I want to show
$$ \int_0^{2 \pi } \frac{ R^2 – r^2 }{R^2-2rR \cos( \phi – \theta ) + r^2 } d\theta = 2 \pi $$
for any $R > r $ and $\phi $
I dont know if I should use Poisson's formula or try to find a substitution.
[Math] Evaluating an integral over the unit circle
complex-analysis
Related Solutions
Here's a way to do it by using only the simplest instance of Poisson's formula on the disk, namely $f(0)=\frac{1}{2\pi}\int_0^{2\pi} f(e^{i\phi})\,d\phi$.
We have some boundary data $g:\mathbb R\to\mathbb R$ and wish to find a harmonic function $U$ on the upper half-plane $\mathbb H$ for this data. Fix a point $x_0+iy_0\in \mathbb H$ and consider the conformal map $F(z)=x_0+iy_0\frac{1+z}{1-z}$. Note that $F$ maps the unit disk onto $\mathbb H$ and $F(0)=x_0+iy_0$. Pull back the boundary data: $g\circ F$ is defined on the unit circle. Let $u$ be the harmonic extension of $g\circ F$ to the disk. By the above, $$u(0)=\frac{1}{2\pi}\int_0^{2\pi} (g\circ F)(e^{i\phi})\,d\phi\tag1$$ On the other hand, $u=U\circ F$ by conformal invariance of harmonic functions. Therefore, the integral in (1) actually gives us $U(x_0+iy_0)$, which is what we wanted. It remains to change the variable of integration in (1), so that integral is taken over $\mathbb R$, in the original domain of definition of $g$. This yields $$\frac{1}{2\pi}\int_0^{2\pi} (g\circ F)(e^{i\phi})\,d\phi = \frac{1}{2\pi}\int_{\mathbb R} g(x) |G'(x)|\,dx \tag2$$ where $G=F^{-1}$. I leave the computation of $G'$ to you.
Yes, your computation is correct. It would be a little easier if it stayed more in the spirit of complex analysis: the Poisson kernel for the unit disk is $$ P(z, \zeta) = \frac{1}{2\pi}\frac{1-|z|^2}{|z-\zeta|^2} $$ Plugging $1/\bar z$ for $z$ we get $$ P(1/\bar z, \zeta) = \frac{1}{2\pi}\frac{1-1/|z|^2}{|1/\bar z-\zeta|^2} = \frac{1}{2\pi}\frac{|z|^2-1}{|1-\bar z\zeta|^2} = \frac{1}{2\pi}\frac{|z|^2-1}{| \bar z - \bar \zeta |^2} = \frac{1}{2\pi}\frac{|z|^2-1}{|z - \zeta|^2} $$ Since the transformation $z\mapsto 1/\bar z$ is anticonformal, it preserves harmonic functions (same proof as for conformal maps; or just think of it as composition of a conformal map with conjugation; the conjugation preserves harmonic functions). Also, it is the identity on the boundary of the unit disk. Conclusion: for $|z|>1$, $$ u(z) = \frac{1}{2\pi}\int_{|\zeta|=1}\frac{|z|^2-1}{|z - \zeta|^2} \, h(\zeta)\,|d\theta| $$
This is essentially the same that you got, but using $1/\bar z$ is a little simpler: in polar coordinates, it maps $re^{i\theta}$ to $r^{-1}e^{i\theta}$, without flipping the sign of polar angle.
In your notation, un-flipping the sign of $\theta$ makes the final formula nicer:
$$u(re^{i \phi}) = \frac{r^2-1}{2 \pi} \int_0^{2 \pi} \frac{h(e^{i \theta})}{1+r^2-2r \cos(\phi-\theta)} d\theta$$
Best Answer
I offer two suggestions. Firstly, setting $a=r/R$ and $t=\phi-\theta$, the integral reduces to $$ \int_0^{2\pi} \frac{1-a^2}{1+a^2-2a\cos{t}} \, dt, $$ with $a<1$. You could now set $z=e^{it}$, which reduces the integral to $$ -i\int_{|z|=1} \frac{1-a^2}{z(1+a^2)-2a z^2 -2a} \, dz, $$ which is easy to do by finding the poles and using the calculus of residues.
Alternatively, a useful formula is the Fourier expansion of the integrand: it is easy to show, using the complex formula for the cosine and the geometric series formula that if $\lvert a \rvert<1$, $$ 1+2\sum_{n=0}^{\infty} a^n\cos{nt} = \frac{1-a^2}{1+a^2-2a\cos{t}}; $$ this converges absolutely uniformly in $t$, so you can interchange the order of summation and integration: $$ \int_0^{2\pi} \frac{1-a^2}{1+a^2-2a\cos{t}} \, dt = \int_0^{2\pi} \left( 1+2\sum_{n=0}^{\infty} a^n\cos{nt} \right) dt = 2\pi + 0, $$ since the integral of a cosine over a whole period is zero.