5
$\begingroup$

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?

$\endgroup$
1

1 Answer 1

5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Amazing! Thank you very much for this! So clear $\endgroup$
    – noodles
    Commented May 31, 2023 at 4:10
  • $\begingroup$ @noodles You can accept this answer if you are happy with it. $\endgroup$
    – Gary
    Commented May 31, 2023 at 4:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .