MATH3401

Lecture 37

Jordan's Lemma

Lemma: If $$f$$ is analytic on $$\left\{ z : \Im z \ge 0\right\} \cap \left\{ z : |z| \gt R_0 \right\}$$ satisfies $$\big|\,f(z) \le \frac{M}{R^\beta}\big|\;$$ $$(*)\;$$ ($$M,\beta \gt 0$$ constants) on $\Gamma_R=\left\{ Re^{i\theta}: R\gt R_0, 0\le \theta \le \pi \right\}.$ Then $\ds \lim_{R\ra \infty } \int_{\Gamma_R } e^{i\alpha z} f(z)\, dz = 0,\; \forall \alpha \gt 0.$

Jordan's Lemma

Proof: On $$\Gamma_R$$ we have $$z= Re^{i \theta }$$, $$dz = i R e^{i\theta}$$

$$\left| \ds\int_{\Gamma_R } e^{i\alpha z} f(z)\, dz \right|$$ $$= R \left| \ds\int_{0}^{\pi} e^{i\alpha Re^{i \theta } } f\left(Re^{i \theta }\right)\, d\theta \right|$$ $$\le R \ds \int_0^{\pi} \left| e^{i\alpha Re^{i \theta } } f\left(Re^{i \theta }\right)\, d\theta \right|$$

$$\qquad \qquad \qquad = R \ds \int_0^{\pi} \left| e^{i\alpha \left(R \cos \theta + i R \sin \theta \right) } f\left(Re^{i \theta }\right) \right| d\theta$$

$$\qquad \qquad \qquad = R \ds \int_0^{\pi} e^{-\alpha \,R \sin \theta} \left| \, f\left(Re^{i \theta }\right) \right| d\theta$$

$$\qquad \qquad \qquad \overset{(*)}{\le} \ds \frac{M}{R^{\beta -1}} \int_0^{\pi} e^{-\alpha \,R \sin \theta} d\theta$$

$$\qquad \qquad \qquad = \ds \frac{2M}{R^{\beta -1}} \int_0^{\pi/2} e^{-\alpha \,R \sin \theta} d\theta$$ $$\quad (**)$$

Jordan's Lemma

Proof (cont): Note that $$\sin \theta \ge \dfrac{2\theta}{\pi}$$ for $$0\le \theta \le \pi/2$$.

So we have

$$(**)$$ $$\le \ds \frac{2M}{R^{\beta-1}} \int_0^{\pi/2} e^{-\frac{2\alpha R}{\pi} \theta} d\theta$$ $$\qquad\qquad$$

$$= \ds \frac{2M}{R^{\beta-1}} \frac{\pi}{2 \alpha R} \left( 1-e^{-\alpha R} \right)$$ $$\qquad$$

$$\quad \quad \;\;= \ds \frac{M \pi }{\alpha R^{\beta}} \left( 1-e^{-\alpha R} \right)$$ $$\ra 0$$ as $$R\ra \infty$$. $$\;\square$$

Integrals $${\small \int_0^{2\pi}f\left(\sin t, \cos t\right)dt}$$

Try the substitution

$$z= \cos t + i \sin t$$ $$\;\Rightarrow \; z^{-1} = \cos t + i \sin t$$.

Motivation: $$z=e^{it}, 0\le t \le 2 \pi$$. $$\Rightarrow$$

$$\cos t = \ds \frac{1}{2} \left(z+z^{-1}\right)$$

$$\sin t = \ds \frac{1}{2i} \left(z-z^{-1}\right)$$

$$\; \Rightarrow$$ $$\;dz = (-\sin t + i \cos t) dt = i z dt$$.

Integrals $${\small \int_0^{2\pi}f\left(\sin t, \cos t\right)dt}$$

Example: Find $$I = \ds \int_0^{2\pi} \frac{dt}{2+ \cos t}$$.

Put $$z = e^{it}$$, so by above, $$dt = dz/iz$$ and $$\cos t = \frac12 \left(z+z^{-1}\right)$$.

