# Complex Integration using residue theorem

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1. Dec 7, 2016

### arpon

1. The problem statement, all variables and given/known data

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

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

3. 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.

2. Dec 7, 2016

### Ray Vickson

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