Can one use contour integration to evaluate $\int^{\pi}_{0} \frac{1}{1-\rho*sin(\theta)}d\theta$ for $0<\rho<1$? This would be trivial if the upper limit were $2\pi$ as we could let $z=e^{i\theta}$, and do the usual integral around the full unit circle. Here, integral is just over a semi-circle, and due to asymmetry, we cant simply do $\frac{1}{2}\int^{\pi}_{0}f(\theta)d\theta$.

Note: Based on another approach (using properties of multivariate normal distribution), I have reason to believe that this integral equals $\frac{\pi+2\tan^{-1}(\sqrt{\frac{\rho}{1-\rho^2}})}{\sqrt{1-\rho^2}}$, but I would like to compute this from first principles.

  • $\begingroup$ I believe the upper square root should not extend over the numerator (and the correct expression can be simplified using $\arcsin\rho$). $\endgroup$ – Emil Jeřábek supports Monica Jul 28 '14 at 21:46
  • $\begingroup$ Have you tried $z=e^{2i\theta}$ ? $\endgroup$ – Christian Remling Jul 28 '14 at 22:42

The usual $x = \tan \theta/2$ substitution reduces this to a rational function integral from $0$ to $\infty.$


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