Integral of x ln(x)
$$\int_0^1{x\ln(x)dx}$$
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?
$\endgroup$ 35 Answers
$\begingroup$By differentiation under the integral sign, $$ \int_{0}^{1} x\log(x)\,dx = \frac{d}{d\alpha}\left.\int_{0}^{1}x^\alpha\,dx\right|_{\alpha=1} =\frac{d}{d\alpha}\left.\left(\frac{1}{\alpha+1}\right)\right|_{\alpha=1}=-\frac{1}{(1+1)^2}=\color{red}{-\frac{1}{4}}.$$
$\endgroup$ $\begingroup$Here is a fun way:
$$\log(x)=-\int_1^{1/x}\frac1t\ dt$$
Thus,
$$\begin{align}\int_0^1x\log(x)\ dx&=-\int_0^1\int_1^{1/x}\frac xt\ dt\ dx\\\text{interchange of integrals}&=-\int_1^\infty\int_0^{1/t}\frac xt\ dx\ dt\\&=-\int_1^\infty\frac{x^2}{2t}\bigg|_{x=0}^{x=1/t}\ dx\\&=-\int_1^\infty\frac1{2t^3}\ dx\\&=+\frac1{4t^2}\bigg|_{t=1}^\infty\\&=-\frac14\end{align}$$
$\endgroup$ 1 $\begingroup$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
$$t\to-\infty\implies e^{2t}t\to0.$$
$\endgroup$ 3 $\begingroup$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
$$\begin{align} \int_0^1 x\log(x)\,dx&\equiv \lim_{\epsilon\to 0}\int_\epsilon^1 x\log(x)\,dx\\\\ &=-\frac14 \end{align}$$
$\endgroup$ $\begingroup$By parts:
$$\begin{cases}u=\log x&u'=\frac1x\\{}\\v'=x&v=\frac12x^2\end{cases}\implies\int_0^1 x\log x\,dx=\overbrace{\left.\frac12x^2\log x\right|_0^1}^{=0\,,\text{ taking limits}}-\frac12\int_0^1x\,dx=-\frac14$$
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