§4.5 The Residue Theorem

4.5 The Residue Theorem

After Riemann and Weierstrass refined the understanding of analytic functions and the formal characterization of Jordan curves, the Cauchy Residue Theorem was consequently formalized. Cauchy had the informal notion of a residue, which we will now formally introduce.

Definition 4.5.1 (Residue).

For some $𝑟 \in ℝ_{> 0}$ , $𝑎 \in 𝑈$ , suppose $𝑓 : 𝐷^{*} (𝑎, 𝑟) \to ℂ$ is holomorphic. Then the residue of $𝑓$ at $𝑎$ , denoted by ${Res}_{𝑧 = 𝑎} 𝑓 (𝑧)$ or $Res (𝑓, 𝑎)$ , is equal to

\underset{𝑧 = 𝑎}{Res} 𝑓 (𝑧) = \frac{1}{2 π i} \oint_{𝜕 𝐷 (𝑎, 𝜌)} 𝑓 (𝑧) d 𝑧, (4.5 . 1)

where $0 < 𝜌 < 𝑟$ is arbitrary. Since $𝑓$ has a Laurent expansion at $𝑎$ , being

\sum_{𝑛 = - \infty}^{\infty} 𝑐_{𝑛} {(𝑧 - 𝑎)}^{𝑛}, 𝑐_{𝑛} = \frac{1}{2 π i} \oint_{𝜕 𝐷 (𝑎, 𝜌)} \frac{𝑓 (𝑧) d 𝑧}{{(𝑧 - 𝑎)}^{𝑛 + 1}},

we get that the residue of

𝑓

𝑎

is equal to the first term

𝑐_{- 1}

of the principal part of its Laurent expansion.

It then follows that the residue at a removable singularity is 0. As a direct consequence of (4.5.1) $(4.5.1) (4.5.1)$ , we can derive explicit formulas for the calculation of residues at poles. If $𝑈 \subseteq ℂ$ is open, $𝑎 \in 𝑈$ is an isolated singularity (a pole of order $𝑚 \neq \infty$ ) of $𝑓 : 𝑈 ∖ {𝑎} \to ℂ$ that is holomorphic, then locally:

𝑓 (𝑧) = 𝑐_{- 𝑚} {(𝑧 - 𝑎)}^{- 𝑚} + 𝑐_{1 - 𝑚} {(𝑧 - 𝑎)}^{1 - 𝑚} + \dots + 𝑐_{- 1} {(𝑧 - 𝑎)}^{- 1} + \dots .

Multiplying by ${(𝑧 - 𝑎)}^{𝑚}$ , we obtain that

{(𝑧 - 𝑎)}^{𝑚} 𝑓 (𝑧) = 𝑐_{- 𝑚} + 𝑐_{1 - 𝑚} (𝑧 - 𝑎) + \dots + 𝑐_{- 1} {(𝑧 - 𝑎)}^{𝑚 - 1} + \dots .

By the definition of a Taylor series, we find that

𝑐_{- 1} = \underset{𝑧 = 𝑎}{Res} 𝑓 (𝑧) = \frac{1}{(𝑚 - 1)!} lim_{𝑧 \to 𝑎} \frac{d^{𝑚 - 1}}{d 𝑧^{𝑚 - 1}} [{(𝑧 - 𝑎)}^{𝑚} 𝑓 (𝑧)] . (4.5 . 2)

Let $𝑧 = \infty$ be an isolated singularity of $𝑓 (𝑧)$ , which is holomorphic in $ℂ ∖ \overline{𝐷 (0, 𝑅)}$ , for sufficiently large finite $𝑅$ . Then for finite $𝜌 > 𝑅$ , the residue at $𝑧 = \infty$ is defined as (notice the orientation)

\underset{𝑧 = \infty}{Res} 𝑓 (𝑧) = \frac{1}{2 π i} ∲_{𝜕 𝐷 (0, 𝜌)} 𝑓 (𝑧) d 𝑧 .

Let $𝜁 = \frac{1}{𝑧}$ . Then we get that

\begin{aligned} \underset{𝑧 = \infty}{Res} 𝑓 (𝑧) & = - \frac{1}{2 π i} ∳_{𝜕 𝐷 (0, 𝜌)} 𝑓 (\frac{1}{𝜁}) d (\frac{1}{𝜁}) \\ = \frac{1}{2 π i} ∳_{𝜕 𝐷 (0, 1 / 𝜌)} \frac{𝑓 (1 / 𝜁)}{𝜁^{2}} d 𝜁 = - \underset{𝜁 = 0}{Res} \frac{𝑓 (1 / 𝜁)}{𝜁^{2}} . \end{aligned}

In this definition, if

𝑓 (𝑧) = \sum_{𝑛 = - \infty}^{\infty} 𝑐_{𝑛} 𝑧^{𝑛} ⟺ \frac{𝑓 (1 / 𝜁)}{𝜁^{2}} = \sum_{𝑛 = - \infty}^{\infty} 𝑐_{𝑛} 𝜁^{- 𝑛 - 2},

the residue at $𝑧 = \infty$ is equal to $- 𝑐_{- 1}$ . We will later explain the reasoning behind this definition.

Theorem 4.5.2 (Residue Theorem).

Let $𝑈 \subset ℂ$ be an open set with a simple closed boundary curve $𝜕 𝑈$ . Suppose ${𝑧_{𝑛}} \subset 𝑈$ is a finite set and $𝑓 (𝑧)$ is holomorphic on $𝑈 ∖ {𝑧_{𝑛}}$ and continuous on $\overline{𝑈} ∖ {𝑧_{𝑛}}$ . Then,

\oint_{𝜕 𝑈} 𝑓 (𝑧) d 𝑧 = 2 π i \sum_{𝑘 = 1}^{𝑛} \underset{𝑧 = 𝑧_{𝑘}}{Res} 𝑓 (𝑧)

Proof.

