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Math Help - Help with evaluating indefinite integral as power series?

  1. #1
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    Help with evaluating indefinite integral as power series?

    How would I solve this problem using power series?

    Integral of [ln(1-t)]/t dt? I know that if I find the power series for [ln(1-t)]/t, then I could integrate each term separately and then that would be the solution, but how do I find the power series for such an expression? I know that the derivative of ln(x) is 1/x and the integral of 1/x is ln(x), and it seems like that would somehow apply in this problem, but I'm not sure how. Help please?

    Thanks.
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  2. #2
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    Quote Originally Posted by virtuoso735 View Post
    How would I solve this problem using power series?

    Integral of [ln(1-t)]/t dt? I know that if I find the power series for [ln(1-t)]/t, then I could integrate each term separately and then that would be the solution, but how do I find the power series for such an expression? I know that the derivative of ln(x) is 1/x and the integral of 1/x is ln(x), and it seems like that would somehow apply in this problem, but I'm not sure how. Help please?

    Thanks.
    Let

    f(t)=\ln(1-t) so we can take the derivative to get

    f'(t)=\frac{-1}{1-t}=-\sum_{n=0}^{\infty}t^{n}

    now we can integrate this to find a series for f(t)

    f(t)=-\sum_{n=0}^{\infty}\frac{t^{n+1}}{n+1}

    so we can use this to find the series we need

    \frac{\ln(t-1)}{t}=\frac{f(t)}{t}=\frac{-\sum_{n=0}^{\infty}\frac{t^{n+1}}{n+1}}{t}=-\sum_{n=0}^{\infty}\frac{t^{n}}{n+1}

    You should be able to finish from here...
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