So for $$C$$ given by $$\left\{ z=e^{it} , 0\le t \le 2 \pi \right\}$$, there holds

$$I = \ds \int_C\frac{dz\,/\,iz}{2+\frac{1}{2} \left(z+z^{-1}\right)}$$ $$=-i \ds\int_C \frac{dz}{2z + \frac12 \left(z^2+1\right)}$$

$$\;\; =-2i \ds\int_C \frac{dz}{2z^2+4z+1}$$ $$\quad (\text{🌏})$$.

Integrals $${\small \int_0^{2\pi}f\left(\sin t, \cos t\right)dt}$$

Example (cont): To evaluate the integral in $$(\text{🌏})$$ note that the integrand is analytic, except at the zeros of the denominator, i. e. $$-2\pm \sqrt{3}$$.

Only $$-2 + \sqrt{3}$$ is inside the countour $$C$$, so by the Residue Theorem and $$(\text{🌏})$$ we obtain $I= (-2i)(2\pi i)\underset{{\tiny z=-2+\sqrt{3}}}{\res} \frac{1}{z^2+4z+1} \quad (\text{😀})$

Integrals $${\small \int_0^{2\pi}f\left(\sin t, \cos t\right)dt}$$

Example (cont): To calculate the residue, note $\frac{1}{z^2+4z+1} = \frac{\phi(z)}{z - (-2+\sqrt{3})}$ where $\phi(z) = \frac{1}{z-\left(-2-\sqrt{3}\right)}$ is analytic and non-zero near $$z=-2+\sqrt{3}.$$

Integrals $${\small \int_0^{2\pi}f\left(\sin t, \cos t\right)dt}$$

Example (cont): So $$\dfrac{1}{z^2+4z+1}$$ has a pole of order $$1$$ at $$z=-2+\sqrt{3}$$, and Theorem 2 (Lec 33) implies $\underset{{\tiny z=-2+\sqrt{3}}}{\res} \frac{1}{z^2+4z+1} = \phi\left(-2 + \sqrt{3}\right) = \frac{1}{2\sqrt{3}}.$

Hence,$$\;(\text{😀})$$ $$\Rightarrow I = (-2i)(2\pi i)\ds \frac{1}{2\sqrt{3}}$$ $$= \ds \frac{2\pi}{\sqrt{3}}$$. $$\qquad \square$$

Final remarks on Laurent series

Laurent series for $$\dfrac{1}{\sinh z}$$ at $$0$$. Note that $$\sinh z = 0$$ for $$z= n\pi i$$ only with $$n\in \Z$$.

So $$\dfrac{1}{\sinh z}$$ will have a Laurent series on $$0\lt |z|\lt \pi$$. Note

$$\;\;\dfrac{1}{\sinh z} = \dfrac{1}{z+ \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots}$$ $$= \dfrac{1}{z}\left( \dfrac{1}{1+ \frac{z^2}{3!} + \frac{z^4}{5!} + \cdots}\right)$$ $$\;\;(★)$$

using the formula for the sum of geometric series, for $$|z|$$ small,...

Final remarks on Laurent series

... so $$\left| \ds \frac{z^2}{3!} + \frac{z^4}{5!} + \cdots \right|\lt 1$$, there holds

$$(★)$$ $$=\ds \frac{1}{z}\frac{1}{1 + \left( \frac{z^2}{3!} + \frac{z^4}{5!} + \cdots \right) }$$

$$\quad \;\;= \ds \frac{1}{z} \Bigg[ 1 - \left( \frac{z^2}{3!} + \frac{z^4}{5!} + \cdots \right) + \left( \frac{z^2}{3!} + \frac{z^4}{5!} + \cdots \right)^2 - \cdots \Bigg]$$

(via formula for geometric series, as long as $$|z|$$ is sufficiently small)

Final remarks on Laurent series

Thus $$(★)$$ $$= \ds \frac{1}{z} \Bigg[ 1 - \frac{z^2}{3!} + z^4\left( -\frac{1}{5!} + \frac{1}{3!3!} \right) - \cdots \Bigg]$$

$$\qquad \quad \;= \ds \frac{1}{z} -\frac{z}{6} + \frac{7z^3}{360} - \cdots$$

In particular, pole of order $$1$$ at $$0$$, with $\underset{{ z=0}}{\res} \frac{1}{\sinh z} = 1. \qquad \square$

Note that it can also be done via division of power series.