I saw this in a proof (by sos440) from another question, but as I don't have 50 reputation I wasn't able to comment to ask about a particular step in the proof.
\begin{align} \frac{\alpha}{2}PV\int_{-\infty}^{\infty}\frac{\log\cos^2x}{(\alpha\beta)^2-x^2}dx &=\frac{\alpha}{2}\sum_{n = -\infty}^{\infty}PV\int_{\Large-\frac{\pi}{2}}^{\Large\frac{\pi}{2}}{\frac{\log \cos^2{x}}{(\alpha\beta)^2 - (x+n\pi)^2}}dx \end{align}
I'm not sure how the sum is produced from the integral. If someone could point me in the right direction that would be great. Thanks!
From this question: How to find PV $\int_0^\infty \frac{\log \cos^2 \alpha x}{\beta^2-x^2} \, \mathrm dx=\alpha \pi$
All there is to the above expansion is an exploitation of the fact that $\log{\cos^2{x}}$ is periodic with period $\pi$. That means we can split the real line into an infinite set of intervals
$$\left [ -\frac{\pi}{2} + n \pi , \frac{\pi}{2} + n \pi \right ) $$
for all $n \in \mathbb{Z}$. Then
$$\int_{-\infty}^{\infty} dx \frac{\log{\cos^2{x}}}{(\alpha \beta)^2-x^2} = \sum_{n=-\infty}^{\infty} \int_{-\pi/2}^{\pi/2} dx \frac{\log{\cos^2{\left (x+n \pi \right )}}}{(\alpha \beta)^2-\left (x+n \pi \right )^2}$$
Applying the periodicity of $\log{\cos^2{x}}$ produces the desired result.