13:06
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Hints:

$$e^{-ax^2}\cos(2abx)=\frac{1}{2}e^{-ax^2}\left(e^{2iabx}+e^{-2iabx}\right)=\frac{1}{2}e^{-(ax^2-2iabx)}+e^{-(ax^2+2iabx)}$$

But

$$\int\limits_{-\infty}^\infty e^{-(ax^2\pm2aibx)}dx=\int\limits_{-\infty}^\infty e^{-a(x\pm bi)^2-ab^2}dx=\frac{e^{-ab^2}}{\sqrt a}\int\limits_{-\infty}^\infty e^{-u^2}du=\sqrt\frac{\pi}{a}\,e^{-ab^2}$$

where we substituted $\,u=\sqrt a(x\pm bi)\,$ in the second equality above

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Some imaginary $i$s missing here. –  GEdgar Apr 28 '13 at 13:02
    
You are right of course, @GEdgar . Thanks. Edit on its way in some minutes. –  DonAntonio Apr 28 '13 at 13:11
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