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# Posts Tagged ‘pole at origin’

## Evaluating the squared sinc integral.

Posted by peeterjoot on December 10, 2011

[Click here for a PDF of this post with nicer formatting]

# Motivation

In the Fermi’s golden rule lecture we used the result for the integral of the squared $\text{sinc}$ function. Here is a reminder of the contours required to perform this integral.

# Guts

We want to evaluate

\begin{aligned}\int_{-\infty}^\infty \frac{\sin^2 (x\left\lvert {\mu} \right\rvert)}{x^2} dx\end{aligned} \hspace{\stretch{1}}(1.2.1)

We make a few change of variables

\begin{aligned}\begin{aligned}\int_{-\infty}^\infty \frac{\sin^2 (x\left\lvert {\mu} \right\rvert)}{x^2} dx &= \left\lvert {\mu} \right\rvert \int_{-\infty}^\infty \frac{\sin^2 (y)}{y^2} dy \\ &= -i \left\lvert {\mu} \right\rvert \int_{-\infty}^\infty \frac{(e^{iy} - e^{-iy})^2}{(2 i y)^2} i dy \\ &=-\frac{i \left\lvert {\mu} \right\rvert}{4} \int_{-i\infty}^{i\infty} \frac{e^{2z} + e^{-2z} - 2}{z^2} dz\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.2.2)

Now we pick a contour that is distorted to one side of the origin as in fig. 1.1

Fig 1.1: Contour distorted to one side of the double pole at the origin

We employ Jordan’s theorem (section 8.12 [1]) now to pick the contours for each of the integrals since we need to ensure the $e^{\pm z}$ terms converges as $R \rightarrow \infty$ for the $z = R e^{i\theta}$ part of the contour. We can write

\begin{aligned}\int_{-\infty}^\infty \frac{\sin^2 (x\left\lvert {\mu} \right\rvert)}{x^2} dx=-\frac{i \left\lvert {\mu} \right\rvert}{4} \left(\int_{C_0 + C_2} \frac{e^{2z}}{z^2} dz-\int_{C_0 + C_1} \frac{e^{-2z}}{z^2} dz-\int_{C_0 + C_1} \frac{2}{z^2} dz\right)\end{aligned} \hspace{\stretch{1}}(1.2.3)

The second two integrals both surround no poles, so we have only the first to deal with

\begin{aligned}\begin{aligned}\int_{C_0 + C_2} \frac{e^{2z}}{z^2} dz &= 2 \pi i \frac{1}{{1!}} {\left.{{ \frac{d}{dz} e^{2z}}}\right\vert}_{{z=0}} \\ &= 4 \pi i \end{aligned}\end{aligned} \hspace{\stretch{1}}(1.2.4)

Putting everything back together we have

\begin{aligned}\int_{-\infty}^\infty \frac{\sin^2 (x\left\lvert {\mu} \right\rvert)}{x^2} dx &= -\frac{i \left\lvert {\mu} \right\rvert}{4} 4 \pi i \\ &= \pi \left\lvert {\mu} \right\rvert\end{aligned} \hspace{\stretch{1}}(1.2.5)

## On the cavalier choice of contours

The choice of which contours to pick above may seem pretty arbitrary, but they are for good reason. Suppose you picked $C_0 + C_1$ for the first integral. On the big $C_1$ arc, then with a $z = R e^{i \theta}$ substitution we have

\begin{aligned}\left\lvert {\int_{C_1} \frac{e^{2 z}}{z^2} dz} \right\rvert &= \left\lvert {\int_{\theta = \pi/2}^{-\pi/2} \frac{e^{ 2 R (\cos\theta + i \sin\theta) }}{R^2 e^{ 2 i \theta}}R i e^{i \theta} d\theta} \right\rvert \\ &= \frac{1}{R}\left\lvert {\int_{\theta = \pi/2}^{-\pi/2} e^{ 2 R (\cos\theta + i \sin\theta) }e^{-i \theta} d\theta} \right\rvert \\ &\le \frac{1}{R}\int_{\theta = -\pi/2}^{\pi/2} \left\lvert {e^{ 2 R \cos\theta }} \right\rvert d\theta \\ &\le \frac{\pi e^{2 R}}{R}\end{aligned} \hspace{\stretch{1}}(1.2.6)

This clearly doesn’t have the zero convergence property that we desire. We need to pick the $C_2$ contour for the first (positive exponent) integral since in that $[\pi/2, 3\pi/2]$ range, $\cos\theta$ is always negative. We can however, use the $C_1$ contour for the second (negative exponent) integral. Explicitly, again by example, using $C_2$ contour for the first integral, over that portion of the arc we have

\begin{aligned}\left\lvert {\int_{C_2} \frac{e^{2 z}}{z^2} dz} \right\rvert &= \left\lvert {\int_{\theta = \pi/2}^{3 \pi/2} \frac{e^{ 2 R (\cos\theta + i \sin\theta) }}{R^2 e^{ 2 i \theta}}R i e^{i \theta} d\theta} \right\rvert \\ &= \frac{1}{R}\left\lvert {\int_{\theta = \pi/2}^{3 \pi/2} e^{ 2 R (\cos\theta + i \sin\theta) }e^{-i \theta} d\theta} \right\rvert \\ &\le \frac{1}{R}\int_{\theta = \pi/2}^{3 \pi/2} \left\lvert {e^{ 2 R \cos\theta }d\theta} \right\rvert \\ &\approx \frac{1}{R}\int_{\theta = \pi/2}^{3 \pi/2} \left\lvert {e^{ -2 R }d\theta} \right\rvert \\ &= \frac{\pi e^{-2 R} }{R}\end{aligned} \hspace{\stretch{1}}(1.2.7)

# References

[1] W.R. Le Page and W.R. LePage. Complex Variables and the Laplace Transform for Engineers. Courier Dover Publications, 1980.

Posted in Math and Physics Learning. | Tagged: , , , | 5 Comments »