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Residue Theorem

Motivation

Recall: 𝑓 has an isolated singularity at 𝑧₀ if 𝑓 is analytic in {0<|𝑧−𝑧₀|<𝑟} for some 𝑟>0. In that case, 𝑓 has a Laurent series expansion 𝑓(𝑧)=∞∑𝑘=−∞𝑎𝑘(𝑧−𝑧0)𝑘, 0<|𝑧−𝑧0|<𝑟. Observe: If 0<𝜌<𝑟 then  ∫|𝑧−𝑧0|=𝜌𝑓(𝑧)𝑑𝑧=∞∑𝑘=−∞𝑎𝑘 ∫|𝑧−𝑧0|=𝜌(𝑧−𝑧0)𝑘𝑑𝑧

What is  ∫|𝑧−𝑧₀|=𝜌(𝑧−𝑧₀)^𝑘𝑑𝑧? For 𝑘≠−1, the function ℎ(𝑧)=(𝑧−𝑧₀)^𝑘 has a primitive, namely 𝐻(𝑧)=1𝑘+1(𝑧−𝑧₀)^𝑘+1. Therefore,  ∫|𝑧−𝑧₀|=𝜌(𝑧−𝑧₀)^𝑘𝑑𝑧=0 for 𝑘≠−1 For 𝑘=−1, the integral is  ∫|𝑧−𝑧₀|=𝜌(𝑧−𝑧₀)^𝑘𝑑𝑧. We can use the Cauchy Integral Formula (or compute this directly) and find  ∫𝛾𝑓(𝑧)𝑑𝑧=𝑏∫𝑎𝑓(𝛾(𝑡))𝛾′(𝑡)𝑑𝑡  ∫|𝑧−𝑧0|=𝜌1𝑧−𝑧0𝑑𝑧=2𝜋∫01𝑧0+𝜌ℯ𝑖𝑡−𝑧0⋅𝜌ℯ𝑖𝑡𝑑𝑡  =2𝜋∫0𝑖𝑑𝑡=2𝜋𝑖  ∫|𝑧−𝑧₀|=𝜌(𝑧−𝑧₀)^𝑘𝑑𝑧=2𝜋𝑖 for 𝑘=−1 Hence  ∫|𝑧−𝑧0|=𝜌𝑓(𝑧)𝑑𝑧=∞∑𝑘=−∞𝑎𝑘 ∫|𝑧−𝑧0|=𝜌(𝑧−𝑧0)𝑘𝑑𝑧=2𝜋𝑖𝑎−1. Therefore, 𝑎₋₁ gets special attention!

The Residue

DefinitionIf 𝑓 has an isolated singularity at 𝑧0 with Laurent series expansion 𝑓(𝑧)=∞∑𝑘=−∞𝑎𝑘(𝑧−𝑧0)𝑘, 0<|𝑧−𝑧0|<𝑟, then the residue of 𝑓 at 𝑧0 is Res(𝑓,𝑧0)=𝑎−1.

Examples of finding a residue: to find Laurent series centered at the singular point and read off the term 𝑎₋₁. 𝑓(𝑧)=1(𝑧−1)(𝑧−2)=−1𝑧−1+∞∑𝑘=0(−1)(𝑧−1)^𝑘 in 0<|𝑧−1|<1. Therefore, Res(𝑓,1)=−1 𝑓(𝑧)=1(𝑧−1)(𝑧−2)=1𝑧−2−∞∑𝑛=0(−1)^𝑛(𝑧−2)^𝑛 in 0<|𝑧−2|<1. Therefore, Res(𝑓,2)=1