Since $𝑈$ is open, there exists a small disk centered at each isolated singularity $𝑧_{𝑘}$ of radii $𝛿_{𝑘}$ . By the Cauchy–Goursat Theorem (Theorem 3.1.9 $Theorem 3.1.9 Theorem 3.1.9$ ), we get that

\int_{⋃_{𝑘 = 1}^{𝑛} {𝐷 (𝑧_{𝑘}, 𝛿_{𝑘})}^{-} \cup 𝜕 𝑈^{+}} 𝑓 (𝑧) d 𝑧 = 0 .

From rearrangement,

\oint_{𝜕 𝑈} 𝑓 (𝑧) d 𝑧 = \sum_{𝑘 = 1}^{𝑛} \oint_{𝜕 𝐷 (𝑧_{𝑘}, 𝛿_{𝑘})} 𝑓 (𝑧) d 𝑧

, and the conclusion follows. ⁠

□

This result itself is fairly trivial. Now we will explain the significance of the residue at infinity.

Theorem 4.5.3 (Global Residue Theorem).

If ${𝑧_{1}, \dots 𝑧_{𝑛}, \infty}$ is discrete and finite, and $𝑓 : \hat{ℂ} ∖ {𝑧_{1}, \dots 𝑧_{𝑛}, \infty} \to ℂ$ is holomorphic, and these points are the isolated singularities of $𝑓$ , then the sum of the residues at each of these isolated singularities is zero, or

\sum_{𝑘 = 1}^{𝑛} \underset{𝑧 = 𝑧_{𝑘}}{Res} 𝑓 (𝑧) + \underset{𝑧 = \infty}{Res} 𝑓 (𝑧) = 0 .

Proof.

Let $𝑅 > {max}_{𝑗 \in ℕ_{\leq 𝑛}} | 𝑧_{𝑛} |$ be arbitrary. By the Residue Theorem (Theorem 4.5.2 $Theorem 4.5.2 Theorem 4.5.2$ ),

- \underset{𝑧 = \infty}{Res} 𝑓 (𝑧) = \frac{1}{2 π i} \oint_{𝜕 𝐷 (0, 𝑅)} 𝑓 (𝑧) d 𝑧 = \sum_{𝑘 = 1}^{𝑛} \underset{𝑧 = 𝑧_{𝑘}}{Res} 𝑓 (𝑧)

as desired. This is merely a restatement of Theorem 4.5.2

Theorem 4.5.2 Theorem 4.5.2

. ⁠

□

There is not a directly trivial reason for the definition of the residue at $\infty$ , except for the fact that it seemingly “unifies” the Riemann sphere.

Figure 4.2: The orientation of a neighborhood that does not enclose

\infty

after projection.

However, if we take a neighborhood of an arbitrary point in $ℂ$ on the Riemann sphere and traverse its boundary clockwise (from the perspective of outside the sphere), its projection onto $ℂ$ will be counterclockwise (Figure 4.2 $Figure 4.2 Figure 4.2$ ). However, the boundary of a neighborhood of $\infty$ in $𝑆^{2}$ will have a clockwise projection (hence the difference in orientation). We define its equality with the residue of $- \frac{𝑓 (1 / 𝜁)}{𝜁^{2}}$ at $𝜁 = 0$ , rather than $𝑓 (1 / 𝜁)$ , because we compose the differential form $𝑓 (𝑧) d 𝑧$ with the inversion, as opposed to $𝑓 (𝑧)$ .

For any closed rectifiable curve $𝛾 \subset 𝑈$ (here we are not bound under the assumption of simpleness), the Residue Theorem can be generalized into:

\oint_{𝛾} 𝑓 (𝑧) d 𝑧 = 2 π i \sum_{𝑘} {Ind}_{𝛾} (𝑧_{𝑘}) \underset{𝑧 = 𝑧_{𝑘}}{Res} 𝑓 (𝑧)

where $𝑧_{𝑘}$ are the singularities of $𝑓$ in $𝑈$ and ${Ind}_{𝛾}$ is the winding index.

Residues are extremely important as they allow for simple evaluation of definite (most commonly improper) real-valued integrals. This is because oftentimes, residues at poles are generally easy to calculate and have an integral representation. We can integrate over a contour (a smooth closed curve) that contains the important part of the real interval. Oftentimes this is the most non-trivial step.

Example 4.5.4.

Evaluate the improper integral

𝐼 = \int_{- \infty}^{\infty} \frac{1}{{(𝑥^{2} + 1)}^{𝑛 + 1}} d 𝑥

, where

𝑛 \in ℕ

Figure 4.3: A semicircular contour with orientation marked.

Solution to Example 4.5.4 $Example 4.5.4 Example 4.5.4$ .

Consider $𝛾$ to be a closed semicircle with radius $𝑅 \geq 2$ as in Figure 4.3 $Figure 4.3 Figure 4.3$ . Notice that the function $𝑧 \mapsto \frac{1}{{(𝑧^{2} + 1)}^{𝑛 + 1}}$ has singularities at only $𝑧 = i$ and $𝑧 = - i$ , both of which are poles of order $𝑛 + 1$ . By (4.5.2) $(4.5.2) (4.5.2)$ , the residue at $𝑧 = i$ is

\begin{aligned} \underset{𝑧 = i}{Res} \frac{1}{{(𝑧^{2} + 1)}^{𝑛 + 1}} & = {\frac{1}{𝑛!} \frac{d^{𝑛}}{d 𝑧^{𝑛}} {(𝑧 + i)}^{- 𝑛 - 1} |}_{𝑧 = i} = \frac{1}{𝑛!} \frac{{(- 1)}^{𝑛} \prod_{𝑘 = 1}^{𝑛} (𝑛 + 𝑘)}{{(2 i)}^{2 𝑛 + 1}} \\ = \frac{{(- 1)}^{𝑛} (2 𝑛)!}{{(𝑛!)}^{2} {(2 i)}^{2 𝑛 + 1}} = \frac{(2 𝑛)!}{2^{2 𝑛 + 1} i {(𝑛!)}^{2}} . \end{aligned}

