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

  1. Oct 30, 2011 #1
    1. The problem statement, all variables and given/known data
    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. Relevant 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?
     
  2. jcsd
  3. Oct 30, 2011 #2

    Dick

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    Homework Helper

    It's the definition of an indefinite integral with a singularity at one end of the interval of integration.
     
  4. Oct 30, 2011 #3
    Thank you for your help.
     
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