More Residue Examples

More examples: 𝑓(𝑧)=sin 𝑧𝑧⁴=1𝑧³−13!1𝑧+15!𝑧−17!𝑧³+−⋯ in 0<|𝑧|<∞. Therefore, Res(𝑓,0)=−13!=−16. 𝑓(𝑧)=cos1𝑧=1−12!1𝑧²+14!1𝑧⁴−16!1𝑧⁶+−⋯. Therefore Res(𝑓,0)=0 𝑓(𝑧)=sin1𝑧=1𝑧−13!1𝑧³+15!1𝑧⁵−+⋯. Therefore Res(𝑓,0)=1 𝑓(𝑧)=cos 𝑧−1𝑧²=−12!+𝑧²4!−+⋯. Therefore Res(𝑓,0)=0 𝑓(𝑧)=1(𝑧²+1=1(𝑧−𝑖)(𝑧+𝑖)=12!(𝑧+𝑖)−(𝑧−𝑖)(𝑧−𝑖)(𝑧+𝑖)  =12!1(𝑧−𝑖)−1(𝑧+𝑖)=12𝑖⋅1(𝑧−𝑖)+(analytic function near 𝑖). Therefore, Res(𝑓,𝑖)=12𝑖=−12𝑖. Similarly, 𝑓(𝑧)=1(𝑧²+1=−12𝑖⋅1(𝑧+𝑖)+(analytic function near −𝑖). Therefore, Res(𝑓,𝑖)=−12𝑖=12𝑖.

The Residue Theorem

Theorem (Residue Theorem)Let 𝐷 be a simply connected domain, and let 𝑓 be analytic in 𝐷, except for isolated singularities. Let 𝐶 be a simple closed curve in 𝐷 (oriented counterclockwise), and let 𝑧1,⋯,𝑧𝑛 be those isolated singularities of 𝑓 that lie inside of 𝐶. Then  ∫𝐶𝑓(𝑧)𝑑𝑧=2𝜋𝑖𝑛∑𝑘=1Res(𝑓,𝑧𝑘).

Example

𝑓(𝑧)=1(𝑧²+1 is analytic in 𝐷=𝐶, except for isolated singularities at 𝑧=±𝑖.  ∫𝐶₁𝑓(𝑧)𝑑𝑧=2𝜋𝑖Res(𝑓,𝑖)=2𝜋𝑖(−12𝑖)=𝜋  ∫𝐶₂𝑓(𝑧)𝑑𝑧=2𝜋𝑖Res(𝑓,−𝑖)=2𝜋𝑖(12𝑖)=−𝜋  ∫𝐶₃𝑓(𝑧)𝑑𝑧=2𝜋𝑖(Res(𝑓,𝑖)+Res(𝑓,−𝑖))=2𝜋𝑖(−12𝑖+12𝑖)=−𝜋  ∫𝐶₄𝑓(𝑧)𝑑𝑧=0 In order to be able to fully take advantage of this powerful theorem, strategies and techniques that can help calculating residues are needed

Recall the Residue Theorem

Recall: 𝑓 has an isolated singularity at 𝑧₀ if 𝑓 is analytic in the punctured disk {0<|𝑧−𝑧₀|<𝑟}. In that case, 𝑓 has a Laurent series representation 𝑓(𝑧)=∞∑𝑘=−∞𝑎𝑘(𝑧−𝑧0)𝑘 in this punctured disk. The representation is unique. The residue of 𝑓 at 𝑧₀ is Res(𝑓,𝑧₀)=𝑎₋₁, the coefficient of term 1𝑧−𝑧₀.

Residues at Removable Singularities

𝑓(𝑧)=∞∑𝑘=−∞𝑎𝑘(𝑧−𝑧0)𝑘, 0<|𝑧−𝑧0|<𝑟.

Recall: 𝑧₀ is a removable singularity if 𝑎_𝑘=0 for all 𝑘<0. In particular: 𝑎₋₁=0 in that case, so that Res(𝑓,𝑧₀)=0. Example: 𝑓(𝑧)=sin 𝑧𝑧=1𝑧∞∑𝑛=0(−1)𝑛(2𝑛+1)!𝑧2𝑛+1  =1𝑧𝑧−13!𝑧3+15!𝑧5−17!𝑧7+−⋯  =1−13!𝑧2+15!𝑧4−17!𝑧6+−⋯ Thus Res(𝑓,0)=0

Residues at Simple Poles

Example: Res(𝑓,𝑧0)= Lim𝑧→𝑧0(𝑧−𝑧0)𝑓(𝑧).

