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Yuliana Jordan

Answered question

2022-03-19

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

\int_{0}^{2 π} \frac{d θ}{ω - a \sin θ}

where

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PCCNQN4XKhjx

Beginner2022-03-20Added 8 answers

The integral is equal to, upon subbing

z = e^{i θ}

and using

\sin {θ} = \frac{z - z^{- 1}}{2 i}

- \frac{2}{a} \oint_{| z | = 1} \frac{d z}{z^{2} - i 2 \frac{ω}{a} z - 1}

The only pole inside the unit circle is at

z = i (\frac{ω}{a}) - i \sqrt{{(\frac{ω}{a})}^{2} - 1}

. The integral is then

i 2 π

times the residue of the integrand at this pole, or

i 2 π \frac{- \frac{2}{a}}{- i 2 \sqrt{{(\frac{ω}{a})}^{2} - 1}} = \frac{2 π}{\sqrt{ω^{2} - a^{2}}}

Ciara Hoffman

Beginner2022-03-21Added 5 answers

I = \int_{0}^{2 π} \frac{d θ}{ω - a \sin θ} d θ = \int_{0}^{π} \frac{d θ}{ω - a \sin θ} d θ + \int_{0}^{π} \frac{d θ}{ω + a \sin θ} d θ

= \int_{0}^{π} \frac{2 ω}{ω^{2} - a^{2} \sin^{2} θ} d θ

= 4 ω \int_{0}^{\frac{π}{2}} \frac{d θ}{ω^{2} - a^{2} \sin^{2} θ}

and if we use the substitution

θ = \arctan t

we have:

I = 4 ω \int_{0}^{+ \infty} \frac{d t}{(1 + t^{2}) (ω^{2} - a^{2} \frac{t^{2}}{1 + t^{2}})} = 4 ω \int_{0}^{+ \infty} \frac{d t}{ω^{2} + (ω^{2} - a^{2}) t^{2}} = \frac{2 π}{\sqrt{ω^{2} - a^{2}}}

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