ACIT4330-Page/content/Lectures/Lecture 30 - Residue Theorem.md

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30	2025-05-08

Integrals over Real Line - Residue Theorem

(Continuing from the previous lecture)

• Something similar can be done for the lower half plane

Check if the extra contribution goes to 0

• We need a criterion to decide when \int_{H_{R}} f(z) \, dz goes to zero for R \to \infty.

Lemma

Let H_{R} be as above. Suppose that \mid f(z) \mid \leq M_{R} for all z \in H_{R}, where M_{R} depends only on the radius R. If \lim_{ R \to \infty } M_{R} \times R = 0 then \lim_{ R \to \infty } \int_{H_{R}} f(z) \, dz = 0.

Proof

Using \mid f(z) \mid \leq M_{R} we have

\mid \int_{H_{R}} f(z) \, dz \mid \leq \int_{H_{R}} \mid f(z) \mid \, dz \leq M_{R} \int_{H_{R}} 1 \, dz = M_{R} \times \pi R

Then we get \lim_{ R \to \infty } \mid \int_{H_{R}} \, dz \leq pi \times \lim_{ R \to \infty } (incomplete equation)

Example

Consider the function f(x) = \frac{1}{1+x^2}. We want to compute \int_{-\infty}^{+\infty} f(x) \, dx.

We can use the previous results. We extend f to the complex-valued function

f(z) = \frac{1}{1+z^2} = \frac{1}{(z+i)(z-i)}

We have two singularities, but only one in the upper half-plane, that is z=i.

(Quick and dirty computation to check) Let's assume for now that \lim_{ R \to \infty } \int_{H_{R}} f(z) \, dz = 0 then

\int_{-\infty}^{+\infty} \frac{1}{1+x^2} \, dx = 2\pi i \times \text{Res}_{z=i} \frac{1}{1+z^2} = 2\pi i \times \lim_{ z \to i } (z - i) \frac{1}{(z-i)(z+i)} = 2\pi i \times \frac{1}{2i} = \pi

Now we check that the integral over H_{R} goes to zero using the lemma.

We can use the Reverse Triangle Inequality | \|u \|- \| v \| \mid \leq \|u - v \|, valid for any Norm \| \cdot \|. we obtain

\mid z^2 + 1 \mid = \mid z^2 - (-1) \mid \geq | |z|^{2} - 1 | = R^{2} - 1 \ \text{(For}\ R \gt 1\text{)}

Then

| f(z) | = \frac{1}{| 1 + z^2 |} \leq \frac{1}{R^2 -1}

We take M_{R} = \frac{1}{R^2 - 1}. We get

\lim_{ R \to \infty } M_{R} \times R = \lim_{ R \to \infty } \frac{R}{R^2 - 1} = 0

This gives the result for our original integral \int_{-\infty}^{+\infty} f(x) \, dx.

Fourier Transform (Applications)

These correspond to integrals of the following type

\widetilde{f}(\omega) = \int_{-\infty}^{+\infty} f(t) e^{i\omega t} \, dt

We can use the Residue Theorem to compute some of these integrals.

Example

Consider f(t) = \frac{1}{a^2 + t^2} where a \gt 0. We want to compute the Fourier transform

\widetilde{f}(\omega) = \int_{-\infty}^{+\infty} \frac{e^{i \omega t}}{a^2 + t^2} \, dt

We apply the Residue Theorem to

g(z) = \frac{e^{i \omega z}}{a^2 + z^2}

We need to distinguish between the two cases where \omega \geq 0 and \omega \lt 0. Write z = x + iy and consider

|e^{i \omega g z} | = | e^{i \omega x} e^{- \omega y} | = e^{-\omega y}

This either decays or grows depending on the sign of \omega.

For \omega \geq 0 we get

| g(z) | = \frac{e^{-\omega y}}{| a^2 + z^2 |}

Which goes to zero in the upper half-plane (where y \geq 0). Then the path gives the result:

\widetilde{f}(\omega) = \int_{-\infty}^{+\infty} \frac{e^{i \omega t}}{a^2 + t^2} \, dt = 2 \pi i \times \text{Res}_{z = ia}\ g(z)

[!note] Note that z^2 + a^2 = (z + ia)(z-ia) with z = ia in the upper half-plane

= 2 \pi i \times \lim_{ z \to ia } (z-ia) \times \frac{e^{i \omega z}}{(z-ia)(z+ia)} = 2\pi i \times \frac{e^{-\omega a}}{2ia} = \frac{\pi}{a}e^{-\omega a}

This is valid for \omega \geq 0.

For \omega \lt 0, we want to close the path in the lower half-plane (that is y \leq 0). This is so that | e^{i \omega z} | = e^{- \omega y} does not blow up.

Then we consider that the lower half has to go in a clockwise direction (which is the opposite direction to what angles on a complex plane usually go). Note that this has negative orientation (clockwise) to apply the Residue Theorem (which needs positive orientation) we add a minus sign.

Then we compute \widetilde{f}(\omega) = \int_{-\infty}^{+\infty} \frac{e^{i \omega t}}{a^2 + t^2} \, dt Here we need the poles of g(z) in the lower half-plane (which is z = -ia). Also to compensate for going in the clockwise direction we add the extra minus:

= -2\pi i \times \text{Res}_{z = -ia} g(z) = - 2 \pi i \times \lim_{ z \to -ia } (z + ia) \frac{e^{i \omega z}}{(z + ia)(z-ia)} = -2 \pi i \times \frac{e^{\omega a}}{-2 i a} = \frac{\pi}{a} e^{\omega a}

The two cases can be combined to give the result:

\widetilde{f}(\omega) = \frac{\pi}{a}e^{-a|\omega|}