Example: 𝑓(𝑧)=1𝑧²+1 has a simple pole at 𝑧₀=𝑖 (and another one at −𝑖). Res1𝑧2+1,𝑖= Lim𝑧→𝑖(𝑧−𝑖)1𝑧2+1  = Lim𝑧→𝑖(𝑧−𝑖)1(𝑧−𝑖)(𝑧+𝑖)  = Lim𝑧→𝑖(𝑧−𝑖)1(𝑧+𝑖)=12𝑖=−𝑖2

Residues at Double Poles

𝑓(𝑧)=∞∑𝑘=−∞𝑎𝑘(𝑧−𝑧0)𝑘, 0<|𝑧−𝑧0|<𝑟.

Recall: 𝑧₀ is a double pole if 𝑎₋₂≠0 and 𝑎_𝑘=0 for all 𝑘≤−3. So 𝑓(𝑧)=𝑎−2(𝑧−𝑧0)2+𝑎−𝑧−𝑧0+𝑎0+𝑎1(𝑧−𝑧0)+𝑎2(𝑧−𝑧0)2+⋯ How do we isolate 𝑎₋? Idea: (𝑧−𝑧0)2𝑓(𝑧)=𝑎−2+𝑎−(𝑧−𝑧0)+𝑎0(𝑧−𝑧0)2+⋯, so that 𝑑𝑑𝑧(𝑧−𝑧0)2𝑓(𝑧)=𝑎−+2𝑎0(𝑧−𝑧0)+⋯, Hence Res(𝑓,𝑧0)=𝑎−= Lim𝑧→𝑧0𝑑𝑑𝑧(𝑧−𝑧0)2𝑓(𝑧)

Example

𝑓(𝑧)=1(𝑧−1)²(𝑧−3) has a double pole at 𝑧₀=1 (and a simple one at 3). Res1(𝑧−1)2(𝑧−3),1= Lim𝑧→1𝑑𝑑𝑧(𝑧−1)21(𝑧−1)2(𝑧−3)  = Lim𝑧→1𝑑𝑑𝑧1(𝑧−3)  = Lim𝑧→1−1(𝑧−3)2=−14.

Residues at Poles of Order 𝑛

𝑓(𝑧)=∞∑𝑘=−∞𝑎𝑘(𝑧−𝑧0)𝑘, 0<|𝑧−𝑧0|<𝑟.

Recall: 𝑧₀ is a pole of order 𝑛 if 𝑎_−𝑛≠0 and 𝑎_𝑘=0 for all 𝑘≤−(𝑛+1). 𝑓(𝑧)=𝑎−𝑛(𝑧−𝑧0)𝑛+⋯+𝑎−𝑧−𝑧0+𝑎0+𝑎1(𝑧−𝑧0)+𝑎2(𝑧−𝑧0)2+⋯ Then Res(𝑓,𝑧0)=𝑎−=1(𝑛−<)!Lim𝑧→𝑧0𝑑𝑛−𝑑𝑧𝑛−(𝑧−𝑧0)𝑛𝑓(𝑧)

More On Residues

RemarkIf 𝑓(𝑧)=𝑔(𝑧)ℎ(𝑧), where 𝑔 and ℎ are analytic near 𝑧0, and ℎ has a simple zero at 𝑧0, then Res(𝑓(𝑧),𝑧0)=𝑔(𝑧0)ℎ′(𝑧0).

Example: 𝑓(𝑧)=1(𝑧−1)²(𝑧−3), choose 𝑔(𝑧)=1(𝑧−1)² and ℎ(𝑧)=(𝑧−3). Then 𝑔 and ℎ are analytic near 𝑧₀=3, and ℎ has a simple zero at 𝑧₀=3. Thus Res(𝑓,𝑧0)=𝑔(3)ℎ′(3)=1(3−1)21
 =14

Evaluating Integrals via the Residue Theorem

Recall: The Residue Theorem  ∫𝐶𝑓(𝑧)𝑑𝑧=2𝜋𝑖𝑛∑𝑘=1Res(𝑓,𝑧𝑘)

