Integral $\int_{0}^{1}\ln x \, dx$

Question

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? – Fabian Dec 25 '12 at 09:23 — Fabian, Dec 25 '12 at 09:23

score 22 · Answer 1 · edited Oct 15 '22 at 04:52

22

$\ln x$ 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?

edited Oct 15 '22 at 04:52

Rócherz

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answered Dec 25 '12 at 09:23

Nameless

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Thanks for the quick response. Yes i can do that :) – user54188 Dec 25 '12 at 09:31
@nameless The integral is perfectly defined as a Lebesgue integral. – Mark Viola Oct 15 '22 at 05:06

score 21 · Answer 2 · answered Dec 25 '12 at 09:53

21

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

answered Dec 25 '12 at 09:53

WimC

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+1 though I see $$- \int_{-\infty}^{0} e^x , dx =-1$$ – Henry Dec 25 '12 at 10:05
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@Henry Did you use a mirror? ;-) But sure, that works too. – WimC Dec 25 '12 at 11:06
quick question why is it negative? – EM4 Sep 03 '23 at 04:09

Integral $\int_{0}^{1}\ln x \, dx$

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