Obtaining Poisson's formula from a integral of summation

Question

If $$u(r,\theta)=\frac1{\pi}\int_0^{2\pi}\Bigg[ \frac12+\sum_{n=1}^{\infty}r^n\cos n(\theta-\phi) \Bigg]f(\phi)d\phi,$$ can anyone help show me why this implies $$u(r,\theta)=\frac{(1-r^2)}{2\pi}\int_0^{2\pi} \frac{f(\phi)}{1-2r\cos(\theta-\phi)+r^2}d\phi \,?$$ I had previously derived the first equation, but am rather stuck on this step. I've been trying to show equivalently that: $$\frac12+\sum_{n=1}^{\infty}r^n\cos n(\theta-\phi)=\frac{1-r^2}{2(1-2r\cos(\theta-\phi)+r^2)}.$$

Answer 1

As we know $|r|<1$

$$\frac12+\sum_{n=1}^{\infty}r^n\cos n(\theta-\phi)=\frac12+\Re\left(\sum_{n=1}^{\infty}(r\exp\,i(\theta-\phi))^n\right)=\frac12+\Re \Bigg( \frac{1}{1-r\exp\,i(\theta-\phi)}-1\Bigg) $$ $$=\frac{1-r^2}{2(1-2r\cos(\theta-\phi)+r^2)}$$

By the geometric series summation and after a lot of algebra to find the real part. This result is equivalent to the desired result