The singularity at $𝑧 = - i$ is not relevant, as it is not enclosed by the contour. By the Residue Theorem (Theorem 4.5.2 $Theorem 4.5.2 Theorem 4.5.2$ ), we have

\begin{aligned} \oint_{𝛾} \frac{1}{{(𝑧^{2} + 1)}^{𝑛 + 1}} d 𝑧 & = \int_{- 𝑅}^{𝑅} \frac{1}{{(𝑥^{2} + 1)}^{𝑛 + 1}} d 𝑥 + \int_{0}^{π} \frac{𝑅 i}{{(𝑅^{2} e^{2 i 𝜃} + 1)}^{𝑛 + 1}} e^{i 𝜃} d 𝜃 \\ = 2 π i \underset{𝑧 = i}{Res} \frac{1}{{(𝑧^{2} + 1)}^{𝑛 + 1}} = \frac{(2 𝑛)! π}{2^{2 𝑛} {(𝑛!)}^{2}} . \end{aligned}

We will now show that the integral over the semicircle vanishes as $𝑅 \to \infty$ . Under the assumption that $𝑅 \geq 2$ , since

| \frac{𝑅 i e^{i 𝜃}}{{(𝑅^{2} e^{2 i 𝜃} + 1)}^{𝑛 + 1}} | = \frac{𝑅}{{| 𝑅^{2} e^{2 i 𝜃} + 1 |}^{𝑛 + 1}} \leq \frac{𝑅}{| 𝑅^{2} - 1 |} \leq \frac{2}{3},

which is integrable over $[0, π]$ , and we can commute the limit with the integral. Therefore, we have

\begin{aligned} \int_{- \infty}^{\infty} \frac{1}{{(𝑥^{2} + 1)}^{𝑛 + 1}} d 𝑥 & = lim_{𝑅 \to \infty} \oint_{𝛾} \frac{1}{{(𝑧^{2} + 1)}^{𝑛 + 1}} d 𝑧 \\ - \int_{0}^{π} lim_{𝑅 \to \infty} \frac{𝑅 i}{{(𝑅^{2} e^{2 i 𝜃} + 1)}^{𝑛 + 1}} e^{i 𝜃} d 𝜃 = \frac{(2 𝑛)! π}{2^{2 𝑛} {(𝑛!)}^{2}} . □ \end{aligned}

Example 4.5.5 (Dirichlet Integral).

Evaluate the integral

\int_{0}^{\infty} \frac{sin 𝑥}{𝑥} d 𝑥

Figure 4.4: An indented semicircular contour with orientation marked.

Solution to Example 4.5.5 $Example 4.5.5 Example 4.5.5$ .

It is common to use integration with parameters to approach this integral. However, we will now provide a solution via contour integration.

Let $𝑓 (𝑧) = \frac{e^{i 𝑧}}{𝑧}$ . Consider a closed contour $𝛾$ in the form of Figure 4.4 $Figure 4.4 Figure 4.4$ , consisting of a semicircle of radius $𝑅$ in $\overline{ℍ^{+}}$ ( $𝐶_{𝑅}$ ), a line segment from $- 𝑅$ to $- 𝜀$ , a smaller semicircle of radius $𝜀$ in the upper half-plane ( $𝐶_{𝜀}$ ), and a line segment from $𝜀$ to $𝑅$ .

By the Cauchy–Goursat Theorem (Theorem 3.1.9 $Theorem 3.1.9 Theorem 3.1.9$ ), we have that

\oint_{𝛾} 𝑓 (𝑧) d 𝑧 = \int_{𝐶_{𝑅}} 𝑓 (𝑧) d 𝑧 + \int_{- 𝑅}^{- 𝜀} 𝑓 (𝑧) d 𝑧 + \int_{𝐶_{𝜀}} 𝑓 (𝑧) d 𝑧 + \int_{𝜀}^{𝑅} 𝑓 (𝑧) d 𝑧 = 0 .

We will now analyze each integral. The first integral is

\int_{𝐶_{𝑅}} 𝑓 (𝑧) d 𝑧 = \int_{0}^{π} \frac{exp (i 𝑅 e^{i 𝜃})}{𝑅 e^{i 𝜃}} 𝑅 i e^{i 𝜃} d 𝜃 = i \int_{0}^{π} e^{i 𝑅 cos 𝜃} e^{- 𝑅 sin 𝜃} d 𝜃 .

Notice that $\frac{2}{π} 𝜃 \leq sin (𝜃) \leq 𝜃$ over the integration range. We want to observe the behavior as $𝑅 \to \infty$ :

\begin{aligned} | i \int_{0}^{π} e^{i 𝑅 cos 𝜃} e^{- 𝑅 sin 𝜃} d 𝜃 | & \leq \int_{0}^{π} e^{- 𝑅 sin 𝜃} d 𝜃 = 2 \int_{0}^{π / 2} e^{- 𝑅 sin 𝜃} d 𝜃 \\ < 2 \int_{0}^{π / 2} e^{- 𝑅 \frac{2}{π} 𝜃} d 𝜃 = {- \frac{π}{𝑅} e^{- 𝑅 \frac{2}{π} 𝜃} |}_{0}^{π / 2} \\ = \frac{π}{𝑅} (1 - e^{- 𝑅}) \to 0 . \end{aligned}

Let us evaluate the integral on $𝛾_{𝜀}$ as $𝜀 \to 0$ :

\int_{𝐶_{𝜀}} 𝑓 (𝑧) d 𝑧 = i \int_{π}^{0} exp (𝜀 (i cos 𝜃 - sin 𝜃)) d 𝜃 = i \int_{π}^{0} e^{- 𝜀 sin 𝜃} e^{i 𝜀 cos 𝜃} d 𝜃 .

Obviously,

| e^{- 𝜀 sin 𝜃} e^{i 𝜀 cos 𝜃} | \leq 1,

and therefore, the integral and the limit may commute:

lim_{𝜀 \to 0} \int_{𝐶_{𝜀}} 𝑓 (𝑧) d 𝑧 = i \int_{π}^{0} lim_{𝜀 \to 0^{+}} e^{- 𝜀 sin 𝜃} e^{i 𝜀 cos 𝜃} d 𝜃 = i \int_{π}^{0} d 𝜃 = - i π .

