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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.

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  • $\begingroup$ How did you conclude that the second part is indeterminate? $\endgroup$
    – Fabian
    Commented Dec 25, 2012 at 9:23

2 Answers 2

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$\ln x$ 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?

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    $\begingroup$ Thanks for the quick response. Yes i can do that :) $\endgroup$
    – user54188
    Commented Dec 25, 2012 at 9:31
  • $\begingroup$ @nameless The integral is perfectly defined as a Lebesgue integral. $\endgroup$
    – Mark Viola
    Commented Oct 15, 2022 at 5:06
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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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    $\begingroup$ +1 though I see $$- \int_{-\infty}^{0} e^x \, dx =-1$$ $\endgroup$
    – Henry
    Commented Dec 25, 2012 at 10:05
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    $\begingroup$ @Henry Did you use a mirror? ;-) But sure, that works too. $\endgroup$
    – WimC
    Commented Dec 25, 2012 at 11:06
  • $\begingroup$ quick question why is it negative? $\endgroup$
    – EM4
    Commented Sep 3, 2023 at 4:09

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