- #1

arpon

- 235

- 16

## Homework Statement

##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.