representation of a power series

**lllll** · April 10th 2008, 03:30 PM

express the following as a power series:

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

so far I have

$\mbox{let} \ g(t) = ln(1-t) \rightarrow g(t) = \int g'(t) dt = \int -\frac{1}{1-t} \ dt= \int - \sum_{n=0}^{\infty} t^n \ dt$

$- \sum_{n=0}^{\infty} \frac{t^{n-1}}{n} = - \sum_{n=1}^{\infty} \frac{t^{n}}{n}$

now putting it back into the original:

$\int - \sum_{n=1}^{\infty} \frac{t^n}{n} \cdot \frac{1}{t} \ dt = \int -\frac{1}{n} \sum_{n=1}^{\infty} t^{n-1} \ dt = -\frac{1}{n} \int \sum_{n=1}^{\infty} t^{n-1} \ dt$

$-\frac{1}{n} \int \sum_{n=1}^{\infty} t^{n-1} \ dt = -\frac{1}{n} \sum_{n=1}^{\infty} \frac{t^{n-2}}{n-1} = -\sum_{n=1}^{\infty} \frac{t^{n-2}}{n^2-n}$

is this correct?

**Mathstud28** · April 10th 2008, 03:33 PM

Originally Posted by lllll

express the following as a power series:

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

so far I have

$\mbox{let} \ g(t) = ln(1-t) \rightarrow g(t) = \int g'(t) dt = \int -\frac{1}{1-t} \ dt= \int - \sum_{n=0}^{\infty} t^n \ dt$

$- \sum_{n=0}^{\infty} \frac{t^{n-1}}{n} = - \sum_{n=1}^{\infty} \frac{t^{n}}{n}$

now putting it back into the original:

$\int - \sum_{n=1}^{\infty} \frac{t^n}{n} \cdot \frac{1}{t} \ dt = \int -\frac{1}{n} \sum_{n=1}^{\infty} t^{n-1} \ dt = -\frac{1}{n} \int \sum_{n=1}^{\infty} t^{n-1} \ dt$

$-\frac{1}{n} \int \sum_{n=1}^{\infty} t^{n-1} \ dt = -\frac{1}{n} \sum_{n=1}^{\infty} \frac{t^{n-2}}{n-1} = -\sum_{n=1}^{\infty} \frac{t^{n-2}}{n^2-n}$

is this correct?

Yes as far as I can see it is correct good job

**Mathstud28** · April 10th 2008, 03:36 PM

Originally Posted by lllll

express the following as a power series:

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

so far I have

$\mbox{let} \ g(t) = ln(1-t) \rightarrow g(t) = \int g'(t) dt = \int -\frac{1}{1-t} \ dt= \int - \sum_{n=0}^{\infty} t^n \ dt$

$- \sum_{n=0}^{\infty} \frac{t^{n-1}}{n} = - \sum_{n=1}^{\infty} \frac{t^{n}}{n}$

now putting it back into the original:

$\int - \sum_{n=1}^{\infty} \frac{t^n}{n} \cdot \frac{1}{t} \ dt = \int -\frac{1}{n} \sum_{n=1}^{\infty} t^{n-1} \ dt = -\frac{1}{n} \int \sum_{n=1}^{\infty} t^{n-1} \ dt$

$-\frac{1}{n} \int \sum_{n=1}^{\infty} t^{n-1} \ dt = -\frac{1}{n} \sum_{n=1}^{\infty} \frac{t^{n-2}}{n-1} = -\sum_{n=1}^{\infty} \frac{t^{n-2}}{n^2-n}$

is this correct?

I do see one problem $\int{t^{n-1}}dt=\frac{t^n}{n}$

**Krizalid** · April 10th 2008, 03:56 PM

Originally Posted by lllll

$- \sum_{n=0}^{\infty} \frac{t^{n-1}}{n} = - \sum_{n=1}^{\infty} \frac{t^{n}}{n}$

Careful here.

The thing is, you should have memorized the power series representation for $\ln(1-x),$ that is $-\sum_{k\,=\,1}^\infty\frac{x^k}k$ for $|x|<1.$ From here you can easily find $\frac{\ln(1-x)}x$ as a series, then interchange sum by integral and you're done.

There's no much here, simple calculations, it's easy.

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