Evaluating the integral over the line segments, we have

\begin{aligned} \int_{- 𝑅}^{- 𝜀} 𝑓 (𝑧) d 𝑧 + \int_{𝜀}^{𝑅} 𝑓 (𝑧) d 𝑧 & = \int_{- 𝑅}^{- 𝜀} \frac{e^{i 𝑧}}{𝑧} d 𝑧 + \int_{𝜀}^{𝑅} \frac{e^{i 𝑧}}{𝑧} d 𝑧 \\ = \int_{- 𝑅}^{- 𝜀} \frac{e^{i 𝑧}}{𝑧} d 𝑧 - \int_{- 𝑅}^{- 𝜀} \frac{e^{- i 𝑧}}{𝑧} d 𝑧 \\ \to 2 i \int_{- \infty}^{0} \frac{sin 𝑧}{𝑧} d 𝑧 = 2 i \int_{0}^{\infty} \frac{sin 𝑧}{𝑧} d 𝑧 . \end{aligned}

Hence,

- i π + 2 i \int_{0}^{\infty} \frac{sin (𝑧)}{𝑧} d 𝑧 = 0 ⟺ \int_{0}^{\infty} \frac{sin 𝑧}{𝑧} d 𝑧 = \frac{π}{2} . □

Example 4.5.6 (Fresnel Integrals).

Evaluate the improper integrals

𝐼_{1} = \int_{0}^{\infty} cos (𝑥^{2}) d 𝑥, 𝐼_{2} = \int_{0}^{\infty} sin (𝑥^{2}) d 𝑥 .

Figure 4.5: A wedge contour with orientation marked.

Solution to Example 4.5.6 $Example 4.5.6 Example 4.5.6$ .

Let $𝑓 (𝑧) = e^{i 𝑧^{2}}$ . Choose the wedge contour composed of

\begin{matrix} Γ_{1} = {𝑥 \in ℝ : 0 \leq 𝑥 \leq 𝑅}, Γ_{2} = {𝑟 e^{i π / 4} : 0 \leq 𝑟 \leq 𝑅}, \\ 𝐶_{𝑅} = {𝑅 e^{i 𝜃} : 0 \leq 𝜃 \leq \frac{π}{4}} \end{matrix}

as in Figure 4.5 $Figure 4.5 Figure 4.5$ . By the Cauchy–Goursat Theorem (Theorem 3.1.9 $Theorem 3.1.9 Theorem 3.1.9$ ), we have that

\int_{Γ_{1}} 𝑓 (𝑧) d 𝑧 + \int_{Γ_{2}} 𝑓 (𝑧) d 𝑧 + \int_{𝐶_{𝑅}} 𝑓 (𝑧) d 𝑧 = 0 . (4.5 . 3)

The third integral can be written as

\int_{𝐶_{𝑅}} 𝑓 (𝑧) d 𝑧 = 𝑅 i \int_{0}^{π / 4} exp [i {(𝑅 e^{i 𝜃})}^{2}] e^{i 𝜃} d 𝜃 .

Using the fact that $\frac{4}{π} 𝜃 < sin (2 𝜃)$ on the integration range, it can be bounded as

\begin{aligned} | \int_{𝐶_{𝑅}} 𝑓 (𝑧) d 𝑧 | & \leq 𝑅 \int_{0}^{π / 4} e^{- 𝑅^{2} sin (2 𝜃)} d 𝜃 < 𝑅 \int_{0}^{π / 4} e^{- \frac{4}{π} 𝑅^{2} 𝜃} d 𝜃 \\ = - \frac{π}{4 𝑅} {e^{- \frac{4}{π} 𝑅^{2} 𝜃} |}_{0}^{π / 4} = \frac{π}{4 𝑅} (1 - e^{- 𝑅^{2}}) . \end{aligned}

As $𝑅 \to \infty$ , this integral tends to $0$ . Let $𝑧 = 𝑟 e^{i π / 4}$ on $Γ_{2}$ . Then, we have

lim_{𝑅 \to \infty} \int_{Γ_{2}} 𝑓 (𝑧) d 𝑧 = \int_{\infty}^{0} exp [i {(𝑟 e^{i π / 4})}^{2}] e^{i π / 4} d 𝑟 = e^{i π / 4} \int_{\infty}^{0} exp (- 𝑟^{2}) d 𝑟 .

From (4.5.3) $(4.5.3) (4.5.3)$ , we have that

\int_{0}^{\infty} e^{i 𝑟^{2}} d 𝑟 = e^{i π / 4} \int_{0}^{\infty} e^{- 𝑟^{2}} d 𝑟 .

Since $\int_{0}^{\infty} e^{- 𝑟^{2}} d 𝑟 = \frac{\sqrt{π}}{2}$ , we have

\int_{0}^{\infty} e^{i 𝑟^{2}} d 𝑟 = (\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}) \frac{\sqrt{π}}{2} .

Since $e^{i 𝑟^{2}} = cos (𝑟^{2}) + i sin (𝑟^{2})$ , we have

\begin{matrix} \int_{0}^{\infty} cos (𝑟^{2}) d 𝑟 = ℜ𝔢 [\int_{0}^{\infty} e^{i 𝑟^{2}} d 𝑟] = \frac{\sqrt{2 π}}{4}, \\ \int_{0}^{\infty} sin (𝑟^{2}) d 𝑟 = ℑ𝔪 [\int_{0}^{\infty} e^{i 𝑟^{2}} d 𝑟] = \frac{\sqrt{2 π}}{4}, \end{matrix}

as desired. ⁠

□

Example 4.5.7.

Evaluate the integrals

\int_{0}^{2 π} Φ (cos 𝜃, sin 𝜃) d 𝜃

, where

Φ (𝜉, 𝜂)

is a rational function of

𝜉

and

𝜂

that is continuous on

𝜃 \in [0, 2 π]

Solution to Example 4.5.7 $Example 4.5.7 Example 4.5.7$ .

