0
$\begingroup$

I want to use the residue theorem to solve this integral: $$\int_0^{2\pi} \frac{1}{13+5\sin(\theta)}d\theta.$$ When applying the residue theorem, $\sin(\theta)$ becomes $\frac{z-\frac{1}z}{2i}$. When plugging this substitution into my integrand, the integrand simplifies as: $$\frac{2}i\int_C \frac{dz}{-5iz^2+26z-5i}.$$ However, I am having difficulty finding where the singularity is so that I can apply the residue theorem to this problem. Is there an easier way?

$\endgroup$
  • $\begingroup$ This is a quadratic polynomial. Its zeros are the singularities. BTW, I get $+5i$ instead of $-5i$. $\endgroup$ – Friedrich Philipp Mar 24 '17 at 5:28
  • $\begingroup$ My result is $\pi/6$. $\endgroup$ – Friedrich Philipp Mar 24 '17 at 5:36
  • $\begingroup$ I understand that it's considerably easier to just do the integral in the way it's presented, but I'm trying to apply the residue theorem. $\endgroup$ – Evan Mar 24 '17 at 5:38
  • $\begingroup$ I got my result using the residue theorem (or rather Cauchy's formula). $\endgroup$ – Friedrich Philipp Mar 24 '17 at 5:41
1
$\begingroup$

First, $$\sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i}$$Now, let $z=e^{i\theta}$, then $\text{d}z=iz\text{d}\theta$, i.e the integral becoms $$I=\frac{1}{i}\int\limits_{|z|=1}\frac{1}{z\left(13+5\dfrac{z-z^{-1}}{2i}\right)}\text{d}z=\int\limits_{|z|=1}\underbrace{\frac{2}{5z^2+26iz-5}}_{f(z)}\text{d}z$$The poles are $$z_{1,2}=\dfrac{-26i\pm\sqrt{-676+100}}{10}=\dfrac{-26i\pm 24i}{10}$$ and only $z=-\dfrac{i}{5}$ is in the unit circle, i.e $$I=2\pi i\cdot\text{res}\left(f(z),-\dfrac{i}{5}\right)=\frac{\pi}{6}$$

$\endgroup$

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.