---
lecture: 30
date: 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:
$$\widetilde{f}(\omega) = \frac{\pi}{a}e^{-a|\omega|}$$