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 $\int_0^\infty \frac{\log \cos^2 \alpha x}{\beta^2-x^2} \, \mathrm dx=\alpha \pi$

  • 3
    $\begingroup$ I give you $15$ reputation then. $\endgroup$
    – Tunk-Fey
    May 31 '14 at 15:38

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.