Integral $\int_{0}^{1}\ln x \, dx$ I have a question about the integral of $\ln x$.
When I try to calculate the integral of $\ln x$ from 0 to 1, I always get the following result.


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*$\int_0^1 \ln x = x(\ln x -1) |_0^1 = 1(\ln 1 -1) - 0 (\ln 0 -1)$
Is the second part of the calculation indeterminate or 0?
What am I doing wrong?
Thanks
Joachim G.

Looking sideways at the graph of $\log(x)$ you can also see that $$\int_0^1\log(x)dx = -\int_0^\infty e^{-x}dx = -1.$$
