1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Jul 28 '14 at 21:46
  • $\begingroup$ Have you tried $z=e^{2i\theta}$ ? $\endgroup$ – Christian Remling Jul 28 '14 at 22:42
3
$\begingroup$

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

$\endgroup$

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.