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 and $0<\phi<\pi$ How do I do this with a keyhole contour?
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 and $0<\phi<\pi$ How do I do this with a keyhole contour?
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.