<s>[INST] 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.
 [/INST] 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! 
</s>