This question was part of mock test of masters exam for which I am preparing and I am unable to solve it.
Show that $\int_{0}^{2\pi } \frac{ \sin(\theta – \phi) } { R^2 – 2rR \cos(\theta -\phi) +r^2 } d{\phi} =0$ .
The part ${ R^2 – 2rR \cos(\theta -\phi) +r^2 }$ seems to be from Poisson's integral fomula for harmonic functions butI am not sure how to use it here. (Due to this reason i am using harmonic functions tag )
If u(z) is harmonic in a domain consisting of disc $|z|\leq R$ . Then for $z=r e^{i\theta}$ we have ,
$u(r e^{i \theta})=\frac{1}{2\pi}\int_{0}^{2\pi} \frac{R^2 -r^2} { R^2 – 2rR \cos(\theta – \phi) +r^2 } u(R e^{i \phi})d {\phi}$.
But $\sin(\theta -\phi)$ is an implicit function.So, I am not able to find u(z) and $R^2 -r^2$.
Can you please help me in proving it?
Best Answer
Since the function that you are integrating is periodic with period $2\pi$, your integral is equal to$$\int_{\theta-\pi}^{\theta+\pi}\frac{\sin(\theta-\phi)}{R^2-2rR\cos(\theta-\phi)+r^2}\,\mathrm d\phi$$andd now the substitution $\theta-\phi=t$ and $\mathrm d\phi=-\mathrm dt$ leads to$$\int_{-\pi}^\pi\frac{\sin t}{R^2-2rR\cos(t)+r^2}\,\mathrm dt,$$which is $0$, since you are integrating an odd function.