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.
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