Convergence of $\int_0^1 x\log(x)\,dx$

Question

$$\int_0^1 x\log(x)\,dx$$

I wonder if this integral is convergent or not.

If we evaluate the integral directly, it is

$$\left[\frac{x^2\log x}{2}-\frac{x^2}{4}\right]_0^1= -\frac{1}{4}-\lim_{a \to 0}\left(\frac{a^2\log a}{2}-\frac{a^2}{4}\right)=-\frac{1}{4}$$

Am I correct with this calculation? If I am, integral converges. What I wonder more is that are there any other methods to find whether it converges or not, such as limit or Dirichlet test?

Answer 1

Yes, the integral is convergent because the integrand can be extended at $0$ to a continuous function in $[0,1]$: $$\lim_{x\to 0^+}x\log(x)\stackrel{x=1/t}{=}\lim_{t\to +\infty}-\frac{\log(t)}{t}=0.$$ Your computation is correct.

Answer 2

Another way to prove the convergence is to prolongate the function by:

$f: x \rightarrow f(x)=xlog(x)$ if $x>0$

$f(0)=0$

$f$ is continuous on the segment [0,1] (can you see why ?), hence its integral converges.

Answer 3

$$\forall \alpha>-1,\qquad g(\alpha)=\int_{0}^{1}x^\alpha\,dx = \frac{1}{\alpha+1}$$ hence by differentiating with respect to $\alpha$ $$\forall\alpha>-1,\qquad g'(\alpha) = \int_{0}^{1} x^\alpha\log(x)\,dx = -\frac{1}{(\alpha+1)^2} $$ and by evaluating at $\alpha=1$ $$ \int_{0}^{1} x\log(x)\,dx = -\frac{1}{(1+1)^2} = \color{red}{-\frac{1}{4}} .$$