Integral of x ln(x)

Question

$$\int_0^1{x\ln(x)dx}$$

I tried to evaluate it but I got stuck at this part - $\lim\limits_{x\mapsto 0}\log(x)$ I believe that is limit doesn't exist and hence it is impossible to evaluate the integral. Am I correct?

Youre looking at the limit from right side($x\to 0^{+}$). And that limit exists — kingW3, Mar 07 '17 at 22:22
$\int_0^1x\log(x)\mathrm dx$ is very easily shown to be finite... just look at the graph http://www.wolframalpha.com/input/?i=int_0%5E1+(x%5Clog(x)) — Brevan Ellefsen, Mar 07 '17 at 22:29
Recall the definition for improper integrals: $\int_{0}^{1}x\ln{x}dx=lim_{r\rightarrow0^+}\int_r^1x\ln{x}dx$, and that this integral exists (by definition) if the limit exists. You do have to take a limit that involves $\ln{x}$ but it should have some other terms with it.... — Jake, Mar 07 '17 at 22:31

score 3 · Answer 1 · answered Mar 07 '17 at 23:39

By differentiation under the integral sign, $$ \int_{0}^{1} x\log(x)\,dx = \frac{d}{d\alpha}\left.\int_{0}^{1}x^\alpha\,dx\right|_{\alpha=1} =\frac{d}{d\alpha}\left.\left(\frac{1}{\alpha+1}\right)\right|_{\alpha=1}=-\frac{1}{(1+1)^2}=\color{red}{-\frac{1}{4}}.$$

score 2 · Answer 2 · answered Mar 07 '17 at 23:38

2

Here is a fun way:

$$\log(x)=-\int_1^{1/x}\frac1t\ dt$$

Thus,

$$\begin{align}\int_0^1x\log(x)\ dx&=-\int_0^1\int_1^{1/x}\frac xt\ dt\ dx\\\text{interchange of integrals}&=-\int_1^\infty\int_0^{1/t}\frac xt\ dx\ dt\\&=-\int_1^\infty\frac{x^2}{2t}\bigg|_{x=0}^{x=1/t}\ dx\\&=-\int_1^\infty\frac1{2t^3}\ dx\\&=+\frac1{4t^2}\bigg|_{t=1}^\infty\\&=-\frac14\end{align}$$

answered Mar 07 '17 at 23:38

Simply Beautiful Art

74,685

you, i like you. – Dev Aggarwal Mar 09 '17 at 01:20

score 1 · Accepted Answer · 2017-03-07T23:00:46.137

1

By parts,

$$\int_0^1 x\log x\,dx=\left.\frac12x^2\log x\right|_0^1-\frac12\int_0^1 x\,dx=0-\frac14.$$

The only special term is $x^2\log x$ evaluated at $x=0$, giving an undeterminate form $0\cdot\infty$.

By L'Hospital,

$$\frac{\log x}{\dfrac1{x^2}}\to\frac{\dfrac1{x}}{-\dfrac2{x^3}}\to 0.$$

Alternatively, one may use the change of variable $x=e^t$ and

$$\int_0^1 x\log x\,dx=\int_{-\infty}^0 e^{2t}t\,dt=\left.\frac12e^{2t}t\right|_{-\infty}^0-\int_{-\infty}^0 e^{2t}\,dt=-\left.\frac14e^{2t}\right|_{-\infty}^0.$$

Here we need

$$t\to-\infty\implies e^{2t}t\to0.$$

edited Mar 07 '17 at 23:00

answered Mar 07 '17 at 22:50

1

@Jake Observe the difference between both posts: 1 minute...or even less. This is so obvious a solution that it isn't strange two or more post the same almost at the same time. – DonAntonio Mar 07 '17 at 22:53
@jake: as often arises, the answers crossed. – Mar 07 '17 at 22:53
@DonAntonio Which is all the reason I love unique posts more and more. ;) – Simply Beautiful Art Mar 08 '17 at 00:28

score 1 · Answer 4 · answered Mar 07 '17 at 22:53

Note that for every $\epsilon>0$, $x\log(x)$ is continuous and hence integrable on $[\epsilon,1]$. Therefore, we have

$$\begin{align} \int_\epsilon^1 x\log(x)\,dx&=\left.\left(\frac14x^2(\log(x)-1)\right)\right|_{\epsilon}^{1}\\\\ &=\frac14 (\epsilon^2-2\epsilon^2\log(\epsilon)-1) \end{align}$$

Taking the limit as $\epsilon\to 0$ yields

$$\begin{align} \int_0^1 x\log(x)\,dx&\equiv \lim_{\epsilon\to 0}\int_\epsilon^1 x\log(x)\,dx\\\\ &=-\frac14 \end{align}$$

score 0 · Answer 5 · answered Mar 07 '17 at 22:49

0

By parts:

$$\begin{cases}u=\log x&u'=\frac1x\\{}\\v'=x&v=\frac12x^2\end{cases}\implies\int_0^1 x\log x\,dx=\overbrace{\left.\frac12x^2\log x\right|_0^1}^{=0\,,\text{ taking limits}}-\frac12\int_0^1x\,dx=-\frac14$$

answered Mar 07 '17 at 22:49

DonAntonio

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Integral of x ln(x)

5 Answers5