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Homework Help: 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]
    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.

  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:
    \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}
    For the integral of the second term, use integration by parts:
    -\int_{0}^{\infty}{x \, \frac{x}{(1 + x^2)^n} \, dx}
    u = x \Rightarrow du = dx
    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}}
    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?
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