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Math Help - indefinate integral/power series

  1. #1
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    indefinate integral/power series

    integral (x - arctan^x)/(x^3) dx

    need to evaluate the integral and what is the radius of convergence?

    thanks if you can show this one.
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  2. #2
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    Quote Originally Posted by rcmango View Post
    integral (x - arctan^x)/(x^3) dx

    need to evaluate the integral and what is the radius of convergence?

    thanks if you can show this one.
    Power series for arctan

    \tan^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + -\cdots

    so

    \frac{x -\tan^{-1} x}{x^3} = \frac{ \frac{x^3}{3} - \frac{x^5}{5} + \frac{x^7}{7} - + \cdots}{x^3}

    = \frac{1}{3} - \frac{x^2}{5} + \frac{x^4}{7} - \frac{x^6}{9} \cdots

    so

    \int \frac{x -\tan^{-1} x}{x^3}\,dx = \int \frac{1}{3} - \frac{x^2}{5} + \frac{x^4}{7} - \frac{x^6}{9} \cdots \,dx= \frac{x}{3} - \frac{x^3}{3 \cdot 5} + \frac{x^5}{5 \cdot 7} - \frac{x^7}{7 \cdot 9} \cdots

    or

    \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{2n-1}}{(2n-1)(2n+1)}

    Then use the ratio test on this series to find the radius of converge. For the interval, check the endpoints.
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  3. #3
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    Hey thanks, your explanation was very nice, and it helped me understand what was really going on.
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