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 ?.
 A: Use L'Hopital's Rule to resolve the indeterminate form: $$\lim_{x\to 0^+}x\ln x=\lim_{x\to 0^+}{\ln x\over x^{-1}}=\lim_{x\to 0^+}{x^{-1}\over -x^{-2}}=\lim_{x\to 0^+}(-x)=0$$
A: Hint: Use L'Hopital's rule to evaluate $\lim_{x\to0^+}x\ln x$.
A: Or alternatively, use the substitution $x=e^y$.
A: interpret $\int_0^1 \ln x dx$ as the signed area of the region bounded by $x = 0, y= \ln(x)$ and $y = 0$ this is the negative of the area of the region bounded by $x = e^y, y = 0$ and $x = 0$(interchange $x$ and $y$ in the previous boundary). the area of latter region is $\int_{-\infty}^0 e^y dy = 1.$  so $\int_0^1 \ln(x) dx = -1.$
another way to think of the same thing is in the first area computation you have the infinitesimal rectangles parallel to the $y$-axis in the latter the infinitesimal rectangles parallel to $x$-axis.
A: We have
$$\begin{gathered}
  \int\limits_0^1 {\ln xdx}  = \mathop {\lim }\limits_{\varepsilon  \to {0^ + }} \int\limits_\varepsilon ^1 {\ln xdx}  = \mathop {\lim }\limits_{\varepsilon  \to {0^ + }} \left( {\left. {\left( {x\ln x} \right)} \right|_{x = \varepsilon }^{x = 1} - \int\limits_\varepsilon ^1 {dx} } \right) = \mathop {\lim }\limits_{\varepsilon  \to {0^ + }} \left( { - \varepsilon \ln \varepsilon  - 1 + \varepsilon } \right) \hfill \\
   =  - \mathop {\lim }\limits_{\varepsilon  \to {0^ + }} \varepsilon \ln \varepsilon  - 1. \hfill \\ 
\end{gathered}$$
On the other hand, by the L'Hospital rule, we have
$$\mathop {\lim }\limits_{\varepsilon  \to {0^ + }} \varepsilon \ln \varepsilon  = \mathop {\lim }\limits_{\varepsilon  \to {0^ + }} \frac{{\ln \varepsilon }}
{{\frac{1}
{\varepsilon }}} = \mathop {\lim }\limits_{\varepsilon  \to {0^ + }} \frac{{\frac{1}
{\varepsilon }}}
{{ - \frac{1}
{{{\varepsilon ^2}}}}} = \mathop {\lim }\limits_{\varepsilon  \to {0^ + }} \left( { - \varepsilon } \right) = 0.$$
So
$$\int\limits_0^1 {\ln xdx}  =  - 1.$$
