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How to integrate (x^2-1)^n

  1. Aug 20, 2009 #1
    1. The problem statement, all variables and given/known data

    The whole expression is
    [tex] \frac{(-1)^n (2n)!}{2^{2n} (n!)^2} \int^{1}_{-1} (x^2-1)^n dx [/tex]
    and the answer should be
    [tex] \frac{2}{2n+1} [/tex]
    but I don't know how to get there.

    I came across this while checking the orthogonality of the associated Legendre functions.

    2. Relevant equations

    3. The attempt at a solution

    First I tried integrating by parts.
    [tex] \int^{1}_{-1} (x^2-1)^n dx = \int^{1}_{-1} (x^2-1)^{n-1} (x^2-1) dx = [/tex]
    [tex] = [(x^2-1)^{n-1} (x^3/3-x)]^{1}_{-1} - 2(n-1) \int^{1}_{-1} (x^2-1)^{n-2} x(x^3/3-x) dx = [/tex]
    [tex] = - 2(n-1) \int^{1}_{-1} (x^2-1)^{n-2} x(x^3/3-x) dx [/tex]

    I think that by integrating by parts I would eventually get rid of n under the integral sign which is good but the integrand itself gets more and more complicated so I'm not sure whether I should continue doing this.

    Then I tried making the substitution [tex] x \rightarrow cos(x) [/tex]

    [tex] \int^{1}_{-1} (x^2-1)^n dx = - \int^{0}_{\pi} (cos^2 (x)-1)^{n} sin(x) dx = [/tex]
    [tex] = \int^{\pi}_{0} (-1)^n sin^2 (x)^{2n-1} dx [/tex]

    and again, I'm not sure whether that will lead me anywhere or not.

    And guidance would be appreciated.
     
  2. jcsd
  3. Aug 20, 2009 #2
    Try the binomial expansion of [itex](x^2-1)^n[/itex] and exploit the fact that the interval of integration is [-1, 1]. Hint: even vs. odd terms. See if that gets anywhere.

    --Elucidus
     
    Last edited: Aug 20, 2009
  4. Aug 20, 2009 #3
    Yes, I forgot to mention earlier that I had also tried the binomial formula.

    [tex] \int^{1}_{-1} (x^2-1)^n dx = \int^{1}_{-1} \sum^n_{k=0} \frac{n!}{k!(n-k)!} x^{2n-2k} (-1)^k dx = \sum^n_{k=0} \frac{(-1)^k n!}{k!(n-k)!} \int^{1}_{-1} x^{2n-2k} dx = \sum^n_{k=0} \frac{(-1)^k n!}{k!(n-k)!} \frac{2}{2n-2k+1} [/tex]

    So far the expression has become

    [tex] \frac{(-1)^n (2n)!}{2^{2n} (n!)^2} \sum^n_{k=0} \frac{(-1)^k n!}{k!(n-k)!} \frac{2}{2n-2k+1} [/tex]

    which has to be equal to [tex] \frac{2}{2n+1} [/tex]

    I still don't understand how.
    Perhaps there are some formulas that could be used but I'm unaware of.

    Most of the derivation is given in this book: http://physics.bgu.ac.il/~gedalin/Teaching/Mater/mmp.pdf
    I understand all of it except the very end (page 311).
     
  5. Aug 20, 2009 #4

    Dick

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  6. Aug 20, 2009 #5
    Okay, I can use
    [tex] \int sin^n (x) dx = -\frac{1}{n}sin^{n-1}(x) cos (x) + \frac{n-1}{n} \int sin^{n-2}(x)dx [/tex]

    which in my case becomes
    [tex] \int^{\pi}_0 sin^n (x) dx = \frac{n-1}{n} \int^{\pi}_0 sin^{n-2}(x)dx [/tex]

    Now
    [tex] \frac{(-1)^n (2n)!}{2^{2n} (n!)^2} \int^{\pi}_{0} (-1)^n sin^{2n+1}(x) dx = [/tex]

    [tex] = \frac{(2n)!}{2^{2n} (n!)^2} \frac{2n}{2n+1} \int^{\pi}_{0} sin^{2n-1}(x) dx = [/tex]

    [tex] = \frac{(2n)!}{2^{2n} (n!)^2} \frac{2n(2n-2)}{(2n+1)(2n-1)} \int^{\pi}_{0} sin^{2n-3}(x) dx = [/tex]

    [tex] = \frac{(2n)!}{2^{2n} (n!)^2} \frac{2n(2n-2)...(2n-2n+2)}{(2n+1)(2n-1)...(2n-2n+1)} \int^{\pi}_{0} sin(x) dx = [/tex]

    [tex] = \frac{(2n)!}{2^{2n} (n!)^2} \frac{2^n n(n-1)...1}{(2n+1)(2n-1)...(2n-2n+1)} 2 = [/tex]

    [tex] = \frac{(2n)!n!}{2^n (n!)^2} 2 \frac{2n(2n-2)...2}{(2n+1)2n(2n-1)...1} = [/tex]

    [tex] = \frac{(2n)!}{2^n n!} 2 \frac{2^n n(n-1)...1}{(2n+1)!} = [/tex]

    [tex] = \frac{2}{2n+1} [/tex]

    Problem solved. Thank you.
     
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