contour integration, complex triangle inequality, residue theorem

Question

Source: https://www.maths.ed.ac.uk/~jmf/Teaching/MT3.html by Jose Figueroa-O'Farrill

Background:

p.v. is Cauchy's principal value

which can be "complexified" to the form,

To make use Cauchy's residue theorem, we close the contour with the following curve,

My confusion:

How the integral over the $C^+_\rho$ part of the contour vanishes...

Most of my confusion here seems to lie in understanding the referenced equation 2.28, which reads,

where equation 2.24 reads,

for completeness, equation 2.36 reads,

Any pointers are appreciated :(

Answer 1

The inequality (2.28) states that the absolute value of the integral $\int_\gamma f(z)\,\mathrm dz$ of a function $f$ along a path $\gamma\colon[a,b]\longrightarrow\mathbb C$ is smaller than or equal to the product of two numbers:

• the maximum of $\lvert f\rvert$ restricted to $\gamma\bigl([a,b]\bigr)$;
• the length of $\gamma$.

Let us now apply this to your problem. The length of $C_\rho^+$ is $\pi\rho$. Furthermore, if $z$ belongs to the semicircle, then $\lvert z^2+4\rvert\geqslant\rho^2-4$ and therefore (if $\rho>1$), $\max\left\lvert\frac1{z^2+4}\right\rvert\leqslant\frac1{\rho^2-4}$. So,$$\left\lvert\int_{C_\rho^+}\frac{\mathrm dz}{z^2+4}\right\rvert\leqslant\frac{\pi\rho}{\rho^2-4}.$$Since $\lim_{\rho\to\infty}\frac{\pi\rho}{\rho^2-4}=0$,$$\lim_{\rho\to\infty}\left\lvert\int_{C_\rho^+}\frac{\mathrm dz}{z^2+4}\right\rvert=0.$$