Integrating a logarithm

  • Thread starter Bunting
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  • #1
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Hello

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

[tex]\int^{1}_{0}(1-x) ln (1-x) dx[/tex]

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

[tex]\int^{1}_{0}ln (1-x) dx - \int^{1}_{0}x ln (1-x) dx[/tex]

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

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

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

Thanks :)
 

Answers and Replies

  • #2
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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
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Integral is proper because the function [tex](1-x)\ln(1-x)[/tex] is continuous and bounded on [tex][0,1)[/tex] (it has finite limit when [tex]x\rightarrow 1[/tex]).

[tex]\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}.[/tex]

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

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