Let $𝑧 = e^{i 𝜃}$ . Consequently, we have $cos 𝜃 = \frac{𝑧 + 𝑧^{- 1}}{2}$ , $sin 𝜃 = \frac{𝑧 - 𝑧^{- 1}}{2 i}$ , and $d 𝑧 = i e^{i 𝜃} d 𝜃$ , implying that $d 𝜃 = \frac{d 𝑧}{i 𝑧}$ . Therefore, by the Residue Theorem (Theorem 4.5.2 $Theorem 4.5.2 Theorem 4.5.2$ ), letting $𝑓 (𝑧) = \frac{1}{i 𝑧} Φ (\frac{𝑧 + 𝑧^{- 1}}{2}, \frac{𝑧 - 𝑧^{- 1}}{2 i})$ , we have

\int_{0}^{2 π} Φ (cos 𝜃, sin 𝜃) d 𝜃 = \oint_{𝜕 𝔻} 𝑓 (𝑧) d 𝑧 = 2 π i \sum_{𝑘 = 1}^{𝑛} \underset{𝑧 = 𝑧_{𝑘}}{Res} 𝑓 (𝑧),

where

𝑧_{𝑘}

where

𝑘 = 1, \dots, 𝑛

are the isolated singularities of

𝑓

𝔻

. ⁠

□

Example 4.5.8.

Evaluate

𝐼 = \int_{0}^{\infty} \frac{𝑥^{𝛼}}{1 + 𝑥^{𝛽}} d 𝑥

, where

0 < 𝛼 + 1 < 𝛽

Figure 4.6: An indented wedge contour with orientation marked.

Solution to Example 4.5.8 $Example 4.5.8 Example 4.5.8$ .

Let $𝑓 (𝑧) = \frac{𝑧^{𝛼}}{1 + 𝑧^{𝛽}}$ and let $- π < Arg (𝑧) \leq π$ in the principal branches of $𝑧^{𝛼} = e^{𝛼 Log (𝑧)}$ and $𝑧^{𝛽} = e^{𝛽 Log (𝑧)}$ . Then except for at the zeros of $1 + 𝑧^{𝛽}$ , $𝑓$ is holomorphic.

The solutions to $𝑧^{𝛽} = - 1$ are $𝑧 = exp (i \frac{π}{𝛽} + 2 i 𝑘 \frac{π}{𝛽})$ . Choose an indented wedge contour (as there is a logarithmic branch point singularity at the origin) with an angle of $2 \frac{π}{𝛽}$ (as in Figure 4.5 $Figure 4.5 Figure 4.5$ ). The only singularity it encloses is $exp (i \frac{π}{𝛽})$ . Since it is a simple zero of $\frac{1}{𝑓}$ , this singularity is a simple pole.

The contour is the union of the following curves:

\begin{matrix} Γ_{1} = {𝑥 \in ℝ : 𝜀 \leq 𝑥 \leq 𝑅}, Γ_{2} = {𝑟 exp (i 2 π / 𝛽) : 𝜀 \leq 𝑟 \leq 𝑅}, \\ 𝐶_{𝑅} = {𝑅 e^{i 𝜃} : 0 \leq 𝜃 \leq 2 π / 𝛽}, 𝐶_{𝜀} = {𝜀 e^{i 𝜃} : 0 \leq 𝜃 \leq 2 π / 𝛽} \end{matrix}

where $𝑅 > 1$ and $0 < 𝜀 < 1$ . By the Residue Theorem (Theorem 4.5.2 $Theorem 4.5.2 Theorem 4.5.2$ ), we get that

lim_{𝜀 \to 0} lim_{𝑅 \to \infty} (\int_{Γ_{1}} + \int_{Γ_{2}} + \int_{𝐶_{𝑅}} + \int_{𝐶_{𝜀}}) 𝑓 (𝑧) d 𝑧 = 2 π i Res [𝑓, exp (i \frac{π}{𝛽})] .

By (4.5.2) $(4.5.2) (4.5.2)$ , it follows that

\begin{aligned} Res [𝑓, exp (i π / 𝛽)] & = lim_{𝑧 \to exp (i π / 𝛽)} \frac{𝑧 - exp (i π / 𝛽)}{(1 + 𝑧^{𝛽}) / 𝑧^{𝛼}} \\ = lim_{𝑧 \to exp (i π / 𝛽)} {(\frac{d}{d 𝑧} (𝑧^{- 𝛼} + 𝑧^{𝛽 - 𝛼}))}^{- 1} \\ = lim_{𝑧 \to exp (i π / 𝛽)} \frac{𝑧^{𝛼 + 1}}{(𝛽 - 𝛼) 𝑧^{𝛽} - 𝛼} \\ = - \frac{1}{𝛽} exp (i \frac{π}{𝛽} (𝛼 + 1)) . \end{aligned}

We can write the integral on $Γ_{2}$ in terms of $𝐼$ :

\begin{aligned} lim_{\begin{matrix} 𝑅 \to \infty \\ 𝜀 \to 0 \end{matrix}} \int_{Γ_{2}} 𝑓 (𝑧) d 𝑧 & = lim_{\begin{matrix} 𝑅 \to \infty \\ 𝜀 \to 0 \end{matrix}} \int_{𝑅}^{0} 𝑓 [𝑟 exp (i 2 π / 𝛽)] exp (i 2 π / 𝛽) d 𝑟 \\ = - exp [i 2 \frac{π}{𝛽} (1 + 𝛼)] \int_{0}^{\infty} \frac{𝑟^{𝛼}}{1 + 𝑟^{𝛽}} d 𝑟 . \end{aligned}

We also have

\int_{𝐶_{𝑅}} 𝑓 (𝑧) d 𝑧 = 𝑅 i \int_{0}^{2 π / 𝛽} 𝑓 (𝑅 e^{i 𝜃}) e^{i 𝜃} d 𝜃 = i \int_{0}^{2 π / 𝛽} \frac{𝑅^{𝛼 + 1}}{1 + 𝑅^{𝛽} e^{i 𝛽 𝜃}} exp [i 𝜃 (1 + 𝛼)] d 𝜃 .

