Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm supposed to verify that this: $$\int_0^\infty \frac{dx}{x^p(x^2+2x\cos{\phi}+1)}=\pi\frac{\sin{p\phi}}{\sin{p\pi}\sin{\phi}}$$

where $0<p<1$ and $0<\phi<\pi$

How do I do this with a keyhole contour?

share|cite|improve this question
Are you sure the limits of integration are $0$ to $+\infty$ and not $-\infty$ to $+\infty$? – Nameless Dec 8 '12 at 19:13
I'm sure they're right. – Desperate Fluffy Dec 8 '12 at 20:02
Have a look at (for example):… and see if you can't do it yourself afterwards. Writing a complete answer to these kinds of questions is a lot of work. – mrf Dec 8 '12 at 20:32
Thank you so much! – Desperate Fluffy Dec 8 '12 at 22:38

The basic procedure is very simple. Suppose we are using the standard keyhole contour of radius $R$ and with the branch cut of the logarithm along the positive real axis and its argument between $0$ and $2\pi.$

Let $$f(x) = \frac{e^{-p \log x}}{x^2+ 2x\cos\phi+1}$$ and let $I$ be the integral we are looking for. Writing $$ I = \int_0^\infty f(x) dx = \int_0^\infty \frac{e^{-p \log x}}{x^2+ 2x\cos\phi+1} dx$$ and integrating counterclockwise we see that the segment just above the real axis goes to $I,$ and the one below to $-I e^{-2\pi i p}.$

Now note that the only additional two poles are at $$ \rho_{0,1} = -\cos\phi \pm \sqrt{\cos^2\phi -1} = -\cos\phi \pm i\sin\phi = - e^{\mp i\phi} = e^{\pi i} e^{\mp i\phi}.$$

It follows by the Cauchy Residue Theorem that $$ I \left(1- e^{-2\pi i p} \right) = 2\pi i \left(\operatorname{Res}_{x=\rho_0} f(x) + \operatorname{Res}_{x=\rho_1} f(x)\right).$$

By definition we have $$ \operatorname{Res}_{x=\rho_0} f(x) = \lim_{x\to\rho_0} \frac{x^{-p}}{x-\rho_1} = \frac{e^{-\pi i p} e^{p i\phi}}{2i\sin\phi} $$ and $$ \operatorname{Res}_{x=\rho_1} f(x) = \lim_{x\to\rho_1} \frac{x^{-p}}{x-\rho_0} = -\frac{e^{-\pi i p} e^{-p i\phi}}{2i\sin\phi} $$

Putting it all together, we find $$I \left(1 - e^{-2\pi i p} \right) = 2\pi i \, e^{-\pi i p} \frac{e^{p i\phi} - e^{-p i\phi}}{2i\sin\phi} = 2\pi i \, e^{-\pi i p} \frac{\sin (p\phi)}{\sin\phi}$$ or $$ I = \frac{2\pi i \, e^{-\pi i p}}{1 - e^{-2\pi i p}} \frac{\sin (p\phi)}{\sin\phi} = \frac{2\pi i}{e^{\pi i p} - e^{-\pi i p}} \frac{\sin (p\phi)}{\sin\phi} = \frac{\pi}{\sin(\pi p)} \frac{\sin (p\phi)}{\sin\phi}.$$ It remains to verify that the integral along the outer circle of radius $R$ disappears as $R$ goes to infinity. But $f(x)$ is $O(1/R^{2+p})$ so the integral is $O(1/R^{1+p})$ which disappears as claimed.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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