# Notes - Complex Analysis MT23, Jordan's lemma

> Source: https://ollybritton.com/notes/uni/part-a/mt23/complex-analysis/notes/jordans-lemma/ · Updated: 2024-04-28 · Tags: uni, notes

- [Course - Complex Analysis MT23](https://ollybritton.com/notes/uni/part-a/mt23/complex-analysis/)
	- [Notes - Complex Analysis MT23, Complex integration](https://ollybritton.com/notes/uni/part-a/mt23/complex-analysis/notes/complex-integration/)
	- [Notes - Complex Analysis MT23, Residue theorem](https://ollybritton.com/notes/uni/part-a/mt23/complex-analysis/notes/residue-theorem/)

### Flashcards
Can you state Jordan's lemma, which lets you discard the semicircular part of a contour integral for a particular class of functions?::
Suppose:

- $f : \mathbb H \to \mathbb C$ is a meromorphic function on upper half-plane
- $f(z) \to 0$ as $z \to \infty$ on $\mathbb H$
- $\gamma_R(t) = Re^{it}$, $t \in [0, \pi]$
- $\alpha \in \mathbb R$, $\alpha > 0$.

Then

$$
\int_{\gamma_R} f(z)e^{i a z} \text dz \to 0
$$
as $R \to \infty$.

Quickly prove Jordan's lemma, i.e. that if

- $f : \mathbb H \to \mathbb C$ is a meromorphic function on upper half-plane
- $f(z) \to 0$ as $z \to \infty$ on $\mathbb H$
- $\gamma_R(t) = Re^{it}$, $t \in [0, \pi]$
- $\alpha \in \mathbb R$, $\alpha > 0$.

then

$$
\int_{\gamma_R} f(z)e^{i a z} \text dz \to 0
$$
as $R \to \infty$, and actually establish in the process the more general result that even if $f(z) \not\to 0$ as $z \to \infty$ on $\mathbb H$, we have
$$
\left|\int_
{\gamma_
R} f(z) e^{iaz} \text d z \right| \le \frac{\pi}{a} \cdot \max_
{\theta \in [0, \pi]} \big|f(Re^{i\theta})\big|
$$::
$$
\begin{aligned}
\left|\int_
{\gamma_R} f(z) e^{iaz} \text d z \right| &= \left|\int^\pi_
0 f(Re^{i\theta}) e^{iaR(\cos \theta + i \sin \theta)} iRe^{i\theta} \text d \theta\right| \\\\
&= R \left| \int^\pi_
0 f(Re^{i\theta}) e^{aR(i\cos \theta - \sin \theta)} \text d \theta\right| \\\\
&\le R \int^\pi_
0 \left| f(Re^{i\theta}) e^{aR(i\cos \theta - \sin \theta)} \right|  \text d \theta \\\\
&= R \int^\pi
_0 |f(Re^{i\theta})| e^{-aR\sin \theta} \text d \theta \\\\
&\le R \cdot \left(\max_
{\theta \in [0, \pi]} \left|f(Re^{i \theta})\right|\right) \cdot \int^\pi_
a e^{-a R\sin \theta} \text d \theta \\\\
&= 2R \cdot \left(\max_
{\theta \in [0, \pi]} \left|f(Re^{i \theta})\right|\right) \cdot \int^{\pi/2}_
0 e^{-a R \sin \theta} \text d \theta \quad (\star1) \\\\
&\le 2R \cdot \left(\max_
{\theta \in [0, \pi]} \left|f(Re^{i \theta})\right|\right) \cdot \int^{\pi/2}_
{0} e^{-2aR\theta/\pi} \text d\theta \quad (\star2) \\\\
&= \frac \pi a\cdot \left(\max_
{\theta \in [0, \pi]} \left|f(Re^{i \theta})\right|\right) \cdot (1 - e^{-aR}) \\\\
&\le \frac \pi a \left(\max_
{\theta \in [0, \pi]} \left|f(Re^{i \theta})\right|\right)
\end{aligned}
$$
There are 2 non-obvious steps above.

- $(1\star)$: This follows from the identity that $\sin \theta = \sin (\pi - \theta)$, so

$$
\begin{aligned}
\int^\pi_0 e^{-aR\sin\theta} \text d \theta &= \int^{\pi/2}_0e^{-aR\sin\theta} \text d\theta + \int^\pi_{\pi/2} e^{-aR\sin\theta} \text d\theta \\\\
&= \int^{\pi/2}_0e^{-aR\sin\theta} \text d\theta + \int^0_{\pi/2}e^{-aR\sin\theta} \cdot (-1) \text d\theta \\\\
&= 2 \int^{\pi/2}_{0} e^{-aR\sin\theta} \text d\theta
\end{aligned}
$$

- $(2\star)$: $\sin(\theta)$ is concave on $[0, \pi / 2]$, so will be bounded below by the straight line from the two endpoints, i.e. $\frac{2\theta}{\pi} \le \sin \theta$. Then negating gives $-\sin \theta \le -\frac{2\theta}{\pi}$.

The final result, about tending to zero, follows from the assumption that $f(z) \to 0$ for sufficiently large $z$.

Suppose you are considering Jordan's lemma, i.e. that if

- $f : \mathbb H \to \mathbb C$ is a meromorphic function on upper half-plane
- $f(z) \to 0$ as $z \to \infty$ on $\mathbb H$
- $\gamma_R(t) = Re^{it}$, $t \in [0, \pi]$
- $\alpha \in \mathbb R$, $\alpha > 0$.

then $\int_{\gamma_R} f(z)e^{i a z} \text dz \to 0$ as $R \to \infty$. However, you have some $\alpha < 0$. What "version" of Jordan's lemma works instead?::
Considering $f$ as a meromorphic function on the lower half-plane, and using a semicircular contour on the lower half-plane.

---
Olly Britton — https://ollybritton.com. Machine-readable index: https://ollybritton.com/llms.txt