It can also be shown that the integral is bounded by a vanishing function as $𝑅 \to \infty$ :

| \int_{0}^{2 π / 𝛽} \frac{𝑅^{𝛼 + 1}}{1 + 𝑅^{𝛽} e^{i 𝛽 𝜃}} exp [i 𝜃 (1 + 𝛼)] d 𝜃 | \leq \int_{0}^{2 π / 𝛽} \frac{𝑅^{𝛼 + 1}}{𝑅^{𝛽} - 1} d 𝜃 = 2 \frac{π}{𝛽} \frac{𝑅^{𝛼 + 1}}{𝑅^{𝛽} - 1} \to 0 .

Similarly, as $𝜀 \to 0$ ,

| \int_{𝐶_{𝜀}} 𝑓 (𝑧) d 𝑧 | \leq 𝜀 \int_{0}^{2 π / 𝛽} | 𝑓 (𝜀 e^{i 𝜃}) | d 𝜃 = \int_{0}^{2 π / 𝛽} \frac{𝜀^{𝛼 + 1}}{1 - 𝜀^{𝛽}} d 𝜃 = 2 \frac{π}{𝛽} \frac{𝜀^{𝛼 + 1}}{1 - 𝜀^{𝛽}} \to 0 .

By letting $𝑅 \to \infty$ and $𝜀 \to 0$ , we have

[1 - exp (i 2 \frac{π}{𝛽} (1 + 𝛼))] 𝐼 = - \frac{2 π i}{𝛽} exp (i \frac{π}{𝛽} (𝛼 + 1)) .

It follows that

𝐼 = \frac{2 π i}{𝛽} {[exp (i \frac{π}{𝛽} (𝛼 + 1)) - exp (- i \frac{π}{𝛽} (𝛼 + 1))]}^{- 1} = \frac{π}{𝛽} csc (\frac{π}{𝛽} (𝛼 + 1)) . □

Example 4.5.9.

Prove that the Fourier transform of $sech (π 𝑥)$ is itself, or that

𝐼 (𝜉) = \int_{- \infty}^{\infty} exp (- 2 π i 𝑥 𝜉) sech (π 𝑥) d 𝑥 = sech (π 𝜉) .

Figure 4.7: A rectangular contour with orientation marked.

Solution to Example 4.5.9 $Example 4.5.9 Example 4.5.9$ .

Fix $𝜉 \in ℝ$ and let $𝑓 (𝑧) = \frac{exp (- 2 π i 𝑧 𝜉)}{cosh (π 𝑧)}$ . Its poles in $ℂ$ occur when $e^{π 𝑧} + e^{- π 𝑧} = 0$ , or equivalently, when $𝑧 = i (𝑛 + \frac{1}{2})$ , where $𝑛 \in ℤ$ .

Since

cosh (π (𝑧 + i)) = - cosh (π 𝑧), exp (- 2 π i (𝑧 + i) 𝜉) = exp (2 π 𝜉) exp (- 2 π i 𝑧 𝜉),

we have that $𝑓 (𝑧)$ is a constant multiple of $𝑓 (𝑧 + i)$ . In particular, $𝑓 (𝑧 + i) = - exp (2 π 𝜉) 𝑓 (𝑧)$ . Therefore, we can use a rectangular contour as shown in Figure 4.7 $Figure 4.7 Figure 4.7$ . Let the sides be denoted by

\begin{matrix} Γ^{\leftarrow} = {𝑥 + i : - 𝑅 \leq 𝑥 \leq 𝑅, 𝑥 \in ℝ}, Γ^{\to} = {𝑥 \in ℝ : - 𝑅 \leq 𝑥 \leq 𝑅} \\ Γ^{↓} = {- 𝑅 + i 𝑦 : 𝑦 \in [0, 1]}, Γ^{↑} = {𝑅 + i 𝑦 : 𝑦 \in [0, 1]} . \end{matrix}

The only enclosed singularity is a simple pole at $𝑧 = \frac{i}{2}$ (simple by evaluation of the Taylor expansion of the denominator). By the Residue Theorem (Theorem 4.5.2 $Theorem 4.5.2 Theorem 4.5.2$ ), we get that

(\int_{Γ^{\to}} + \int_{Γ^{↑}} + \int_{Γ^{\leftarrow}} + \int_{Γ^{↓}}) 𝑓 (𝑧) d 𝑧 = 2 π i Res (𝑓, \frac{i}{2}) . (4.5 . 4)

By (4.5.2) $(4.5.2) (4.5.2)$ , we have

\begin{aligned} Res (𝑓, \frac{i}{2}) & = lim_{𝑧 \to i / 2} (𝑧 - \frac{i}{2}) \frac{exp (- 2 π i 𝑧 𝜉)}{cosh (π 𝑧)} \\ = lim_{𝑧 \to i / 2} \frac{d}{d 𝑧} {(\frac{cosh (π 𝑧)}{exp (- 2 π i 𝑧 𝜉)})}^{- 1} \\ = lim_{𝑧 \to i / 2} \frac{exp (- 2 π i 𝑧 𝜉)}{π sinh (π 𝑧) + 2 π i 𝜉 cosh (π 𝑧)} \\ = \frac{exp (π 𝜉)}{π i} . \end{aligned}

The sum of the horizontal line integrals is equal to

\begin{aligned} \int_{- 𝑅}^{𝑅} 𝑓 (𝑧) d 𝑧 + \int_{𝑅}^{- 𝑅} 𝑓 (𝑧 + i) d 𝑧 & = \int_{- 𝑅}^{𝑅} 𝑓 (𝑧) d 𝑧 - \int_{𝑅}^{- 𝑅} e^{2 π 𝜉} 𝑓 (𝑧) d 𝑧 \\ = (1 + e^{2 π 𝜉}) \int_{- 𝑅}^{𝑅} 𝑓 (𝑧) d 𝑧 . \end{aligned}

