1
$\begingroup$

I don't know how to calculate this complex integral: $$\int_{0}^ {\pi} \frac{(x \sin x)dx }{1-2a \cos x+a^2}, a>0$$

$\endgroup$
  • $\begingroup$ Make it an integral from $-\pi$ to $\pi$, since the function is even. Look for a way to turn this into an integral over unit circle. $\endgroup$ – user147263 May 25 '15 at 4:39
  • $\begingroup$ Consider: $a^2 - 2a \cos(x) + 1 = (a + e^{ix})(a + e^{-ix})$ $\endgroup$ – user230734 May 25 '15 at 5:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.