Yuliana Jordan

2022-03-19

Can anyone suggest a simple method to solve this integral by complex variables?
${\int }_{0}^{2\pi }\frac{d\theta }{\omega -a\mathrm{sin}\theta }$
where $|a|<|w|$.

PCCNQN4XKhjx

The integral is equal to, upon subbing $z={e}^{i\theta }$ and using $\mathrm{sin}\left\{\theta \right\}=\frac{z-{z}^{-1}}{2i}$
$-\frac{2}{a}{\oint }_{|z|=1}\frac{dz}{{z}^{2}-i2\frac{\omega }{a}z-1}$
The only pole inside the unit circle is at $z=i\left(\frac{\omega }{a}\right)-i\sqrt{{\left(\frac{\omega }{a}\right)}^{2}-1}$. The integral is then $i2\pi$ times the residue of the integrand at this pole, or
$i2\pi \frac{-\frac{2}{a}}{-i2\sqrt{{\left(\frac{\omega }{a}\right)}^{2}-1}}=\frac{2\pi }{\sqrt{{\omega }^{2}-{a}^{2}}}$

Ciara Hoffman

$I={\int }_{0}^{2\pi }\frac{d\theta }{\omega -a\mathrm{sin}\theta }d\theta ={\int }_{0}^{\pi }\frac{d\theta }{\omega -a\mathrm{sin}\theta }d\theta +{\int }_{0}^{\pi }\frac{d\theta }{\omega +a\mathrm{sin}\theta }d\theta$
$={\int }_{0}^{\pi }\frac{2\omega }{{\omega }^{2}-{a}^{2}{\mathrm{sin}}^{2}\theta }d\theta$
$=4\omega {\int }_{0}^{\frac{\pi }{2}}\frac{d\theta }{{\omega }^{2}-{a}^{2}{\mathrm{sin}}^{2}\theta }$
and if we use the substitution $\theta =\mathrm{arctan}t$ we have:
$I=4\omega {\int }_{0}^{+\mathrm{\infty }}\frac{dt}{\left(1+{t}^{2}\right)\left({\omega }^{2}-{a}^{2}\frac{{t}^{2}}{1+{t}^{2}}\right)}=4\omega {\int }_{0}^{+\mathrm{\infty }}\frac{dt}{{\omega }^{2}+\left({\omega }^{2}-{a}^{2}\right){t}^{2}}=\frac{2\pi }{\sqrt{{\omega }^{2}-{a}^{2}}}$

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