Examples:  ∫|𝑧|=1ℯ^3𝑧𝑑𝑧=2𝜋𝑖Res(𝑓,0), where 𝑓(𝑧)=ℯ^3𝑧=∞∑𝑘=11𝑘!3𝑧^𝑘. Thus Res(𝑓,0)=3, so that  ∫|𝑧|=1ℯ3𝑧𝑑𝑧=6𝜋𝑖  ∫|𝑧|=2tan 𝑧𝑑𝑧=2𝜋𝑖Res(𝑓,𝜋2)+Res(𝑓,−𝜋2), where 𝑓(𝑧)=tan 𝑧=sin 𝑧cos 𝑧. To find Res(𝑓,𝜋2): Note that 𝑓(𝑧)=𝑔(𝑧)ℎ(𝑧), where 𝑔(𝑧)=sin(𝑧) and ℎ(𝑧)=cos(𝑧) are analytic near 𝜋2 and ℎ(𝜋2)=0. Thus Res(𝑓,𝜋2)=𝑔(𝜋2)ℎ′(𝜋2)=sin(𝜋2)−sin(𝜋2)=−1 Similarly, Res(𝑓,−𝜋2)=𝑔(−𝜋2)ℎ′(−𝜋2)=sin(−𝜋2)−sin(−𝜋2)=−1−(−1)=−1 Thus  ∫|𝑧|=2tan 𝑧𝑑𝑧=2𝜋𝑖(−1−1)=−4𝜋𝑖.  ∫𝐶₁1(𝑧−1)²(𝑧−3)𝑑𝑧=2𝜋𝑖Res(𝑓,1). Res(𝑓,1)= Lim𝑧→1𝑑𝑑𝑧(𝑧−1)2(𝑧−1)2(𝑧−3)  = Lim𝑧→1𝑑𝑑𝑧1𝑧−3= Lim𝑧→1−1(𝑧−3)2=−14 Thus  ∫𝐶₁1(𝑧−1)²(𝑧−3)𝑑𝑧=−142𝜋𝑖=−𝜋𝑖2.  ∫𝐶₂1(𝑧−1)²(𝑧−3)𝑑𝑧=2𝜋𝑖(Res(𝑓,1)+Res(𝑓,3)) Res(𝑓,3)= Lim𝑧→3(𝑧−3)(𝑧−1)2(𝑧−3)= Lim𝑧→31(𝑧−1)2=14

More Examples

The Residue Theorem can also be used to evaluate real integrals, for example of the following forms: 2𝜋∫0𝑅(cos 𝑡,sin 𝑡)𝑑𝑡, where 𝑅(𝑥,𝑦) is a rational function of the real variables 𝑥 and 𝑦. ∞∫−∞𝑓(𝑥)𝑑𝑥, where 𝑓 is a rational function of 𝑥. ∞∫−∞𝑓(𝑥)cos(𝛼𝑥)𝑑𝑥, where 𝑓 is a rational function of 𝑥. ∞∫−∞𝑓(𝑥)sin(𝛼𝑥)𝑑𝑥, where 𝑓 is a rational function of 𝑥.

Evaluating an Improper Integral via the Residue Theorem

GoalEvaluate ∞∫0cos 𝑥1+𝑥2𝑑𝑥.

