Show that $\int_0^1 \frac{x\log x}{(1+x)^2} dx$ is Lebesgue integrable

Question

I want to show that $\int_0^1 \frac{x\log x}{(1+x)^2} dx$ is Lebesgue integrable. I pick some $\delta\in (0,1)$ and rewrite

$$\int_0^1 \frac{x\log x}{(1+x)^2} dx=\int_0^\delta \frac{x\log x}{(1+x)^2} dx+ \int_\delta^1 \frac{x\log x}{(1+x)^2} dx.$$

Since $f(x)=\frac{x\log x}{(1+x)^2}$ is continuous on the compact interval $[\delta, 1]$, then there exists some $M \in \mathbb{R}$ such that $f(x) \leq M$ for all $x \in [\delta, 1]$. Thus, it follows that $\int_\delta^1 \frac{x\log x}{(1+x)^2} dx$ is integrable.

I am having trouble with $\int_0^\delta \frac{x\log x}{(1+x)^2} dx$. The issue is that $\log x\rightarrow -\infty$ as $x \rightarrow 0$. Some help, please? Thank you in advance.

Answer 1

Use the fact that $x\log x\to 0$ as $x\to 0^+$. You can prove this using Hopital's rule.

Answer 2

This is Riemann integrable, and thus Lebesgue integrable with the same value.

Note that on the given interval, $x\log x\leq 0$, thus $$ x\log x\leq \frac{x\log x}{(1+x)^2}\leq \frac{x\log x}{2}\\ \implies \int_0^1 x\log x \mathrm dx \leq \int_0^1 \frac{x\log x}{(1+x)^2}\mathrm dx\leq \int_0^1\frac{x\log x}{2}\mathrm dx $$ by the monotonicity of the integral. Furthermore, we have $$ \int_0^1x\log x\mathrm dx\stackrel{\text{parts}}{=}\frac{x^2}{2}\log x\vert_0^1-1/4=\frac{1}{2}\lim_{x\rightarrow 0}x^2\log x-1/4=-1/4 $$ and so the Lebesgue integral exists and has value bounded between $[-1/4,-1/8]$