1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Poisson integral formula to solve other integrals

  1. Feb 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Use

    1) [itex]\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[/itex]

    to compute the integral:

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


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

    [itex]u = tan\left(\frac{x}{2}\right)[/itex]

    and simplified

    [itex]\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[/itex]

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

    Thanks in advance!
     
  2. jcsd
  3. Feb 6, 2012 #2
    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
     
  4. Feb 7, 2012 #3
    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,

    [itex]\int \frac{f(x)}{1−acos(x)}dx[/itex]

    for
    [itex]f(\theta) = sin(\theta)[/itex]
     
  5. Feb 7, 2012 #4
    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
    [itex]
    \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
    [/itex]
    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Poisson integral formula to solve other integrals
  1. Integration formula (Replies: 5)

  2. Solve Integration (Replies: 3)

Loading...