# $\int_0^\infty\frac{x\log(x)}{(1+x)^3}dx$ converges in making use of the sequence

Show that the integral $$\int_0^\infty\frac{x\log(x)}{(1+x)^3}dx$$ converges in making use of the sequence $$f_n(x)=\begin{cases} 0 & \text{if } x\in[0,1/n) \\ \frac{x\log(x)}{(1+x)^3} & \text{if } x \in [1/n, \infty) \end{cases}$$

Could anyone give a strategy to solve this question? The first thing that comes to my mind is using the Lebesgue Dominated Convergence Theorem (LDCT). But, I couldn't see how to use it. Thanks!

• $\log(x)$ is not defined for $x<0$. You should type a correct problem statement and add your attempts. – Jack D'Aurizio Jan 12 at 0:01
• Thank you very much for your correction! – Ergin Suer Jan 12 at 0:03
• You may just use the fact that $x\log x$ is integrable over $(0,1)$ and $0\leq\log(x)\leq\sqrt{x}$ for any $x\geq 1$. – Jack D'Aurizio Jan 12 at 0:07
• Then the exact value of the integral can be found by differentiating $\int_{0}^{+\infty}\frac{x^\alpha\,dx}{(1+x)^3}\,dx = \frac{\pi(1-\alpha)}{2\text{sinc}(\pi\alpha)}$ at $\alpha=1$, for instance. – Jack D'Aurizio Jan 12 at 0:10
• I wrote an answer by using your hint. I would be glad if you could check it. Thanks! – Ergin Suer Jan 14 at 23:47

Note that the sequence $$(f_n)$$ point-wise converges to the function $$f$$ defined by $$f(x):=\frac{x\log(x)}{(1+x)^3}$$ for $$x\in(0,\infty)$$. If we find an integrable function $$g$$ on $$(0,\infty)$$ such that $$|f_n(x)|\leq g(x)$$ for all $$n\in\Bbb N, x\in (0,\infty)$$, then by the aid of LDCT, we will have $$\int_0^\infty f(x)dx=\int_0^\infty\lim_{n\to\infty}f_n(x)dx=\lim_{n\to\infty}\int_0^\infty f_n(x)=\lim_{n\to\infty}\int_0^\infty |f_n(x)|\leq \int_0^\infty g(x)dx<\infty.$$
First note that $$\log(x)<\sqrt{x}$$ for any $$x\geq 1$$ and we observe that, for any $$n\in \Bbb N$$, if $$x\in (1/n,1]$$ then \begin{align} |f_n(x)|=\left|\frac{x\log(x)}{(1+x)^3}\right|=\frac{x|\log(x)|}{(1+x)^3}=\frac{x(-\log(x))}{(1+x)^3}=\frac{x\log(1/x)}{(1+x)^3} < \frac{x\sqrt{1/x}}{x^3}=x^{-5/2}. \end{align} Also, if $$x\in [1,\infty)$$ then $$|f_n(x)|=\frac{x\log(x)}{(1+x)^3}<\frac{x\sqrt{x}}{x^3}=x^{-3/2}$$. So, for any $$n\in \Bbb N$$, we have $$|f_n(x)| which implies that $$|f_n(x)|