As $𝑅 \to \infty$ , we have $\int_{Γ^{\to}} 𝑓 (𝑧) d 𝑧 + \int_{Γ^{\leftarrow}} 𝑓 (𝑧) d 𝑧 \to (1 + e^{2 π 𝜉}) 𝐼 (𝜉)$ . The remaining two integrals can be written as

\begin{matrix} \int_{Γ^{↑}} 𝑓 (𝑧) d 𝑧 = \int_{0}^{1} \frac{exp (- 2 π i (𝑅 + i 𝑧) 𝜉)}{cosh (π (𝑅 + i 𝑧))} d 𝑧 \\ \int_{Γ^{↓}} 𝑓 (𝑧) d 𝑧 = \int_{1}^{0} \frac{exp (2 π i (𝑅 - i 𝑧) 𝜉)}{cosh (π (- 𝑅 + i 𝑧))} d 𝑧 . \end{matrix}

They can be bounded with

\begin{aligned} | \int_{0}^{1} \frac{exp (- 2 π i (𝑅 + i 𝑧) 𝜉)}{cosh (π (𝑅 + i 𝑧))} d 𝑧 | & \leq 2 \int_{0}^{1} \frac{exp (2 π 𝑧 𝜉)}{| e^{π 𝑅} e^{π i 𝑧} + e^{- π 𝑅} e^{- π i 𝑧} |} d 𝑧 \\ \leq 2 \int_{0}^{1} \frac{exp (2 π 𝑧 𝜉)}{| e^{π 𝑅} - e^{- π 𝑅} |} d 𝑧 \end{aligned}

and

\begin{aligned} | \int_{1}^{0} \frac{exp (2 π i (𝑅 - i 𝑧) 𝜉)}{cosh (π (- 𝑅 + i 𝑧))} d 𝑧 | & \leq 2 \int_{0}^{1} \frac{exp (2 π 𝑧 𝜉)}{| e^{- π 𝑅} e^{π i 𝑧} + e^{π 𝑅} e^{- π i 𝑧} |} d 𝑧 \\ \leq 2 \int_{0}^{1} \frac{exp (2 π 𝑧 𝜉)}{| e^{π 𝑅} - e^{- π 𝑅} |} d 𝑧 . \end{aligned}

Since the integrands are continuous and uniformly convergent to $0$ with respect to $𝑧$ , we have

\int_{Γ^{↑}} 𝑓 (𝑧) d 𝑧 + \int_{Γ^{↓}} 𝑓 (𝑧) d 𝑧 \to 0

as $𝑅 \to \infty$ . By rearrangement of (4.5.4) $(4.5.4) (4.5.4)$ ,

𝐼 (𝜉) (1 + e^{2 π 𝜉}) = 2 exp (π 𝜉),

or that

𝐼 (𝜉) = \frac{2}{e^{- π 𝜉} + e^{π 𝜉}} = sech (π 𝜉),

which proves the result. ⁠ $□$

Contour integration provides a powerful method for evaluating real improper integrals by leveraging the Residue Theorem (Theorem 4.5.2 $Theorem 4.5.2 Theorem 4.5.2$ ). The primary challenge often lies in constructing a suitable contour in the complex plane that encloses the relevant singularities of the integrand $𝑓$ while ensuring that the contribution from the contributions from the remaining segments of the contour either vanishes or can be calculated with ease.

If the function $𝑓$ is even and integrated on a domain such as $ℝ_{\geq 0}$ , then the integral can be extended to the entire real axis. If $𝑓$ decays sufficiently rapidly in the upper half plane $ℍ^{+}$ , a semicircular contour is generally preferable, as illustrated in Figure 4.3 $Figure 4.3 Figure 4.3$ . In the presence of singularities on the contour itself, we can insert arc indentations around them, as shown in Figure 4.4 $Figure 4.4 Figure 4.4$ .

If $𝑓 (𝑧)$ is a constant multiple of $𝑓 (𝑧 + i 𝑦)$ (a type of quasi-periodicity) for some $𝑦 \in ℝ$ , it is a strong indication to use a rectangular contour. If $𝑓 (𝑧)$ is a constant multiple of $𝑓 (𝑧 e^{i 𝜏})$ for some $𝜏 \in ℝ$ , a wedge-shaped contour is an appropriate choice.

In the case that there are indentations along the contour, we have

Theorem 4.5.10.

Let $𝜆 > 0$ and let $𝑎 \in ℂ$ . Suppose $𝑓 (𝑧)$ is a holomorphic function on $𝐷^{*} (𝑎, 𝜆)$ with a simple pole at $𝑧 = 𝑎 \in 𝑈$ . Let $0 < 𝜀 < 𝜆$ and define $𝛾_{𝜀} \subseteq 𝜕 𝐷 (𝑎, 𝜀)$ be a counterclockwise-oriented, connected arc subtending an angle $𝜗$ . Then,

lim_{𝜀 \to 0} \int_{𝛾_{𝜀}} 𝑓 (𝑧) d 𝑧 = i 𝜗 \cdot \underset{𝑧 = 𝑎}{Res} 𝑓 (𝑧) .

Proof.

Parameterize $𝛾_{𝜀}$ with $𝑧 = 𝑎 + 𝜀 e^{i 𝜃}$ , where $𝜃 \in [𝛼, 𝛽]$ and $𝛽 - 𝛼 = 𝜗$ . Then,

\int_{𝛾_{𝜀}} 𝑓 (𝑧) d 𝑧 = \int_{𝛼}^{𝛽} 𝑓 (𝑎 + 𝜀 e^{i 𝜃}) \frac{d 𝑧}{d 𝜃} d 𝜃 = 𝜀 i \int_{𝛼}^{𝛽} 𝑓 (𝑎 + 𝜀 e^{i 𝜃}) e^{i 𝜃} d 𝜃 .

Since $𝑓$ has a simple pole at $𝑧 = 𝑎$ , we can write a Laurent expansion around $𝑎$ as

𝑓 (𝑧) = \frac{𝑐_{- 1}}{𝑧 - 𝑎} + 𝜑 (𝑧),

where $𝜑 (𝑧)$ is holomorphic in a neighborhood of $𝑎$ and $𝑐_{- 1} = {Res}_{𝑧 = 𝑎} 𝑓 (𝑧)$ .

Then for $𝑧 = 𝑎 + 𝜀 e^{i 𝜃}$ ,

𝑓 (𝑎 + 𝜀 e^{i 𝜃}) = \frac{𝑐_{- 1}}{𝜀 e^{i 𝜃}} + 𝜑 (𝑎 + 𝜀 e^{i 𝜃}) .

