2
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ $\log(x)$ 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
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.