TITLE: Show that $\int_0^{2\pi}\frac{R^2-r^2}{R^2 - 2Rr\cos (\varphi-\vartheta) + r^2}d\vartheta$ is independent of $R>r>0$, using only real numbers.
QUESTION [6 upvotes]: The poisson kernel is sometimes written as
$$
\frac{1}{2\pi}\int_0^{2\pi} \frac{R^2-r^2}{R^2 - 2Rr\cos(\varphi-\vartheta) + r^2} \mathrm{d}\vartheta = 1 \ , \ \ R>r>0
$$
Where $\varphi$ is some arbitary angle. 
This much used in complex analysis amongst other fields. Is there some basic, elementary way of showing that the integral is independent on $R$ and $r$?
Splitting the integral at $\int_0^\pi + \int_\pi^{2\pi}$ and using the Weierstrass substitution seems like somewhat ugly and a  messy way to approach the problem.

REPLY [6 votes]: $\newcommand{\+}{^{\dagger}}%
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$\ds{{1 \over 2\pi}\int_{0}^{2\pi}{R^{2} - r^{2} \over R^{2} - 2Rr\cos\pars{\phi-\theta} + r^{2}}\,\dd\theta = 1\,,\quad R > r > 0}$

The integral is obviously independent of $\phi$: Just derive the integral respect of $\phi$ and use $\ds{\partiald{}{\phi} = - \partiald{}{\theta}}$. Since the integrand is a periodic function of $\theta$ it turns out that the $\phi$-derivative of the integral is zero. Then, we just need to consider the integral:
$$
{\cal I} \equiv {1 \over \pi}\int_{0}^{\pi}{1 - h^{2}
\over 1 - 2h\cos\pars{\theta} + h^{2}}\,\dd\theta\,,\qquad 0< h \equiv {r \over R} < 1
$$

With Weierstrass substitution $\ds{t \equiv \tan\pars{\theta \over 2}}$:
\begin{align}
{\cal I} &= {1 - h^{2} \over \pi}\int_{0}^{\infty}
{1 \over
 1 - 2h\pars{1 - t^{2}}/\pars{1 + t^{2}} + h^{2}}\,{2\,\dd t \over 1 + t^{2}}
\\[3mm]&=
2\,{1 - h^{2} \over \pi}\int_{0}^{\infty}
{\dd t \over \pars{h^{2} + 2h + 1}t^{2} + h^{2} - 2h + 1}
=
{2 \over \pi}\,\int_{0}^{\infty}
{\bracks{\pars{1 + h}/\pars{1 - h}}\,\dd t \over \bracks{\pars{1 + h}t/\pars{1 - h}}^{2} + 1}
\\[3mm]&=
{2 \over \pi}\
\overbrace{\int_{0}^{\infty}{1 \over t^{2} + 1}}^{\ds{=\ {\pi/2}}}
=\color{#00f}{\Large 1}
\end{align}