Hint for the integral by complex variable

Question

Can anyone suggest a simple method to solve this integral by complex variables?

$$\int_{0}^{2\pi }\frac{d\theta }{\omega -a \sin\theta }$$

where $|a|<|w|$. I am actually trying to find out the time period for non-uniform oscillator.

Answer 1

The integral is equal to, upon subbing $z=e^{i \theta}$ and using $\sin{\theta} = (z-z^{-1})/(2 i)$:

$$-\frac{2}{a} \oint_{|z|=1} \frac{dz}{z^2-i 2 \frac{\omega}{a} z-1} $$

The only pole inside the unit circle is at $z=i (\omega/a) - i \sqrt{(\omega/a)^2-1}$. The integral is then $i 2 \pi$ times the residue of the integrand at this pole, or

$$i 2 \pi \frac{-2/a}{-i 2 \sqrt{(\omega/a)^2-1}} = \frac{2 \pi}{\sqrt{\omega^2-a^2}}$$

Answer 2

$$\begin{eqnarray*}I=\int_{0}^{2\pi}\frac{d\theta}{\omega-a\sin\theta}\,d\theta &=& \int_{0}^{\pi}\frac{d\theta}{\omega-a\sin\theta}\,d\theta+\int_{0}^{\pi}\frac{d\theta}{\omega+a\sin\theta}\,d\theta\\&=&\int_{0}^{\pi}\frac{2\omega}{\omega^2-a^2\sin^2\theta}\,d\theta\\&=&4\omega\int_{0}^{\pi/2}\frac{d\theta}{\omega^2-a^2\sin^2\theta}\end{eqnarray*}$$ and if we use the substitution $\theta=\arctan t$ we have: $$ I = 4\omega\int_{0}^{+\infty}\frac{dt}{(1+t^2)\left(\omega^2-a^2\frac{t^2}{1+t^2}\right)}=4\omega\int_{0}^{+\infty}\frac{dt}{\omega^2+(\omega^2-a^2) t^2}=\color{red}{\frac{2\pi}{\sqrt{\omega^2-a^2}}}.$$

Answer 3

The inequality $|\omega| > |a|$ implies $\omega\neq 0$. If $a = 0$, then the integral evaluates to $\frac{2\pi}{\omega}$. So suppose $a \neq 0$. Using the parametrization $z = e^{i\theta}$, $0 \le \theta \le 2\pi$ for the unit circle, and letting $\lambda = \frac{\omega}{a}$, we can write

$$\int_0^{2\pi} \frac{d\theta}{\omega - a\sin \theta} = \frac{1}{a}\oint_{|z| = 1} \frac{1}{\lambda - \frac{z - z^{-1}}{2i}}\frac{dz}{iz} = -\frac{2}{a}\oint_{|z| = 1} \frac{dz}{z^2 - 2\lambda i z + 1}.$$

The roots of $z^2 - 2i\lambda z - 1$ are $i(\lambda \pm \sqrt{\lambda^2 - 1})$, and the root $i(\lambda - \sqrt{\lambda^2 - 1})$ is the only one that lies inside the circle. So by Cauchy's integral formula, the integral is equal to

$$-\frac{2}{a} \cdot 2\pi i \frac{1}{z - i(\lambda +\sqrt{\lambda^2 - 1})}\bigg|_{z = i(\lambda - \sqrt{\lambda^2 - 1})} = \frac{2\pi}{a\sqrt{\lambda^2 - 1}} = \frac{2\pi}{\sqrt{\omega^2 - a^2}}.$$