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!

Validity of Reduction Formula

  1. May 11, 2012 #1
    1. The problem statement, all variables and given/known data

    If In denotes [tex]\int_0^∞ \! \frac{1}{(1+x^2)^n} \, \mathrm{d} x[/tex]
    Prove that [itex]2nI_{n+1} = (2n-1)I_n[/itex], and state the values of n for which this reduction formula is valid.

    2. Relevant equations



    3. The attempt at a solution

    [tex]I_n=\int_0^∞ \! \frac{1}{(1+x^2)^n} \, \mathrm{d} x[/tex]
    [tex]=\int_0^∞ \! (1+x^2)^{-n} \, \mathrm{d} x[/tex]
    By parts:
    [tex]=\left[ x(1+x^2)^{-n} \right]_0^∞ + 2n\int_0^∞ \! \frac{x^2}{(1+x^2)^{n+1}} \, \mathrm{d} x[/tex]
    [tex]=0 + 2n\int_0^∞ \! \frac{(1+x^2)-1}{(1+x^2)^{n+1}} \, \mathrm{d} x[/tex]
    [tex]=2n\int_0^∞ \! \frac{(1+x^2)}{(1+x^2)^{n+1}} \, \mathrm{d} x - 2n\int_0^∞ \! \frac{1}{(1+x^2)^{n+1}} \, \mathrm{d} x[/tex]
    [tex]=2nI_{n}-2nI_{n+1}[/tex]
    [tex]2nI_{n+1}=(2n-1)I_{n}[/tex]
    as required.

    It's the next bit where I'm stuck - the range of values for which n is valid. Obviously when part of the integral has been evaluated (following parts), this requires that n>0 otherwise the expression doesn't converge. I can't see anywhere else in the method where there is a restriction for n to be a specific value so I went with n>0 as my answer, but my book says n>1/2, can anyway shed some light on this for me.

    Thanks
     
  2. jcsd
  3. May 11, 2012 #2
    the expression inside the integral is always positive and so does the integral.at n=1/2 the integral does not converge and observe that at n=1/2 ,I(n+1) is zero and for 0<n<1/2, I(n+1) is negative.
     
  4. May 11, 2012 #3
    you should use:
    [tex]
    \frac{1}{(1 + x^2)^n} = \frac{1 + x^2 - x^2}{(1 + x^2)^n} = \frac{1}{(1 + x^2)^{n - 1}} - \frac{x^2}{(1 + x^2)^n}
    [/tex]
    For the integral of the second term, use integration by parts:
    [tex]
    -\int_{0}^{\infty}{x \, \frac{x}{(1 + x^2)^n} \, dx}
    [/tex]
    [tex]
    u = x \Rightarrow du = dx
    [/tex]
    [tex]
    dv = \frac{x}{(1 + x^2)^n} \, dx \Rightarrow v = \int{ \frac{x}{(1 + x^2)^n} \, dx} \stackrel{t = 1 + x^2}{=} \frac{1}{2} \, \int{t^{-n} \, dt} = \frac{t^{1- n}}{2(1 - n)} = -\frac{1}{2 (n - 1) (1 + x^2)^{n - 1}}
    [/tex]
    Combine everything, identify the relevant integrals with [itex]I_n[/itex], and [itex]I_{n - 1}[/itex], and see what you get.
     
  5. May 11, 2012 #4
    Oh, I see you already did the steps. As for the range of validity, answer these questions:

    1) What is the value of the integral [itex]\int{t^{-n} \, dt}[/itex] for [itex]n = 1[/itex]?

    2) When does the integrated out part [itex]\frac{x}{2(n - 1)(1 + x^2)^{n - 1}}[/itex] converge when [itex]x \rightarrow \infty[/itex]?
     
    Last edited: May 11, 2012
  6. May 11, 2012 #5
    Please preview your posts, LaTeX errors happen all the time.
     
  7. May 11, 2012 #6
    Thanks for your response.

    1) ln(t), so this would suggest it doesn't converge for n=1?

    2) Firstly I'm not sure where you've got this fraction from, I can't find it in any working of yours or mine? It would converge for n>1 for sure. For n=1 it won't converge, but for n<1 I have no idea?
     
  8. May 11, 2012 #7
    what is [itex]u v[/itex] in the integration by parts?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Validity of Reduction Formula
  1. Reduction Formula (Replies: 9)

  2. Reduction formula (Replies: 3)

  3. Reduction Formulae (Replies: 1)

  4. Reduction formula (Replies: 5)

Loading...