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. 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.
 A: Step I. Using integration by parts one can easily show that
$$
J_n=\int_0^{2\pi} \cos^{2n}x\,dx=\frac{2\pi}{4^n}\binom{2n}{n},
$$
as for the cosine integral we make use of a recurrence,
\begin{align}
J_{n+1}&=\int_0^{2\pi} \cos^{2(n+1)}x\,dx=\int_0^{2\pi} (\cos^{2n+1}x)(\sin x)'\,dx =0+(2n+1)\int_0^{2\pi} \cos^{2n}x\,\sin^2 x\,dx \\&=
(2n+1)\int_0^{2\pi} \cos^{2n}x\,(1-\cos^2 x)\,dx=(2n+1)J_n-(2n+1)J_{n+1},
\end{align}
and thus $J_{n+1}=\frac{2n+1}{2n+2}J_n=2^{-n-1}\prod_{j=0}^{n+1}\frac{2j-1}{j+1}J_0=\frac{2\pi}{4^{n+1}}\frac{(2n+2)!}{(n+1)!(n+1)!}$.
Step II. If $|a|<1$, then
$$
\frac{1}{\sqrt{1-a^2}}=(1-a^2)^{-1/2}=\sum_{n=0}^\infty \binom{-\frac{1}{2}}{n}a^{2n}=\cdots=\sum_{n=0}^\infty \binom{2n}{n}\frac{a^{2n}}{2^{2n}}.
$$
Step III. Then
$$
I=\frac{1}{2\pi}\int_0^{2\pi} \frac{R^2-r^2}{R^2 - 2Rr\cos(\phi-\theta) + r^2} \mathrm{d}\theta
=\frac{R^2-r^2}{2\pi(R^2+r^2)}\int_0^{2\pi}\frac{dx}{1-a\cos x},
$$
where $a=\dfrac{2rR}{r^2+R^2}.$
Also
$$
\int_0^{2\pi}\frac{dx}{1-a\cos x}=\sum_{n=0}^\infty\int_0^{2\pi} a^{2n}\cos^{2n}x\,dx=\sum_{n=0}^\infty \frac{2\pi\,a^{2n}}{4^n}\binom{2n}{n}=\frac{2\pi}{\sqrt{1-a^2}}
$$
Step IV. Finally,
\begin{align}
I&=\frac{1}{2\pi}\int_0^{2\pi} \frac{R^2-r^2}{R^2 - 2Rr\cos(\phi-\theta) + r^2} \mathrm{d}\theta
=\frac{R^2-r^2}{2\pi(R^2+r^2)}\int_0^{2\pi}\frac{dx}{1-a\cos x} \\
&=\frac{R^2-r^2}{2\pi(R^2+r^2)}\cdot
\frac{2\pi}{\sqrt{1-\left(\frac{2rR}{r^2+R^2}\right)^2}}
=\cdots=1.
\end{align}
Note. I prefer the proof of T.A.E, but the question was asking of a proof without the use of complex numbers! 
A: $\newcommand{\+}{^{\dagger}}%
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$\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}
A: Because you're integrating over a full period of $\cos$ and because $R > 0$, your expression becomes
$$
             \frac{1}{2\pi}\int_{0}^{2\pi}\frac{1-(r/R)^{2}}{1-2(r/R)\cos\theta+(r/R)^{2}}\,d\theta.
$$
Let $\rho = r/R$. Then, by assumption, $0 < \rho < 1$, and 
$$
\begin{align}
      \frac{1-(r/R)^{2}}{1-2(r/R)\cos\theta+(r/R)^{2}}
          & =\frac{1-\rho^{2}}{(1-\rho e^{i\theta})(1-\rho e^{-i\theta})} \\
          & = \frac{1}{1-\rho e^{i\theta}}+\frac{\rho e^{-i\theta}}{1-\rho e^{-i\theta}} \\
          & = \sum_{n=-\infty}^{\infty}\rho^{|n|} e^{in\theta} \\
          & = 1 + 2\sum_{n=1}^{\infty}\rho^{n}\cos(n\theta).
\end{align}
$$
When you integrate the terms in this sum over $0 \le \theta \le 2\pi$, only the integral of the constant term is non-zero. The final integral value is 1, regardless of $0 \le \rho < 1$.
If you don't like the complex exponential in between, just verify directly that
$$
      F(\rho) = \frac{1}{2\pi}\int_{0}^{2\pi}\frac{1-\rho^{2}}{1-2\rho\cos\theta+\rho^{2}}d\theta
$$
satisfies the differential equation
$$
                     F''(\rho)+\frac{1}{\rho}F'(\rho)=0.
$$
This a separable equation whose solutions have the form
$$
             F'(\rho) = C\frac{1}{\rho} \\
             F(\rho) = C\ln \rho + D.
$$
The constant $C$ must be $0$ in this case for obvious reasons.
