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!

  • $\begingroup$ $\log(x)$ is not defined for $x<0$. You should type a correct problem statement and add your attempts. $\endgroup$ – Jack D'Aurizio Jan 12 at 0:01
  • $\begingroup$ Thank you very much for your correction! $\endgroup$ – Ergin Suer Jan 12 at 0:03
  • $\begingroup$ 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 $x\geq 1$. $\endgroup$ – Jack D'Aurizio Jan 12 at 0:07
  • $\begingroup$ 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. $\endgroup$ – Jack D'Aurizio Jan 12 at 0:10
  • $\begingroup$ I wrote an answer by using your hint. I would be glad if you could check it. Thanks! $\endgroup$ – 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 $\int_0^\infty f(x)dx=\int_0^\infty\lim_{n\to\infty}f_n(x)dx=\lim_{n\to\infty}\int_0^\infty f_n(x)=\lim_{n\to\infty}\int_0^\infty |f_n(x)|\leq \int_0^\infty g(x)dx<\infty.$

Here we go:

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 $$|f_n(x)|<g_n(x):=\begin{cases} \frac{1}{x^{5/2}} &, x\in (\frac{1}{n},1] \\ \frac{1}{x^{3/2}} &, x\in[1,\infty) \end{cases}$$ which implies that $$|f_n(x)|<g(x):=\lim_{n\to\infty}g_n(x)=\begin{cases} \frac{1}{x^{5/2}} &, x\in (0,1] \\ \frac{1}{x^{3/2}} &, x\in[1,\infty) \end{cases}$$


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