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

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.

Answer 1

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!

Answer 2

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.

Answer 3

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{{1 \over 2\pi}\int_{0}^{2\pi}{R^{2} - r^{2} \over R^{2} - 2Rr\cos\pars{\phi-\theta} + r^{2}}\,\dd\theta = 1\,,\quad R > r > 0}$

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}