Poisson integral formula to solve other integrals

1. Feb 5, 2012

slugbunny

1. The problem statement, all variables and given/known data
Use

1) $\frac{1}{2\pi}\int\limits_{-\pi}^{\pi} \frac{r_0^2 - r^2}{r_0^2 - 2rr_0cos(\theta-t) + r^2} dt = 1$

to compute the integral:

2) $\int\limits_{-\pi}^{\pi} \left[1 - acos(x) \right]^{-1} dx$
for $0<a<1$
[/itex].

3. The attempt at a solution
I looked on Wolfram for help. I did a tan substitution,

$u = tan\left(\frac{x}{2}\right)$

and simplified

$\int\limits_{-\pi}^{\pi} \left[1 - acos(x) \right]^{-1} dx \rightarrow \int\limits_{-\pi}^{\pi} \frac{2}{u^2+1-a+au^2}dx$

It kind of looks like 1)...? I am pretty confused.

2. Feb 6, 2012

susskind_leon

Mate, you're supposed to use formula 1), so likely tan-substitution is not the correct way. Think about how you need to relate r, r0 to a

3. Feb 7, 2012

slugbunny

Thanks for the reply. Using the substitution, I got the correct answer (you base it off of the equality in 1) or something). But I can't figure out how to use integrals to solve integrals!!

The second part to the question is to use the PIF to solve 2)...when the integral of 2) is multiplied by an f(x).

so,

$\int \frac{f(x)}{1−acos(x)}dx$

for
$f(\theta) = sin(\theta)$

4. Feb 7, 2012

susskind_leon

When you take a hard look at the integral 1), you realize that the integral you're supposed to carry out is actually the same thing as integral 1), when you choose a appropriately (and multiply with a constant). Try rewriting it like this
$\frac{1}{2\pi}(r_0^2-r^2)\int_{-\pi}^{\pi}\frac{1}{r_0^2+r^2-2rr_0\cos(\theta-t)}dt=\frac{1}{2\pi}\frac{r_0^2-r^2}{r_0^2+r^2}\int_{-\pi}^{\pi}\frac{1}{1-\frac{2rr_0}{r_0^2+r^2}\cos(\theta-t)}dt$
When you multiply by sin(theta) and integrate over theta, you will get zero because of the oddness of the integrand.

Last edited: Feb 7, 2012