How can I integrate (1-x)ln(1-x) using integration by parts?

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

Thanks :)

 

[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.

 

[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.