# Integrating a logarithm

1. May 4, 2008

### Bunting

Hello

So I have a problem, which is to use integration by parts to integrate...

$$\int^{1}_{0}(1-x) ln (1-x) dx$$

The way I have been working is it to seperate it out into just...

$$\int^{1}_{0}ln (1-x) dx - \int^{1}_{0}x ln (1-x) dx$$

and then integrating by parts on each of these seperatele, but for instance if I integrate by parts the first bit, I get...

$$[xln(1-x)]^{1}_{0} + \int^{1}_{0}x \frac{1}{1-x}$$

And im thinking the first part to this doesn't make sense, because $$ln (0)$$ is a mathematical nono. So im confused with regard to this problem - has anybody any decent suggestions on how to do this?

Thanks :)

2. May 4, 2008

### Santa1

The integral isn't proper (you're working right up to a pole!) so you must take limit to 1, not just put in the values.

3. May 4, 2008

### Nedeljko

Integral is proper because the function $$(1-x)\ln(1-x)$$ is continuous and bounded on $$[0,1)$$ (it has finite limit when $$x\rightarrow 1$$).

$$\int_0^1(1-x)\ln(1-x)\,dx=\left.-\frac{(1-x)^2\ln(1-x)}{2}\right|_0^1-\int_0^1\frac{1-x}{2}\,dx=\left.\frac{(1-x)^2}{4}\right|_0^1=-\frac{1}{4}.$$

Your problem is in decomposing the finite value in the form of substraction of two infinite values.