# Complex Integration using residue theorem

• arpon
This gives the result ##\pi i##.In summary, the conversation discusses the calculation of a limit involving the integral of a complex function over a semicircular contour. The solution involves using Cauchy's Residue Theorem and evaluating the integral explicitly to obtain the desired result.
arpon

## Homework Statement

[/B]
##C_\rho## is a semicircle of radius ##\rho## in the upper-half plane.
What is
$$\lim_{\rho\rightarrow 0} \int_{C_{\rho}} \frac{e^{iaz}-e^{ibz}}{z^2} \,dz$$

## Homework Equations

If ##C## is a closed loop and ##z_1, z_2 ... z_n## are the singular points inside ##C##,
$$\int_C f(z) \,dz =2\pi i \sum_{k=1}^n Res_{z=z_k}f(z)$$

## The Attempt at a Solution

According to the solution,
$$--------------$$
\begin{align} f(z)&= \frac{e^{iaz}-e^{ibz}}{z^2}\\ &= \frac{1}{z^2}\left[\left(1+\frac{iaz}{1!}+\frac{(iaz)^2}{2!}+\frac{(iaz)^3}{3!}+...\right)-\left(1+\frac{ibz}{1!}+\frac{(ibz)^2}{2!}+\frac{(ibz)^3}{3!}+...\right)\right]\\ &=\frac{i(a-b)}{z}+...~~~(0<|z|<\infty) \end{align}
So ##z=0## is a simple pole of ##f(z)##, with residue ##B_0=i(a-b)##. Thus
\begin{align} \lim_{\rho\rightarrow 0} \int_{C_{\rho}}f(z) \,dz &=-B_0\pi i\\ &=-i(a-b)\pi i\\ &=\pi(a-b)\end{align}
$$--------------$$
I do not understand equation (4).
I guess Cauchy's Residue Theorem is applied here. According to that theorem, integral over a closed contour is equal to ##2\pi i## times the residue. I guess, it was assumed that for semicircular contour, the integral would be ##\pi i## times the residue, may be because of integrals over an upper semi-circle and over a lower semi-circle being equal.
But I do not understand why the integrals over an upper semi-circle and over a lower semi-circle are equal.
##f(z)## is not symmetric about real-axis, because
\begin{align} f(z)=&f(re^{i\theta})\\ &=\frac{\exp(ia\cdot re^{i\theta})-\exp(ib\cdot re^{i\theta})}{r^2 e^{i2\theta}}\\ &=\frac{\exp(-ar\sin{\theta+iar\cos{\theta}})-\exp(-br\sin{\theta+ibr\cos{\theta}})}{r^2e^{i2\theta}}\end{align}
Now, if ##\theta## is substituted by ##-\theta##, the expression is not the same.

arpon said:

## Homework Statement

View attachment 110058 [/B]
##C_\rho## is a semicircle of radius ##\rho## in the upper-half plane.
What is
$$\lim_{\rho\rightarrow 0} \int_{C_{\rho}} \frac{e^{iaz}-e^{ibz}}{z^2} \,dz$$

.

To obtain eq. (4), just evaluate the integral ##\int_{C_{\rho}} (1/z) dz## explicitly; there is no need to use Theorems.

arpon

## 1. What is the residue theorem in complex integration?

The residue theorem is a mathematical tool used in complex analysis to evaluate integrals along closed curves. It states that the value of an integral around a closed curve is equal to the sum of the residues of the function inside the curve.

## 2. How is the residue of a function calculated?

The residue of a function at a particular point is equal to the coefficient of the term with a negative power in the Laurent series expansion of the function about that point. It can also be calculated using the formula Res(f,z0) = limz→z0 (z-z0)f(z).

## 3. What are the applications of the residue theorem?

The residue theorem has many applications in mathematics and physics, including solving real integrals, calculating complex logarithms and trigonometric functions, and computing the values of improper integrals. It is also used in the study of complex functions, such as meromorphic functions and analytic continuation.

## 4. Can the residue theorem be used to evaluate any integral?

No, the residue theorem can only be used to evaluate integrals along closed curves that enclose singularities of the function. If the closed curve does not enclose any singularities, the integral will be equal to zero.

## 5. Are there any limitations to using the residue theorem?

The residue theorem can only be used for integrals over closed curves. It also requires knowledge of complex analysis and the behavior of complex functions near singularities. In some cases, the residue theorem may not be the most efficient method for evaluating an integral, and other techniques may be more suitable.

Replies
3
Views
1K
Replies
3
Views
630
Replies
14
Views
2K
Replies
5
Views
505
Replies
32
Views
2K
Replies
23
Views
2K
Replies
13
Views
1K
Replies
8
Views
1K
Replies
9
Views
2K
Replies
12
Views
2K