Notes - Complex Analysis MT23, Jordan's lemma
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
\[\left|\int_
{\gamma_
R} f(z) e^{iaz} \text d z \right| \le \frac{\pi}{a} \cdot \max_
{\theta \in [0, \pi]} \big|f(Re^{i\theta})\big|\]
There are 2 non-obvious steps above.
- $(1\star)$: This follows from the identity that $\sin \theta = \sin (\pi - \theta)$, so
- $(2\star)$: $\sin(\theta)$ 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.