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Power Series of Integral xarctan(3x)

  1. Oct 25, 2009 #1
    1. The problem statement, all variables and given/known data

    Evaluate the integral of xarctan(3x) from 0 to 0.1 by expressing the integral in terms of a power series.

    2. Relevant equations



    3. The attempt at a solution

    I differentiated xarctan(3x) until I got two functions that I could turn into power series (arctan(3x) and x(1/(1+9x^2)).

    After I found the power series of these two functions I integrated them as many times as was necessary to arrive back at the integral of the original function and got (both as sums from one to infinity):
    [(-1)^(n-1)*9^(n-1)*x^(2n+1)]/[(2n+1)(2n)(2n-1)] + [(-1)^(n-1)*9^(n-1)*x^(2n)]/[(2n)(2n-1)]

    When I evaluated it from 0 to 0.1, I ended up with 0.0050927 which is not the correct answer, but I don't see where I would have done it wrong. Any help would be appreciated.
     
  2. jcsd
  3. Oct 25, 2009 #2

    zcd

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    A faster way to find the power series would be knowing [tex]\arctan{u}=\sum_{n=0}^{\infty}(-1)^{n}\frac{u^{2n+1}}{2n+1}[/tex] convergent for [tex]|x|\leq{1}[/tex]. [tex]x\arctan{3x}=\sum_{n=0}^{\infty}(-1)^{n}\frac{3^{2n+1}}{2n+1}x^{2(n+1)}[/tex]. I haven't checked your work but maybe you made a mistake while differentiating.

    On a separate note, [tex]\int \frac{x}{1+9x^{2}}\,dx=\frac{1}{18}\ln{|1+9x^{2}|}+C[/tex], not the arctan you were looking for.
     
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