How to integrate $\int_{0}^{1}\ln\left(\, x\,\right)\,{\rm d}x$? I encountered this integral in the quantum field theory calculation. Can I do this: 
$$
\left. \int_{0}^{1}\ln\left(\, x\,\right)\,{\rm d}x
=x\ln\left(\, x\,\right)\right\vert_{0}^{1}
-\int_{0}^{1}\,{\rm d}x
=\left. x\ln\left(\, x\,\right)\right\vert_{\, x\ =\ 0}\ -\ 1
$$
So the first term looks divergent. But Mathematica gives finite result and the integral is $-1$. Why isn't the first term divergent ?.

Use L'Hopital's Rule to resolve the indeterminate form: $$\lim_{x\to 0^+}x\ln x=\lim_{x\to 0^+}{\ln x\over x^{-1}}=\lim_{x\to 0^+}{x^{-1}\over -x^{-2}}=\lim_{x\to 0^+}(-x)=0$$
