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$\int_a^{\infty}\:f(x)dx$ is convergent, $f(x)$ is monotonically decreasing and continuous. Consider: Is $xf(x)$ also a monotonic decreasing function?prove it or give a counterexample, please. Besides, Is $\lim_{x\to +\infty} x\ln(x) f(x)=0$ right?

user350652
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$xf(x)$ need not be monotonically decreasing. Consider:

$f(x)=1$ if $x\in [0,1]$; $f(x)=1/x^2$ if $x\in(1,\infty)$

$\displaystyle\int_0^\infty f(x)dx=\displaystyle\int_0^1 f(x)dx+\displaystyle\int_1^\infty f(x)dx=2$

even though $xf(x)$ is increasing on $(0,1)$.

Edit: (Thank you Winther)

It's actually possible to get infinitely many intervals where $xf(x)$ is increasing. Consider the following function: $$f(x)=\frac{1}{(x-n)^2}~if~x\in[2n,2n+1]~for~some~n\in\mathbb{N}$$ $$f(x)=\frac{1}{(\lceil x\rceil-n)^2}~if~x\in(2n-1,2n)~for~some~n\in\mathbb{N}$$

This puts a series of flat segments into our function. On the flat segments, $xf(x)$ is increasing. The integral, however, converges. In fact,$$\displaystyle\int_1^\infty f(x)dx=\int_1^\infty \frac{dx}{x^2}+\sum_{k=1}^\infty\frac{1}{k^2}=1+\frac{\pi^2}{6}$$ using the famous solution to the Basel Problem (https://en.wikipedia.org/wiki/Basel_problem)

Nathaniel B
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  • But $f(x)$ isn't continuous. – user350652 Jun 28 '16 at 03:53
  • Unless I'm making an error it is continuous for all non-negative real numbers. It is most certainly continuous between 0 and 1 as well as between 1 and infinity, and at the "connecting point" $(x=1)$, $1=1/x^2$ so it is continuous there as well. This makes it continuous for all positive reals. – Nathaniel B Jun 28 '16 at 04:02
  • Sorry, I make a mistake. You are right. – user350652 Jun 28 '16 at 04:13
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    I suspect the problem should be "is eventually monotonically decreasing" (i.e. for large enough $x$). If this is the case then your example can be slightly modified to construct a function that is not eventually monotonically decreasing. Just take $f(x)$ to be constant on $[2n,2n+1)$ and decreasing like $1/x^2$ on intervals $[2n+1,2n+2)$ with $n\in\mathbb{N}$ (matched up to make it continuous). – Winther Jun 28 '16 at 04:36
  • That' right. $xf(x)$ is eventually monotonically decreasing. Can it be proved? What about the other question? – user350652 Jun 28 '16 at 05:34