Lecture 30 - Residue Theorem

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 $\mid f(z)\mid \le M_R$ 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 $\mid f(z)\mid \le M_R$ we have
$\mid {\int }_{H_R}f(z)dz\mid \le {\int }_{H_R}\mid f(z)\mid dz$
$\le M_R{\int }_{H_R}1dz=M_R\times \pi R$
Then we get
${\lim }_{R\to \infty }\mid {\int }_{H_R}dz\le pi\times {\lim }_{R\to \infty }$ (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 $|\Vert u\Vert -\Vert v\Vert \mid \le \Vert u-v\Vert$, valid for any Norm $\Vert \cdot \Vert$. we obtain
$\mid z^2+1\mid =\mid z^2-(-1)\mid$
$\ge ||z{|}^2-1|$
$=R^2-1\text{(For}R>1\text{)}$
Then
$|f(z)|=\frac{1}{|1+z^2|}\le \frac{1}{R^2-1}$
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
$\tilde{f}(\omega )={\int }_{-\infty }^{+\infty }f(t)e^{i\omega t}dt$
We can use the Residue Theorem to compute some of these integrals.

Example

Consider $f(t)=\frac{1}{a^2+t^2}$ where $a>0$.
We want to compute the Fourier transform
$\tilde{f}(\omega )={\int }_{-\infty }^{+\infty }\frac{e^{i\omega t}}{a^2+t^2}dt$
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 $\omega \ge 0$ and $\omega <0$.
Write $z=x+iy$ and consider
$|e^{i\omega gz}|=|e^{i\omega x}{e}^{-\omega y}|=e^{-\omega y}$
This either decays or grows depending on the sign of $\omega$.

For $\omega \ge 0$ we get
$|g(z)|=\frac{e^{-\omega y}}{|a^2+z^2|}$
Which goes to zero in the upper half-plane (where $y\ge 0$).
Then the path gives the result:
$\tilde{f}(\omega )={\int }_{-\infty }^{+\infty }\frac{e^{i\omega t}}{a^2+t^2}dt=2\pi i\times {\text{Res}}_{z=ia}g(z)$

Note 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 $\omega \ge 0$.

For $\omega <0$, we want to close the path in the lower half-plane (that is $y\le 0$).
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
$\tilde{f}(\omega )={\int }_{-\infty }^{+\infty }\frac{e^{i\omega t}}{a^2+t^2}dt$
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:
$=-2\pi i\times {\text{Res}}_{z=-ia}g(z)$
$=-2\pi i\times {\lim }_{z\to -ia}(z+ia)\frac{e^{i\omega z}}{(z+ia)(z-ia)}=-2\pi i\times \frac{e^{\omega a}}{-2ia}=\frac{\pi }{a}{e}^{\omega a}$
The two cases can be combined to give the result:
$\tilde{f}(\omega )=\frac{\pi }{a}{e}^{-a|\omega |}$