So,

\begin{aligned} \int_{𝛾_{𝜀}} 𝑓 (𝑧) d 𝑧 & = 𝜀 i \int_{𝛼}^{𝛽} (\frac{𝑐_{- 1}}{𝜀 e^{i 𝜃}} + 𝜑 (𝑎 + 𝜀 e^{i 𝜃})) e^{i 𝜃} d 𝜃 \\ = i 𝑐_{- 1} \int_{𝛼}^{𝛽} d 𝜃 + 𝜀 i \int_{𝛼}^{𝛽} 𝜑 (𝑎 + 𝜀 e^{i 𝜃}) e^{i 𝜃} d 𝜃 \\ = i 𝑐_{- 1} 𝜗 + 𝜀 i \int_{𝛼}^{𝛽} 𝜑 (𝑎 + 𝜀 e^{i 𝜃}) e^{i 𝜃} d 𝜃 . \end{aligned}

Let $𝜀 < \frac{𝜆}{2}$ . Since $𝜑$ is continuous on the disk $\overline{𝐷 (𝑎, \frac{𝜆}{2})}$ , it is bounded. Therefore, letting $𝜀 \to 0$ , we have

lim_{𝜀 \to 0} 𝜀 i \int_{𝛼}^{𝛽} 𝜑 (𝑎 + 𝜀 e^{i 𝜃}) e^{i 𝜃} d 𝜃 = lim_{𝜀 \to 0} 𝜀 i \int_{𝛼}^{𝛽} 𝜑 (𝑎) e^{i 𝜃} d 𝜃 = 0 .

Therefore,

lim_{𝜀 \to 0} \int_{𝛾_{𝜀}} 𝑓 (𝑧) d 𝑧 = i 𝜗 \underset{𝑧 = 𝑎}{Res} 𝑓 (𝑧) . □

In the case that a branch point singularity is present on the contour, we may attempt to rewrite the function in a way such that the branch point is irrelevant. Otherwise, there are two types of “keyhole contours” that can be used to avoid the branch cut.

Example 4.5.11.

Evaluate

𝐼 = \int_{0}^{\infty} \frac{log (𝑥^{2} + 1)}{𝑥^{2} + 1} d 𝑥

Solution to Example 4.5.11 $Example 4.5.11 Example 4.5.11$ .

Notice that the integrand itself has branch points at $𝑧 = \pm i$ coinciding with the poles from the denominator. We can rewrite the integral as

\begin{aligned} 𝐼 & = \frac{1}{2} \int_{- \infty}^{\infty} \frac{log (𝑥^{2} + 1)}{𝑥^{2} + 1} = \int_{- \infty}^{\infty} \frac{log \sqrt{(𝑥 + i) (𝑥 - i)}}{𝑥^{2} + 1} \\ = \int_{- \infty}^{\infty} \frac{log | 𝑥 \pm i |}{𝑥^{2} + 1} = ℜ𝔢 \int_{- \infty}^{\infty} \frac{log (𝑥 + i)}{𝑥^{2} + 1} . (4.5 . 5) \end{aligned}

Let $𝛾 = Γ \cup 𝐶_{𝑅}$ , where concretely,

Γ = {𝑥 \in ℝ : - 𝑅 \leq 𝑥 \leq 𝑅}, 𝐶_{𝑅} = {𝑅 e^{i 𝜃} : 0 \leq 𝜃 \leq π}

and $𝑅 > 2$ , and let $𝑓 (𝑧) = \frac{Log (𝑧 + i)}{𝑧^{2} + 1}$ , where the branch for $Log$ is chosen to satisfy $[0, π] \subset ℑ𝔪 log (ℂ^{*})$ , such as the principal branch. The only singularity of $𝑓$ in the upper half plane is a simple pole at $𝑧 = i$ . By the Residue Theorem (Theorem 4.5.2 $Theorem 4.5.2 Theorem 4.5.2$ ), we have

lim_{𝑅 \to \infty} \oint_{𝛾} 𝑓 (𝑧) d 𝑧 = lim_{𝑅 \to \infty} (\int_{Γ} + \int_{𝐶_{𝑅}}) 𝑓 (𝑧) d 𝑧 = 2 π i \underset{𝑧 = i}{Res} 𝑓 (𝑧) .

By (4.5.2) $(4.5.2) (4.5.2)$ , we have

\underset{𝑧 = i}{Res} 𝑓 (𝑧) = lim_{𝑧 \to i} (𝑧 - i) \frac{log (𝑧 + i)}{𝑧^{2} + 1} = lim_{𝑧 \to i} \frac{log (𝑧 + i)}{𝑧 + i} = \frac{log (2 i)}{2 i} = \frac{π}{4} - i \frac{log (2)}{2} .

Additionally, for $𝑧 \in 𝐶_{𝑅}$ , since as $𝑅 \to \infty$ , $| 𝑓 (𝑧) | = | \frac{Log (𝑧 + i)}{𝑧^{2} + 1} | \leq \frac{| log | 𝑧 + i | | + π}{𝑅^{2} - 1} \leq \frac{log | 𝑅 + 1 | + π}{𝑅^{2} - 1} < \frac{𝑅 + 1 + π}{𝑅^{2} - 1} \to 0$ by virtue of $𝑅 > 2$ , it follows that $\int_{𝐶_{𝑅}} 𝑓 (𝑧) d 𝑧 \to 0$ .

Since ${lim}_{𝑅 \to \infty} \int_{Γ} 𝑓 (𝑧) d 𝑧 = \int_{- \infty}^{\infty} 𝑓 (𝑧) d 𝑧$ and

\int_{- \infty}^{\infty} 𝑓 (𝑧) d 𝑧 = \frac{π^{2} i}{2} + π log (2),

by (4.5.5) $(4.5.5) (4.5.5)$ , we have $𝐼 = ℜ𝔢 \int_{- \infty}^{\infty} 𝑓 (𝑧) d 𝑧 = π log (2)$ . ⁠ $□$