# Thread: Power Series of -ln(1-x)/x?

1. ## Power Series of -ln(1-x)/x?

From the list of basic power series: $\displaystyle -ln(1-x) = x + \frac{x^2}{2} +\frac{x^3}{3} + \frac{x^4}{4} +...$

How do you write a power series for $\displaystyle f(x)= \frac{-ln(1-x)}{x}$ ?

Using the result how do you write a series expression for the $\displaystyle \int_{0}^{1} \frac{-ln(1-x)}{x}dx$ ?

I started out by looking at the theorem for power series expansion and finding the derivatives of f(x), but that turned out to be a mess. Help is much appreciated!

2. Instead of solving for the second power series directly just divide the power series you have by x.

3. Originally Posted by blgonz03
From the list of basic power series: $\displaystyle -ln(1-x) = x + \frac{x^2}{2} +\frac{x^3}{3} + \frac{x^4}{4} +...$

How do you write a power series for $\displaystyle f(x)= \frac{-ln(1-x)}{x}$ ?

Using the result how do you write a series expression for the $\displaystyle \int_{0}^{1} \frac{-ln(1-x)}{x}dx$ ?

I started out by looking at the theorem for power series expansion and finding the derivatives of f(x), but that turned out to be a mess. Help is much appreciated!
$\displaystyle -ln(1-x) = x + \frac{x^2}{2} +\frac{x^3}{3} + \frac{x^4}{4} +...$

$\displaystyle \frac{-ln(1-x)}{x} = 1 + \frac{x}{2} +\frac{x^2}{3} + \frac{x^3}{4} +...$

4. Originally Posted by skeeter
$\displaystyle -ln(1-x) = x + \frac{x^2}{2} +\frac{x^3}{3} + \frac{x^4}{4} +...$

$\displaystyle \frac{-ln(1-x)}{x} = 1 + \frac{x}{2} +\frac{x^2}{3} + \frac{x^3}{4} +...$

but why?

$\displaystyle \displaystyle \int_{0}^{1} \frac{-ln(1-x)}{x}dx = \int_{0}^{1} ln(1-x) \cdot -\frac{1}{x}dx = \int_0^{-\infty} u \, du$

... which diverges.
But, it doesn't diverg. It equals $\displaystyle \frac{-\pi^2}{6}$

5. Originally Posted by Drexel28
But, it doesn't diverg. It equals $\displaystyle \frac{-\pi^2}{6}$
that would be the series ... $\displaystyle \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$

6. Originally Posted by skeeter
that would be the series ... $\displaystyle \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$
Yes, and using some analytic methods one can show that $\displaystyle \displaystyle \int_0^1\frac{\ln(1-x)}{x}dx=\int_0^1\sum_{n=1}^{\infty}\frac{x^{n-1}}{n}=\sum_{n=1}^{\infty}\int_0^1\frac{x^{n-1}}{n}dx=\sum_{n=1}^{\infty}\frac{1}{n^2}$

7. Originally Posted by Drexel28
Yes, and using some analytic methods one can show that $\displaystyle \displaystyle \int_0^1\frac{\ln(1-x)}{x}dx=\int_0^1\sum_{n=1}^{\infty}\frac{x^{n-1}}{n}=\sum_{n=1}^{\infty}\int_0^1\frac{x^{n-1}}{n}dx=\sum_{n=1}^{\infty}\frac{1}{n^2}$
learn something new everday ... saw that right after when I went back and integrated the power series.