# Why only one singularity is involved? $\int_{0}^{2\pi} \frac{1}{13+5\sin(\theta)}~d\theta$

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?

• A more general integral: math.stackexchange.com/questions/2760900/… If $a^2+b^2<c^2$ then $\int_0^{2\pi} \frac1{a\cos(t)+b\sin(t)+c}dt=\frac{2\pi}{\sqrt{c^2-a^2-b^2}}$. Commented May 31, 2023 at 6:12

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.

• Amazing! Thank you very much for this! So clear Commented May 31, 2023 at 4:10
• @noodles You can accept this answer if you are happy with it.
– Gary
Commented May 31, 2023 at 4:17