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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?

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  • $\begingroup$ Are you sure the limits of integration are $0$ to $+\infty$ and not $-\infty$ to $+\infty$? $\endgroup$
    – Nameless
    Commented Dec 8, 2012 at 19:13
  • $\begingroup$ I'm sure they're right. $\endgroup$ Commented Dec 8, 2012 at 20:02
  • $\begingroup$ Have a look at (for example): math.stackexchange.com/questions/114884/… and see if you can't do it yourself afterwards. Writing a complete answer to these kinds of questions is a lot of work. $\endgroup$
    – mrf
    Commented Dec 8, 2012 at 20:32

1 Answer 1

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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.

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