I solved the integral $$\int_{0}^{2\pi} \frac{1}{13+5\sin(\theta)}~d\theta$$ with the residue theorem and Cauchy’s Integral Formula.
The following is the solution, but I am unsure why in the end we consider only one of the residues which is a singularity point in the unit circle.
Making the substitution $z=e^{i\theta}$, $d\theta=\frac{dz}{iz}$, with $z$ being the unit circle in the complex plane:
$$ \int_{0}^{2\pi}\frac{1}{13+5\sin(\theta)}d\theta = 2\oint_{C}\frac{1}{5z^2+26iz-5}dz = \frac{2}{5}\oint_{C}\frac{1}{(z+\frac{i}{5})(z+5i)}dz$$
Therefore, the singular points are $-i/5$ and $-5i$.
Now this is where I’m confused. Since we made the substitution on the unit circle, the only relevant singularity point is $-i/5$, and thus we ignore the other residue when using Cauchy’s Integral Formula.
$$ \frac{2}{5}\oint_{C}\frac{1}{(z+\frac{i}{5})(z+5i)}=\frac{2}{5}\cdot2\pi i\cdot \operatorname{Res}\left(z=-\frac{i}{5}\right)=\frac{\pi}{6}$$
My question is, since we arbitrarily used the unit circle for our substitution, isn’t it just as valid to use a substitution for a circle with a larger radius? Could the larger radius lead to all singular points of $f(z)$ being inside this larger circle, resulting in two residues? Or will the substitution with the inclusion of radius $R$ at the beginning lead to the same outcome?
Best Answer
If you choose a circle with radius $R$, $z=Re^{i\theta}$, then we have
$$\sin\theta=\frac{z}{2iR}-\frac{R}{2iz},~~~~d\theta=\frac{1}{iz}dz$$ plug in $$ \int_{0}^{2\pi}\frac{1}{13+5\sin(\theta)}d\theta=2R\oint_{C}\frac{1}{5z^2+26Riz-5R^2}dz=\frac{2R}{5}\oint_{C}\frac{1}{(z+\frac{R}{5}i)(z+5Ri)}dz$$
You can see, no matter how large you choose $R$ is, it is always only one pole inside this circle, which is $-\frac{R}{5}i$. The other pole, $-5Ri$ is always outside this circle, hence no need to consider it.
Evaluate the residue,
$$\frac{2R}{5}\oint_{C}\frac{1}{(z+\frac{R}{5}i)(z+5Ri)}dz=\frac{2R}{5}\cdot 2\pi i\cdot Res\left(-\frac{R}{5}i\right)=\frac\pi6$$
You get the identical result.