How to prove $\int_{0}^{2\pi } \frac{ \sin(\theta - \phi) } { R^2 - 2rR \cos(\theta -\phi) +r^2 } d{\phi} =0$ 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?
 A: 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.
A: If you want, you can directly calculate the integral. In fact,
\begin{eqnarray}
&&\int_{0}^{2\pi }  \frac{ \sin(\theta - \phi) } { R^2 - 2rR \cos(\theta -\phi) +r^2 } d{\phi} \\
&=&-\frac{1}{2rR}\int_{0}^{2\pi } \frac{1} { R^2 - 2rR \cos(\theta -\phi) +r^2 } d\bigg[R^2 - 2rR \cos(\theta -\phi) +r^2 \bigg] \\
&=&-\frac{1}{2rR}\ln\bigg[ R^2 - 2rR \cos(\theta -\phi) +r^2 \bigg]\bigg|_{\phi=0}^{2\pi}\\
&=&0.
\end{eqnarray}
A: For $0 \le r < R$ and $z = re ^{i \theta}$ we have (from Cauchy's integral formula or the Residue theorem or the definition of the winding number):
$$
 1 = \frac{1}{2 \pi i } \int_{|\zeta| = R} \frac{d\zeta}{\zeta - z}
= \frac{1}{2 \pi } \int_0^{2 \pi} \frac{R}{R - r e ^{i ( \theta - \phi)}} d\phi \\
= \frac{1}{2 \pi } \int_0^{2 \pi} \frac{R(R-r e ^{-i ( \theta - \phi)})}{R^2 - 2rR \cos(\theta-\phi) + r^2} d\phi \, .
$$
Taking the imaginary part gives
$$
0 = \frac{1}{2 \pi } \int_0^{2 \pi} \frac{rR (\sin\theta - \phi)}{R^2 - 2rR \cos(\theta-\phi) + r^2} d\phi \, .
$$

The connection to the Poisson formula is that
$$
\frac{R^2-r^2}{{R^2 - 2rR \cos(\theta-\phi) + r^2} } = \operatorname{Re}\left( \frac{R+r e ^{i ( \theta - \phi)}}{R+r e ^{i ( \theta - \phi)}}\right)
$$
and
$$
\frac{2Rr \sin(\theta - \phi)}{{R^2 - 2rR \cos(\theta-\phi) + r^2} } = \operatorname{Im}\left( \frac{R+r e ^{i ( \theta - \phi)}}{R+r e ^{i ( \theta - \phi)}}\right)
$$
