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!

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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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...
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook