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When solving Laplacian equation, I need to integrate the following integration:

$$ \int_0^{2\pi}\frac{1+3\text{sin}\theta}{a^2+r^2-2ar\text{cos}(\theta-\phi)}\text{d}\theta $$

How to work it out?

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1 Answer 1

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Note that,

$$ \int_0^{2\pi}\frac{1+3\text{sin}\theta}{a^2+r^2-2ar\text{cos}(\theta-\phi)}\text{d}\theta= \int_0^{2\pi}\frac{1+3\text{sin}\theta}{a^2+r^2-2ar\,({\cos}(\theta)\cos(\phi)+\sin(\theta)\sin(\phi))}\text{d}\theta \,.$$

You can use residue theorem to evaluate the above integral using the technique. Also, you can find more detials about the technique here, see example(III).

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I think that the observation $$ a^2 + r^2 - 2ar\cos(\theta-\phi) = \left|ae^{i\phi} - z\right|^2 = (ae^{i\phi} - z)(ae^{-i\phi} - \bar{z})$$ for $z = re^{i\theta}$ will also help apply residue theorem. –  sos440 Nov 25 '12 at 6:06
    
That's beautiful method, thank you! –  hxhxhx88 Nov 26 '12 at 14:03
    
@hxhxhx88: You are welcome. –  Mhenni Benghorbal Nov 26 '12 at 20:11

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