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.
- $\int_0^1 \ln x = x(\ln x -1) |_0^1 = 1(\ln 1 -1) - 0 (\ln 0 -1)$
Is the second part of the calculation indeterminate or 0?
What am I doing wrong?
Thanks Joachim G.
$\endgroup$ 12 Answers
$\begingroup$$\ln x$ is not defined at $0$ so the integral $$\int_0^1\ln x\, dx$$ is improper. Thus, $$\int_0^1\ln x\, dx=(x\ln x -x)|_0^1=1\ln 1-1-\lim_{x\to 0}x\ln x-x=-1+\lim_{x\to 0}x\ln x-x$$ We need to evaluate $$\lim_{x\to 0}x\ln x=\lim_{x\to 0}\frac{\ln x}{\frac1x}$$ Can you do that?
$\endgroup$ 1 $\begingroup$Looking sideways at the graph of $\log(x)$ you can also see that $$\int_0^1\log(x)dx = -\int_0^\infty e^{-x}dx = -1.$$
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