# Help with evaluating indefinite integral as power series?

• Apr 13th 2009, 07:47 PM
virtuoso735
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.
• Apr 13th 2009, 08:04 PM
TheEmptySet
Quote:

Originally Posted by virtuoso735
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...