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Integral: recursion

  1. Jul 29, 2010 #1
    Can someone help me with this? I don't know how to begin

    I(n) = integral 1/(1+x^2)^n dx with n ∈ ℕ0

    Than: ∀ n ∈ ℕ0, n≠1 : I(n) = 1/(2(n-1)) * x/((1+x^2)^(n-1)) + (2n-3)/(2(n-1)) * I(n-1)

    I have to prove this, but I don't know how to start
     
  2. jcsd
  3. Jul 29, 2010 #2
    Learn latex for beggining

    [tex]I_n=\int \frac{dx}{(1+x^2)^n}[/tex]

    Try to do this integral for [tex]n=0,2,3...[/tex]
     
  4. Jul 29, 2010 #3

    arildno

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    Hint:
    Write:
    [tex]I_{n}=\int{1}*\frac{1}{(1+x^{2})^{n}}dx[/tex]
     
    Last edited: Jul 29, 2010
  5. Jul 29, 2010 #4
    I know, than I can do:

    u = 1/((1+x^2)^n) --> du = (-2nx)/((1+x^2)^(n+1))s
    dv = 1 --> v = x

    = x/((1+x^2)^n) + 2n integral x^2/((1+x^2)^(n+1))

    But what now?
     
  6. Jul 29, 2010 #5

    arildno

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    Rewrite:
    [tex]\frac{x^{2}}{(1+x^{2})^{n+1}}=\frac{x^{2}+1-1}{(1+x^{2})^{n+1}}=\frac{1}{(1+x^{2})^{n}}-\frac{1}{(1+x^{2})^{n+1}}[/tex]
    See if that helps..:smile:
     
  7. Jul 29, 2010 #6
    For the moment I have:

    I(n) = x/((1-2n)((1+x^2)^n)) - (2n/(1-2n))*I(n+1)

    What do I have to do next?
     
  8. Jul 29, 2010 #7

    arildno

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    Solve that equation for I(n+1) instead, and report back!
     
  9. Jul 29, 2010 #8
    I(n+1) = x/((x^2+1)^(n+1)) + 2(n+1)*(I(n+1)-I(n+2))
     
  10. Jul 29, 2010 #9

    arildno

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    No!
    Just rearrange, and get:

    [tex]I_{n+1}=\frac{1}{2n}\frac{x}{(1+x^{2})^{n}}+\frac{2n-1}{2n}I_{n}[/tex]
    Verify this, and then see if you manage the last step in the derivation on your own.
     
  11. Jul 29, 2010 #10
    Ok, now I see it, you have to change n into n-1 and than you get the result
    Thank you very much!
     
  12. Jul 29, 2010 #11

    arildno

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    Right!
    Note that in our derivation n=0 is prohibited, and by the index change this entails prohibition on n=1 instead
    My pleasure! :smile:
     
  13. Jul 29, 2010 #12
    n ∈ ℕ0

    with ℕ0 , I meant: 'all the natural numbers except for 0'
     
  14. Jul 29, 2010 #13

    arildno

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    I know.

    Note, however, that with just prior to dividing with 2n, we have the equation:

    2nI(n+1)=x/(1+x^2)^{n}+(2n-1)I(n)

    Note that setting n=0 here yields: 0=x-I(0), which is correct, up to an arbitrary constant C...
     
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