# 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.

• $$\int_0^1 \ln x = x(\ln x -1) |_0^1 = 1(\ln 1 -1) - 0 (\ln 0 -1)$$

Is the second part of the calculation indeterminate or 0?

What am I doing wrong?

Thanks Joachim G.

• How did you conclude that the second part is indeterminate? Commented Dec 25, 2012 at 9:23

$$\ln x$$ is not defined at $$0$$ so the integral $$\int_0^1\ln x\, dx$$ is improper. Thus, \begin{align} \int_0^1\ln x\, dx &= (x\ln x -x)|_0^1 \\ &=1\ln 1-1-\lim_{x\to 0}x\ln x-0 \\ &= -1+\lim_{x\to 0}x\ln x. \end{align} We need to evaluate $$\lim_{x\to 0}x\ln x=\lim_{x\to 0}\frac{\ln x}{\frac1x}.$$ Can you do that?

• Thanks for the quick response. Yes i can do that :) Commented Dec 25, 2012 at 9:31
• @nameless The integral is perfectly defined as a Lebesgue integral. Commented Oct 15, 2022 at 5:06

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

• +1 though I see $$- \int_{-\infty}^{0} e^x \, dx =-1$$ Commented Dec 25, 2012 at 10:05
• @Henry Did you use a mirror? ;-) But sure, that works too.
– WimC
Commented Dec 25, 2012 at 11:06
• quick question why is it negative?
– EM4
Commented Sep 3, 2023 at 4:09