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Math Help - Indefinite Integral as Power Series

  1. #1
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    Indefinite Integral as Power Series

    I have no idea how to do this

    I have to write this integral:



    as a power series. Please help!
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  2. #2
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    Quote Originally Posted by phack View Post
    I have no idea how to do this

    I have to write this integral:



    as a power series. Please help!
    x Log[1+n^2]/n^2 + c

    unless you really want integral log(1+x^2) / x^2 dx

    Then we can write:

    log(1+x^2) = x^2 - x^4/2 + x^6/3 - ... + (-1)^(n+1) x^(2n)/n + ...

    for |x|<1.

    Then:

    integral log(1+x^2) / x^2 dx = sum_{n=1 to infnty} integral (-1)^{n+1}x^{2n-1}/n dx

    .................= sum_{n=1 to infnty} (-1)^{n+1}x^{2n}/(n (2n-1)

    RonL
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  3. #3
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    I'm not sure I'm following your steps. Why did you place an x infront in your first step?

    And how did you get this?...

    log(1+x^2) = x^2 - x^4/2 + x^6/3 - ... + (-1)^(n+1) x^(2n)/n + ...


    Please clarify.
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  4. #4
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    Quote Originally Posted by phack View Post
    I'm not sure I'm following your steps. Why did you place an x infront in your first step?

    And how did you get this?...

    log(1+x^2) = x^2 - x^4/2 + x^6/3 - ... + (-1)^(n+1) x^(2n)/n + ...


    Please clarify.
    Because,

    ln(1+x)= x - x^2/2+x^3/3 -... for -1<x<=1

    But you have,

    ln(1+x^2)

    Thus, instead of x you have x^2.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by phack View Post
    I'm not sure I'm following your steps. Why did you place an x infront in your first step?
    Because in your question you have an integral over x, but no x in the
    integrand. Therefore the integral is x times whatever you had under
    the integral sign (as this was independent of x). Therefore the first step
    is the answer to the question you asked. After that I proceeded to answer
    the question I assumed that you realy wanted to ask.

    And how did you get this?...

    log(1+x^2) = x^2 - x^4/2 + x^6/3 - ... + (-1)^(n+1) x^(2n)/n + ...


    Please clarify.
    As ImPerfectHacker says, I took a standard series representation of
    log(1+u) and evaluated it at u=x^2. You can find this series either by
    just knowing it, or taking the Taylor series of log(1+x) about 0, or by
    integrating term by term the series expansion of 1/(1+x).

    RonL
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