Poisson integral formula to solve other integrals

[slugbunny]

Homework Statement

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].

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.

Thanks in advance!

 

[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

 

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

 

[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.