An Improper Integral

Note: ∞∫0⋯𝑑𝑥 means Lim𝑅→∞𝑅∫0⋯𝑑𝑥, so we need to consider 𝑅∫0cos 𝑥1+𝑥²𝑑𝑥 and then let 𝑅→∞. Idea: 𝑅∫0cos 𝑥1+𝑥2𝑑𝑥=12𝑅∫−𝑅cos 𝑥1+𝑥2𝑑𝑥  =12𝑅∫−𝑅cos 𝑥+𝑖sin 𝑥1+𝑥2𝑑𝑥  =12𝑅∫−𝑅ℯ𝑖𝑥1+𝑥2𝑑𝑥. Idea: 12𝑅∫−𝑅ℯ𝑖𝑥1+𝑥2𝑑𝑥= 12 ∫[−𝑅,𝑅]ℯ𝑖𝑧1+𝑧2𝑑𝑧  =12 ∫𝐶𝑅ℯ𝑖𝑧1+𝑧2𝑑𝑧−12 ∫Γ𝑅ℯ𝑖𝑧1+𝑧2𝑑𝑧  =122𝜋𝑖Resℯ𝑖𝑧1+𝑧2,𝑖−12 ∫Γ𝑅ℯ𝑖𝑧1+𝑧2𝑑𝑧 we thus need to find the residue of 𝑓(𝑧)=ℯ^𝑖𝑧1+𝑧² at 𝑧₀=𝑖 and estimate the integral over Γ_𝑅. Finding the residue of 𝑓(𝑧)=ℯ^𝑖𝑧1+𝑧² at 𝑧₀=𝑖 𝑓 has a simple pole at 𝑧=𝑖. Thus Res(𝑓,𝑖)= Lim𝑧→𝑖(𝑧−𝑖)𝑓(𝑧)= Lim𝑧→𝑖(𝑧−𝑖)ℯ^𝑖𝑧1+𝑧²= Lim𝑧→𝑖ℯ^𝑖𝑧𝑧+𝑖=ℯ^-12𝑖. Hence 12 ∫𝐶_𝑅ℯ^𝑖𝑧1+𝑧²𝑑𝑧=122𝜋𝑖12𝑖ℯ=𝜋2ℯ. Estimating 12 ∫Γ_𝑅ℯ^𝑖𝑧1+𝑧²𝑑𝑧 We are only interested in what happens as 𝑅→∞ Want to show 12 ∫Γ_𝑅ℯ^𝑖𝑧1+𝑧²𝑑𝑧→0 as 𝑅→∞ Therefore, it suffices to show that  ∫Γ_𝑅ℯ^𝑖𝑧1+𝑧²𝑑𝑧≤const(𝑅), where the constant, const(𝑅) goes to zero as 𝑅→∞ Recall:  ∫Γ_𝑅𝑓(𝑧)𝑑𝑧≤length(Γ_𝑅)⋅ max𝑧∈Γ_𝑅|𝑓(𝑧)|. Thus:  ∫Γ_𝑅ℯ^𝑖𝑧1+𝑧²𝑑𝑧≤length(Γ_𝑅)⋅ max𝑧∈Γ_𝑅ℯ^𝑖𝑧1+𝑧². ℯ^𝑖𝑧1+𝑧²=ℯ^Re(𝑖𝑧)|1+𝑧²|=ℯ^−𝑦|1+𝑧²|≤ℯ^−𝑦𝑅²−1≤1𝑅²−1 for 𝑧∈Γ_𝑅, since |𝑧|=𝑅 and 𝑦≥0 on Γ_𝑅. So  ∫Γ_𝑅ℯ^𝑖𝑧1+𝑧²𝑑𝑧≤𝜋𝑅1𝑅²−1→0 as 𝑅→∞ Thus  ∫Γ_𝑅ℯ^𝑖𝑧1+𝑧²𝑑𝑧→0 as 𝑅→∞ To find: ∞∫0cos 𝑥1+𝑥²𝑑𝑥. ∞∫0cos 𝑥1+𝑥²𝑑𝑥=Lim𝑅→∞𝑅∫0cos 𝑥1+𝑥²𝑑𝑥 𝑅∫0cos 𝑥1+𝑥²𝑑𝑥=12 ∫𝐶_𝑅ℯ^𝑖𝑧1+𝑧²𝑑𝑧−12 ∫Γ_𝑅ℯ^𝑖𝑧1+𝑧²𝑑𝑧 12 ∫𝐶_𝑅ℯ^𝑖𝑧1+𝑧²𝑑𝑧=12⋅2𝜋𝑖Resℯ^𝑖𝑧1+𝑧²,𝑖=𝜋2ℯ  ∫Γ_𝑅ℯ^𝑖𝑧1+𝑧²𝑑𝑧→0 as 𝑅→∞ Hence ∞∫0cos 𝑥1+𝑥²𝑑𝑥=𝜋2ℯ.