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I was given this integral:

$$\int\frac{\ln {(1-t)}}{3t}\,\mathrm dt$$

I am having trouble with writing this as a power series, and I'm not sure where to start. I know I need the Maclaurin expansion of $\ln {(1-t)}$, but I'm not sure how to do this. Thanks!

José Carlos Santos
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johnners17
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    There should be no difficulty finding the Maclaurin series for $\ln {(1-t)}$. Its derivative, up to sign, is just $\frac 1{1-t}$ which is the usual Geometric Series, so correct the sign and integrate that term by term. – lulu May 04 '19 at 17:53
  • so then what should i do to find the power series? – johnners17 May 04 '19 at 17:53
  • Go step by step. Start with $\ln {(1-t)}$. Then divide that by $t$. Then integrate term by term. – lulu May 04 '19 at 17:54
  • @lulu so then what would the function look like? How would I set it up after I do that? – johnners17 May 04 '19 at 17:54
  • Step by step, as I say. – lulu May 04 '19 at 17:55
  • @lulu wait, I'm confused. So I take the derivative of $ln(1-t)$, then where would that go in the integral? Would it look like $$\int\frac{dt}{(1-t)(3t)}$$ ? – johnners17 May 04 '19 at 17:56
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    Please go step by step. Don't just demand the full solution at once. I gave you a recipe for finding the Maclaurin expansion of $\ln {(1-t)}$. Just do that first. – lulu May 04 '19 at 17:57
  • @lulu I got the expansion to be:

    $$t+\frac{t^2}{2}+\frac{t^3}{3}$$

     – johnners17 May 04 '19 at 18:00
  • I think you are just taking that from the posted solution below (you also copied the sign error). – lulu May 04 '19 at 18:06

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Since$$\frac1{1-t}=1+t+t^2+t^3+\cdots,$$you have$$\log(1-t)=-\left(t+\frac{t^2}2+\frac{t^3}3+\cdots\right)$$and therefore\begin{align}\int\frac{\log(1-t)}{3t}\,\mathrm dt&=-\frac13\int1+\frac t2+\frac{t^2}3+\cdots\,\mathrm dt\\&=-\frac13\left(t+\frac{t^2}{2^2}+\frac{t^3}{3^2}+\cdots\right).\end{align}

José Carlos Santos
  • 427,504