How to integrate $\int_1^e \ln{x} \, dx$

Question

This seems really tricky to me. I can't figure out how to integrate $\ln x$.

Try to look it up in some integration table. That would help. — Hui Yu, Jan 16 '13 at 16:03

user 1591719 · Accepted Answer · 2013-01-16T18:01:38.047

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Based upon geometrical observation we write that

$$\int_1^e \ln x\space \mathrm{dx}=e-\int_0^1 e^x\space \mathrm{dx}=1$$

The question may also be viewed as a particular case of Young's inequality.

edited Jan 16 '13 at 18:01

answered Jan 16 '13 at 16:13

user 1591719

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That is MUCH nicer (IMHO) then the regular integration by parts trick! – Amihai Zivan Jan 16 '13 at 16:21
@AmihaiZivan: thanks! :-) – user 1591719 Jan 16 '13 at 16:22
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@Chris'ssister I wonder what did you do here? Can you please explain? – xyres Jan 16 '13 at 16:44
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@weasel: I simply used the inverse function of $\ln x$. Make a drawing and see what happens in the rectangle $[1, e] \times [0, 1]$ if you employ the inverse function. – user 1591719 Jan 16 '13 at 16:47
Thank you. I learned a new method, here. – xyres Jan 22 '13 at 15:18

score 7 · Answer 2 · answered Jan 16 '13 at 16:04

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Hint Integration by parts is helpful.

Let $u=$ln$x$ , $dv=1dx$
so $du=\frac{1}{x} dx$ , $v=x$

so that $uv$ - $\int_{1}^{e}vdu$ = $x$ln$x \mid_{1}^{e}- \int_{1}^{e} dx$

answered Jan 16 '13 at 16:04

Rustyn

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score 2 · Answer 3 · answered Jan 16 '13 at 17:57

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$$ \int\ln x\,dx=\int u\,dx = xu-\int x\,du = x\ln x - \int x\left(\frac1x\,dx\right). $$ Now here's the hard part: $x\cdot\dfrac1x$ simplifies to $1$. At least, I've seen lots of students get stuck on that part. Some of them want to antidifferentiate $x$ and also $\dfrac1x$.

So you've got $$ x\ln x-\int 1\,dx. $$ You can probably do the rest.

answered Jan 16 '13 at 17:57

Michael Hardy

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(+1) I remember somewhere some answer was upvoted for the humour in answer. – 007resu Jan 16 '13 at 18:23

score 0 · Answer 4 · answered Jan 16 '13 at 17:01

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Change variable formula $ \int_{\varphi(a)}^{\varphi(b)}f (x) \, dx = \int_a^b f(\varphi(u))\varphi^\prime (u) du$ \begin{align} \int_{1}^{e}\ln (x)\, dx = & \int_{e^{0}}^{e^{1}}\ln (x) \, dx \\ = & \int_0^1\ln( e^u ) (e^u)^\prime du \\ = & \int_0^1 u \cdot e^u du \\ \end{align} Now use integration by parts.

answered Jan 16 '13 at 17:01

Elias Costa

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And what is u here? – xyres Jan 17 '13 at 03:26
Reminds me of my favorite 0=1 proof! Let's do $\int_{}^{}\frac{1}{x}dx$ by partial integration! $\int_{}^{}\frac{1}{x}dx=\frac{1}{x}\cdot x -\int_{}^{}x d\frac{1}{x}=1+\int_{}^{}\frac{x}{x^2}dx=1+\int_{}^{}\frac{1}{x}dx$ Apparently $\int_{}^{}\frac{1}{x}dx=1+\int_{}^{}\frac{1}{x}dx$$$
Therefore $0=1$ I always tell my students that finding the error helps to understand integration by parts. I learned this proof 50 years ago. It is still fun!
– Paul vdVeen Sep 19 '22 at 17:48

How to integrate $\int_1^e \ln{x} \, dx$

4 Answers4