Integral of xln(x) with 0 as a limit of integration

  • #1

Homework Statement


Not exactly a homework problem, but I noticed that wolfram alpha gives
[itex]\displaystyle{\int_{0}^{1}}x\ln{x}\,dx=-\frac{1}{4}[/itex]
I was wondering why this is so.

2. Homework Equations & solution attempt
For [itex]\epsilon>0[/itex]
[itex]\displaystyle{\int_{\epsilon}^{1}}x\ln{x}\,dx=-\frac{1}{4}-\left(\frac{1}{4}\epsilon^2(2\ln{\epsilon}-1)\right)[/itex]
Taking [itex]\epsilon \to 0[/itex] gives the answer provided by wolfram alpha.

My question is what justifies this last step of taking [itex]\epsilon \to 0[/itex]? Continuity?
 

Answers and Replies

  • #2
Dick
Science Advisor
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It's the definition of an indefinite integral with a singularity at one end of the interval of integration.
 
  • #3
It's the definition of an indefinite integral with a singularity at one end of the interval of integration.
Thank you for your help.
 

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