I'm trying to evaluate the integral $$\int_{-\pi}^{\pi} \frac{\sin^2 t}{3+\cos t}dt$$ using complex numbers.

Meaning, instead of calculating $$\int_{-\pi}^{\pi} \frac{\sin^2 t}{3+\cos t}dt,$$ I want to calculate the integral $$\int_\Gamma \frac{\sin^2 z}{3+\cos z}dz$$ where $\Gamma$ is half the circle with center at origin and radius $\pi$ (in the positive direction), and then the straight line on the $x$ axis from $-\pi$ to $\pi$.

We know that this integral is $0$ because $$\frac{\sin^2 z}{3+\cos z}$$ is analytic in the entire area bounded by $\Gamma$. So that's not an issue, but to evaluate our original integral, we need to now calculate $$\int_\gamma \frac{\sin^2 z}{3+\cos z}dz$$ where $\gamma$ is just the upper half of the circle I mentioned above. Without the line that goes from $-\pi$ to $\pi$ on the $x$ axis.

How do I calculate this integral?

Edit: A good parametrization might be $z=\pi e^{i\theta}$ where $0 \leq \theta \leq \pi$. And then we need to calculate the integral $$\int_{0}^{\pi} \frac{\sin^2 (\pi e^{i \theta})}{3+\cos (\pi e^{i\theta})}i\pi e^{i\theta} d\theta.$$ Doesn't seem easy to do.

  • $\begingroup$ You need to find the simple poles inside the region of $\Gamma$. $\endgroup$ – Thomas Andrews Jan 5 '15 at 16:05
  • $\begingroup$ I think you want to use a different contour. Try a change of variable instead: $z = e^{it}$, then $[-\pi,\pi]$ gets mapped to a circular contour. $\endgroup$ – Cameron Williams Jan 5 '15 at 16:06
  • $\begingroup$ I think this is an integral you want to convert to a $z$-integral to use Cauchy's formula or the residue calculus. The point is to identify the integrand as the parametrized form of a particular contour integral, it is not as you attempt it I think. See Section 7.3 of supermath.info/GuideToGamelin.pdf for examples of the technique. $\endgroup$ – James S. Cook Jan 5 '15 at 16:07
  • $\begingroup$ There are no poles. It's analytic in $\Gamma$. $\endgroup$ – Oria Gruber Jan 5 '15 at 16:08
  • $\begingroup$ A simple approach is to calculate the integral over the real interval $[-\pi,\pi]$. The integrand has an antiderivative which can be found explicitly. $\endgroup$ – Hans Engler Jan 5 '15 at 16:51

By setting $z=e^{it}$, we have $dz=i e^{it} dt$, so $dt=-\frac{i}{z}dz$ and:

$$ I = \int_{-\pi}^{+\pi}\frac{\sin^2 t}{3+\cos t}\,dt = -i\oint \frac{\left(\frac{z-1/z}{2i}\right)^2}{z\left(3+\frac{z+1/z}{2}\right)}\,dz=i\oint\frac{(1-z^2)^2}{z^2(1+6z+z^2)}\,dz$$ where the path of integration is the unit circle. The integrand function $f(z)=\frac{(1-z^2)^2}{z^2(1+6z+z^2)}$ has two poles inside the unit disk, in $z=0$ and $z=-3+\sqrt{8}$. The last integral can be computed by applying the residue theorem:

$$ I = -2\pi\cdot\sum_{\xi\in\{0,-3+\sqrt{8}\}}\operatorname{Res}(f(z),z=\xi)=-2\pi(-3+2\sqrt{2})=\color{red}{2\pi(3-\sqrt{8})}.$$


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.