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

\[\left|\int_ {\gamma_ R} f(z) e^{iaz} \text d z \right| \le \frac{\pi}{a} \cdot \max_ {\theta \in [0, \pi]} \big|f(Re^{i\theta})\big|\]

\[\begin{aligned} \left|\int_ {\gamma_R} f(z) e^{iaz} \text d z \right| &= \left|\int^\pi_ 0 f(Re^{i\theta}) e^{iaR(\cos \theta + i \sin \theta)} iRe^{i\theta} \text d \theta\right| \\\\ &= R \left| \int^\pi_ 0 f(Re^{i\theta}) e^{aR(i\cos \theta - \sin \theta)} \text d \theta\right| \\\\ &\le R \int^\pi_ 0 \left| f(Re^{i\theta}) e^{aR(i\cos \theta - \sin \theta)} \right| \text d \theta \\\\ &= R \int^\pi _0 |f(Re^{i\theta})| e^{-aR\sin \theta} \text d \theta \\\\ &\le R \cdot \left(\max_ {\theta \in [0, \pi]} \left|f(Re^{i \theta})\right|\right) \cdot \int^\pi_ a e^{-a R\sin \theta} \text d \theta \\\\ &= 2R \cdot \left(\max_ {\theta \in [0, \pi]} \left|f(Re^{i \theta})\right|\right) \cdot \int^{\pi/2}_ 0 e^{-a R \sin \theta} \text d \theta \quad (\star1) \\\\ &\le 2R \cdot \left(\max_ {\theta \in [0, \pi]} \left|f(Re^{i \theta})\right|\right) \cdot \int^{\pi/2}_ {0} e^{-2aR\theta/\pi} \text d\theta \quad (\star2) \\\\ &= \frac \pi a\cdot \left(\max_ {\theta \in [0, \pi]} \left|f(Re^{i \theta})\right|\right) \cdot (1 - e^{-aR}) \\\\ &\le \frac \pi a \left(\max_ {\theta \in [0, \pi]} \left|f(Re^{i \theta})\right|\right) \end{aligned}\]

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.




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