# $\int_0^1 \log(x) dx = -1$ but lower Riemann sum is $-\infty$?

Consider

$\int_0^1 \log(x) dx = -1$,

where the answer $1$ was obtained by using the fundamental theorem of calculus - taking the antiderivative of $\log(x)$ and then evaluating at the endpoints.

But $\log (x)$ is an unbounded function on the interval $[0,1]$ so the lower Riemann sum will be $-\infty$ and hence the integral doesn't even exist.

So does mean that whenever we are using the fundamental theorem of calculus we are making the assumption that we are dealing with Lebesgue integration instead of Riemann integration?

• In order to apply the Riemann integral theory, you should recognize that this definite integral is improper because of the singularity at the lower limit of integration. – hardmath Apr 1 '17 at 14:21
• Remember that the Riemann integral is only defined for bounded functions on bounded domains, so this must be treated as an improper Riemann integral. – Chappers Apr 1 '17 at 14:23

It should be interpreted as an improper Riemann integral: $$\lim_{\epsilon\to 0^+} \int_\epsilon^1 \ln x \, dx .$$ For each $\epsilon \in (0,1]$, the lower Riemann sum is finite.