How to prove this function is bounded? I've got the following function:
$v(r,\theta) = \frac{1}{2\pi} \int_{0}^{2\pi} \frac{r^{2} - 1}{r^{2} + 1 - 2 r \cos\left(\psi - \theta)\right)}f(\psi)d\psi$
The function $f$ is NOT specified, but it is given that $f$ is continuous. We consider the function for values $r>1$.
How do I prove that $v$ is bounded for all $r>1$?
I thought about taking a derivative with respect to $r$, but got nowhere. I have been pulling my hair out over this one.
NOTE: The above function is the solution to $\Delta v = 0$ for $r>1$, with $v(1,\theta) = f(\theta)$.
 A: $$\frac{r^{2} - 1}{r^{2} + 1 - 2 r \cos\left(\psi - \theta)\right)}\le \frac{r^{2} - 1}{r^{2} + 1 - 2 r } = \frac{r^2-1}{(r-1)^2} = \frac{r+1}{r-1} \to 1$$
when $r\to \infty$. So $v$ is bounded when $r$ is large. 
Another point of view: for $r<1$, the function
$$\begin{split}
u(r, \theta) := v\left(\frac 1r , \theta\right) &= \frac{1}{2\pi}\int_0^{2\pi} \frac{\left(\frac 1r\right)^2  -1}{\left(\frac 1r\right)^2 +1 - 2\frac 1r \cos (\psi -\theta) } f(\psi) d\psi \\
&= \frac{1}{2\pi}\int_0^{2\pi} \frac{1 -r^2}{1 +r^2 - 2r \cos (\psi -\theta) } f(\psi) d\psi
\end{split}$$
can obviously be extended to $r = 0$. Indeed this $u$ satifies $\Delta u =0$ and $u = f$ on the boundary $\{r=1\}$. So your $v$ is formed by "flipping" this $u$ from $\{r<1\}$ to $\{r>1\}$.  
A: First we will evaluate the integral $I(r)$ which is given by
$$\begin{align}
I(r)&\equiv \frac{1}{2\pi}\int_0^{2\pi}\frac{(r^2-1)}{r^2+1-2r\cos (\psi-\theta)}\,d\psi\\\\
&=\frac{1}{2\pi}\int_0^{2\pi}\frac{(r^2-1)}{r^2+1-2r\cos \psi}\,d\psi
\end{align}$$
since the cosine function is $2\pi$-periodic.
We can evaluate $I(r)$ by making the substitution $z=e^{i\psi}$.  Then, for $r>1$ we have
$$\begin{align}
I(r)&=\frac{1}{2\pi}\oint_{|z|=1}\frac{(r^2-1)}{r^2+1-2r\left(\frac{z+z^{-1}}{2}\right)}\,\frac{1}{iz}\,dz\\\\
&=\frac{1}{2\pi}\oint_{|z|=1}\frac{(i/r)(r^2-1)}{(z-r)(z-1/r)}\,dz\\\\
&=1
\end{align}$$
where we used the Residue Theorem to evaluate the contour integral.
Now, the function $v(r,\theta)$ can be written
$$\begin{align}
v(r,\theta)=\frac{1}{2\pi}\int_0^{2\pi}\frac{(r^2-1)}{r^2+1-2r\cos (\psi-\theta)}(f(\psi)-f(\theta))\,d\psi+f(\theta)\\\\
\end{align}$$
Since $f$ is continuous, it is bounded on $[0,2\pi]$.  Denote the supremum of $f$ on $[0,2\pi]$ by $M$.  Then, we have for $r>1$
$$\begin{align}
|v(r,\theta)|&=\left|\frac{1}{2\pi}\int_0^{2\pi}\frac{(r^2-1)}{r^2+1-2r\cos (\psi-\theta)}f(\psi)\,d\psi\right|\\\\
&\le \frac{1}{2\pi}\int_0^{2\pi}\frac{(r^2-1)}{r^2+1-2r\cos (\psi-\theta)}\left|f(\psi)\right|\,d\psi\\\\
&\le M
\end{align}$$
Therefore, $|v|\le M$ and we are done!
