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content/Lectures/Lecture 30 - Residue Theorem.md

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

content/index.md

@@ -27,4 +27,5 @@ title: ACIT4330 Lecture Notes
 - [[Lecture 17 - Lp Spaces]]
 # Complex Analysis
 - [[Lecture 18 - Complex Analysis]]
-- [[Lecture 19 - Derivatives]]
+- [[Lecture 19 - Derivatives]]
+- [[Lecture 30 - Residue Theorem|Last Lecture - Residue Theorem]]