Urgent and important??! integral inequality

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December 17th 2006, 10:33 AM
miss_lolitta

Urgent and important??! integral inequality

Hi guys,

please can someone here help me to prove that:

Assume that:

$\int_1^{\infty}c(x)^{1/2}<\infty$
this implies that:

$\lim_{x \to\infty}{x c(x)^{1/2}}=0$

thanks in advance
December 17th 2006, 01:30 PM
ThePerfectHacker

Quote:

Originally Posted by miss_lolitta

Hi guys,

please can someone here help me to prove that:

Assume that:

$\int_1^{\infty}c(x)^{1/2}<\infty$
this implies that:

$\lim_{x \to\infty}{x c(x)^{1/2}}=0$

thanks in advance

If,
$c(x)$ is a decreasing function and continous. Then I have a proof.
December 17th 2006, 01:57 PM
miss_lolitta

hi..ThePerfectHacker

right..c(x) is a decreasing function and continous:o

thanks:) so much
December 17th 2006, 03:05 PM
ThePerfectHacker

Quote:

Originally Posted by miss_lolitta

hi..ThePerfectHacker

right..c(x) is a decreasing function and continous:o

thanks:) so much

Then it is simple.

$c(x)\geq 0$
Otherwise the expression is undefined.

Now, if $c(x)$ is decreasing function and continous then,
$\sqrt{c(x)}$ is decreasing.
And we are told that,
$\int_1^{\infty} \sqrt{c(x)}dx$
Converges.

That means, by the integral test,
$\sum_{k=1}^{\infty} \sqrt{c(k)}$
Converges.
But that means that,
$\lim_{k\to \infty}\sqrt{c(k)}=0$ .

Thus,
$\lim_{x\to \infty}x\cdot \sqrt{c(x)}=?$
We now that,
$\lim_{x\to \infty}\sqrt{c(x)}=0$ *)
This is a L'Hopital rule problem.
(But then we need to know that the function is differenciable).

*)I am not sure about this. Perhaps this statement works only the other way.