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

[fixed_point]

Homework Statement

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

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

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

 

[Dick]

It's the definition of an indefinite integral with a singularity at one end of the interval of integration.

 

[fixed_point]

It's the definition of an indefinite integral with a singularity at one end of the interval of integration.

Thank you for your help.