MATH529 L24: Residues

Overview

• Residue theorem

• Evaluation of real integrals

Integration of Laurent series

• Consider $f:D\to ℂ$, $f$ analytic in $D:r<|z-z_0|<R.$, with Laurent series

  ${\displaystyle f\left(z\right)=\sum _{k=-\infty }^{\infty }{a}_k{(z-z_0)}^k}$

• Integrate term-by-term on a circle $C$ around $z_0$, $|z-z_0|<\rho$, $r<\rho <R$

  ${\displaystyle \underset{C}{\oint }{a}_k{(z-z_0)}^{k}dz}$

• Recall integrals of monomials on unit circle $z=e^{i\theta }$

  ${\displaystyle \underset{\left|z\right|=1}{\oint }=i\underset{0}{\overset{2\pi }{\int }}{e}^{-2i\theta }{e}^{i\theta }d\theta =i\underset{0}{\overset{2\pi }{\int }}\cos \theta d\theta +\underset{0}{\overset{2\pi }{\int }}\sin \theta d\theta =0,}$ ${\displaystyle \underset{\left|z\right|=1}{\oint }=i\underset{0}{\overset{2\pi }{\int }}{e}^{-i\theta }{e}^{i\theta }d\theta =i\underset{0}{\overset{2\pi }{\int }}1\cdot d\theta =2\pi i.}$

Residues

• Conclude that the only contribution to

  ${\displaystyle \underset{C}{\oint }f\left(z\right)dz=\sum _{k=-\infty }^{\infty }{a}_k\underset{C}{\oint }{(z-z_0)}^{k}dz,}$

  comes from the $k=-1$ term

  ${\displaystyle \underset{C}{\oint }f\left(z\right)dz=2\pi i{a}_{-1}}$

• Define the residue of $f$ at $z=z_0$ a pole of order $n$ by

  ${\displaystyle \mathrm{Res}(f\left(z\right),z_0)=\frac{1}{(n-1)!}{lim}_{z\to z_0}\frac{d^{n-1}}{d{z}^{n-1}}[{(z-z_0)}^{n}f(z)]=a_{-1}}$

Evaluation of real integrals $I={\int }_0^{2\pi }f(\cos \theta ,\sin \theta )d\theta$

• Let $C=\{z,\left|z\right|=1\}$, i.e., $z=e^{i\theta },dz=i{e}^{i\theta }d\theta \Rightarrow d\theta ={(iz)}^{-1}dz\Rightarrow$

  ${\displaystyle I=\underset{C}{\oint }f(\frac{1}{2}(z+z^{-1}),\frac{1}{2i}(z-z^{-1}))\frac{dz}{iz}}$

• Example

  ${\displaystyle I=\underset{0}{\overset{2\pi }{\int }}\frac{d\theta }{{(2+\cos \theta )}^2}=\underset{C}{\oint }\frac{1}{{(2+\frac{1}{2}(z+z^{-1}))}^2}\frac{dz}{iz}\Rightarrow }$ ${\displaystyle I=\frac{4}{i}\underset{C}{\oint }\frac{zdz}{{(z^2+4z+1)}^2}=8\pi \mathrm{Res}\{f\left(z\right),\sqrt{3}-2\}=\frac{4\pi }{3\sqrt{3}}}$ ${\displaystyle \mathrm{Res}\{f\left(z\right),\sqrt{3}-2\}={lim}_{z\to \sqrt{3}-2}\frac{d}{dz}\left[{(z-\sqrt{3}+2)}^2\frac{z}{{(z^2+4z+1)}^2}\right]=\frac{1}{6\sqrt{3}}}$

Evaluation of $I={\int }_{-\infty }^{\infty }f\left(x\right)\cos (ax)dx$, $J={\int }_{-\infty }^{\infty }f\left(x\right)\sin (ax)dx$

• Integrals are scalar products from Fourier series and evaluated as

  ${\displaystyle I={lim}_{R\to \infty }[\underset{C}{\oint }f\left(z\right)\cos (az)dz-\underset{S}{\int }f\left(z\right)\cos (az)dz]}$ ${\displaystyle J={lim}_{R\to \infty }[\underset{C}{\oint }f\left(z\right)\sin (az)dz-\underset{S}{\int }f\left(z\right)\sin (az)dz]}$

  on contour formed of line segment along real axis from $-R$ to $R$ and upper semicircle $S$, $z=R{e}^{i\theta },0⩽\theta ⩽\pi$

• Note that for $z\in ℂ$, it is possible for $\cos z$, $\sin z>1$

• When integrals over $S$ go to zero as $R\to \infty$, with $z_k$ singularities in $\mathrm{Im}z>0$

  ${\displaystyle I=2\pi i\sum _{k=1}^n\mathrm{Res}\{f\left(z\right)\cos (az),z_k\},J=2\pi i\sum _{k=1}^n\mathrm{Res}\{f\left(z\right)\sin (az),z_k\}}$

Bounding theorems, singularities on real axis

• Behavior of integrals over upper half-plane semicircle depends on integrand

Theorem. (19.6.1) $f\left(z\right)=P\left(z\right)/Q\left(z\right)$ a rational function with degrees $n,m$ of $P,Q$. When $m⩾n+1$, ${lim}_{R\to \infty }{\int }_{S}P\left(z\right)/Q\left(z\right)dz=0$

Theorem. (19.6.2) $f\left(z\right)=P\left(z\right)/Q\left(z\right)$ a rational function with degrees $n,m$ of $P,Q$. When $m⩾n+1$, ${lim}_{R\to \infty }{\int }_{S}P\left(z\right)/Q\left(z\right)e^{iaz}dz=0$

• A singularity $z_k\in ℝ$ must be avoided through a semicircle $z-z_k=r{e}^{i\theta }$, $\theta$ from ${\theta }_1=\pi$ to ${\theta }_2=0$

  ${\displaystyle {lim}_{r\to 0}ir\underset{\pi }{\overset{0}{\int }}f(z_k+r{e}^{i\theta })e^{i\theta }d\theta =\pi i\mathrm{Res}\{f\left(z\right),z_k\}}$