Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

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.

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5 Responses to “Evaluating the squared sinc integral.”

  1. Doug said

    How can you justify the contours used in equation 2.2? For the first term, you choose C0+C2 to include the double pole at the origin, whereas in the second term, you choose C0+C1 to omit the double pole at the origin. How can you include or omit the poles in such cavalier fashion? (I believe the third term evaluates to zero because its residue is zero, and not due to the choice of C1 vs. C2.)

  2. There is an error in above calculation.
    In line where variable “y” is changed by “z” before exp(-2z) should be plus and not minus

    • peeterjoot said

      I don’t see where you are talking about (the line before “Now we pick a contour”?). Note that the change of variables was i y \rightarrow z. The minus sign in the equation right before that i y \rightarrow z is due to a factorization of unity dy = -i i dy.

      NOTE: I’ve regenerated the attached pdf with equation numbers for all equations and refreshed this page.

      EDIT2: Ah, I see the rogue sign you are talking about. Thanks.

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