Evaluate the principal value of the integral $$\int\limits_{-\infty}^{\infty} \frac{\cos z}{z-w} \ dz. \ \ \ \ \ |\text{Im} \ z|>0 $$
I could not solve this problem during tutorial class. Upon looking at the solution sheet that was uploaded the next day, I am still not clear about the method that is used to solve this. So the solution given is:
$$ \int\limits_{-\infty}^{\infty} \frac{\cos z}{z-w} \ dz = \begin{cases} 2\pi i \ \text{Res}_{z=w} \left(\frac{e^{iz}}{2(z-w)}\right) = \pi i e^{iw}, & \text{if} \ \text{Im} \ w >0 \\ -2\pi i \ \text{Res}_{z=w} \left(\frac{e^{-iz}}{2(z-w)}\right) = -\pi i e^{iw}, & \text{if} \ \text{Im} \ w <0 \end{cases}$$
The part which I don't understand is the beginning part of the solution where they claim (without giving any reason) that $\int\limits_{-\infty}^{\infty} \frac{\cos z}{z-w} \ dz = \int\limits_{-\infty}^{\infty} \frac{e^{iz}}{z-w} \ dz$ $\text{if}$ $\text{Im} \ w>0$ and $\int\limits_{-\infty}^{\infty} \frac{\cos z}{z-w} \ dz = \int\limits_{-\infty}^{\infty} \frac{e^{-iz}}{z-w} \ dz$ if $\text{Im} \ w<0$.
Evaluating the integrals with the exponential terms is clear to me. But what is the reasoning behind replacing the cosine with the exponential function? Need help understanding this.
Using Euler's Formula, we can write
$$\int_{-\infty}^\infty \frac{\cos(z)}{z-w}\,dz=\int_{-\infty}^\infty \frac{e^{iz}}{z-w}\,dz+\int_{-\infty}^\infty \frac{e^{-iz}}{z-w}\,dz\tag 1$$
For $\text{Im}(w)>0$, the pole at $z=w$ is in the upper-half plane. As a consequence, we assert that the second integral on the right-hand side of $(1)$ is zero. To see this more clearly, we proceed as follows.
Let $C_R$ be the contour comprised of the (i) real line segment from $-R$ to $R$ and (ii) the semicircular arc in the lower-half plane that has radius $R$ and is centered at $0$. Since $\frac{e^{-iz}}{z-w}$ is analytic in and on $C_R$, Cauchy's Integral Theorem guarantees that
$$\begin{align} \oint_{C_R}\frac{e^{-iz}}{z-w}\,dz&=\int_{-R}^R \frac{e^{-iz}}{z-w}\,dz+\int_{2\pi}^\pi \frac{e^{-iRe^{i\phi}}}{Re^{i\phi}-w}\,iRe^{i\phi}\,d\phi\tag 2\\\\ &=0 \end{align}$$
As $R\to 0$, the second integral on the right-hand side of $(2)$ approaches zero. Therefore, we arrive at the coveted result
$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty \frac{e^{-iz}}{z-w}\,dz=0}$$
for $\text{Im}(w)>0$ as was to be shown!
The analysis for the case $\text{Im}(w)<0$ proceeds analogously and is left as an exercise for the reader.