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?
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?
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).