Vtyjkyuk

$F(t)=int^infty_0 x^2e^-tx dx$ defined for $t>0$ is a continuous function? How prove?
I tried to use the definition of continuity, but I could not. Do you have another method?

You can either do integration by parts to find a better expression for the function or use the dominated convergence theorem, If you are familiar with measure theory.

$$displaystyleintx^2mathrme^-tx,mathrmdx$$
Using integration by parts, $int fg' = fg - int f'g$ where $f = x^2$ and $g = e^-tx$, you can show that
$$displaystyleintx^2mathrme^-tx,mathrmdx =-dfracx^2mathrme^-txt-displaystyleint-dfrac2xmathrme^-txt,mathrmdx $$
Notice that
$$displaystyleint-dfrac2xmathrme^-txt,mathrmdx =-dfrac2tdisplaystyleintxmathrme^-tx,mathrmdx$$
Using integration by parts, $int fg' = fg - int f'g$ where $f = x$ and $g = e^-tx$, you can show that

beginequation
int
xmathrme^-tx
=-dfracxmathrme^-txt+displaystyleintdfracmathrme^-txt,mathrmdx
=
-dfracxmathrme^-txt-frac1t^2e^-tx
endequation
Going back to the first equation, we get
$$displaystyleintx^2mathrme^-tx,mathrmdx
=
-dfracleft(t^2x^2+2tx+2right)mathrme^-txt^3+A$$
Including the limits, we get
$$intlimits_0^inftyx^2mathrme^-tx,mathrmdx
=
-dfracleft(t^2x^2+2tx+2right)mathrme^-txt^3Bigvert_0^infty = frac2t^3$$
So, now you have to check if $frac2t^3$ is continuous on $]0,infty[$.

Substituting $u=tx$, we have
$$ F(t) = int_0^infty x^2e^-tx,dx = frac1t^2 int_0^infty (tx)^2e^-tx ,dx
= frac1t^2 int_0^infty u^2e^-u fracdut = fracF(1)t^3 $$
so if you can prove that $F(t)$ always exists, it will be as continuous as $t^-3$.

For fixed $t$, let
$$ï¼£=int^infty_0 x^2e^-txdx.$$
Choose $Delta tin R$ such that $t+Delta t>0$ and
$$ |e^-xDelta t-1|le frac1C|Delta t| $$
and hence
begineqnarray
|F(t+Delta t)-F(t)|&=&left|int^infty_0 x^2(e^-(t+Delta t)x-e^-t) dxright|\
&=&left|int^infty_0 x^2e^-tx(e^-xDelta t-1) dxright|\
&le&int^infty_0 x^2e^-txleft|e^-xDelta t-1right|dx\
&le&C|Delta t|int^infty_0 x^2e^-txdx\
&le&|Delta t|
endeqnarray
which implies
$$ lim_Delta tto0F(t+Delta t)=F(t).$$
So $F(t)$ is continuous.