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For $xin(0,infty)$, let $F(x)=int_0^inftyfrac1-e^-txt^frac32 dt$. Show that $F:(0,infty)to(0,infty)$ is well-defined, bijective and has continuous non-zero derivative.

I don't know what argument to say about bijectivity. And for derivative exists I suppose we use uniform convergence on compact interval which enables us to interchange limit and integration, but does that mean we have to cut $(0,infty)$ into three parts? As for showing derivative is continuous, it just sounds like a lot of calculation...

To see that $F$ is well defined, you have to study separately the two singularities of $$f(t) = frac1-e^-xtt^3/2$$
For $t to 0^+$ you have
$$frac1-e^-xtt^3/2 sim fracxtt^3/2 = fracxt^1/2$$
since $int_0^1 fracxt^1/2 dt$ converges, you have convergence of $int_0^1 f(t) dt$.

For $t to + infty$ you have
$$frac1-e^-xtt^3/2 sim frac1t^3/2$$
since $int_1^+infty frac1t^3/2 dt$ converges, you have convergence of $int_1^+infty f(t) dt$.

This shows that $F$ is well defined for all $x$.

Now, for $0<x_1 < x_2$ and for all $t > 0$ you have
$$frac1-e^-x_1tt^3/2<frac1-e^-x_2tt^3/2$$
thus integrating you have
$$int_0^+ infty frac1-e^-x_1tt^3/2 mathrm dt <int_0^+ infty frac1-e^-x_2tt^3/2 mathrm dt$$

This shows that $F$ is strictly increasing on $(0, + infty)$. In particular it is injective. To show that it is bijective it is enough to prove continuity. To show that $F$ is continuous, it is enough to prove that $F$ is differentiable.
This seems quite complicated.

HOWEVER:
Note that, if you call $u=xt$, then your function becomes:
$$F(x)=sqrtx int_0^+ infty frac1-e^-uu^3/2 mathrm du = F(1) sqrt x$$
which is a well-known function: continuous, strictly increasing, differentiable and bijective.

The derivative of the integrand is
$$fracddxleft(frac1-e^-txt^3/2right) = frace^-txsqrtt$$
We claim that
$$F'(x) = int_0^infty frace^-txsqrtt,dt$$
for all $x > 0$.

Fix $a > 0$ and notice that for $x ge a$ we have
$$frace^-txsqrtt le frace^-tasqrtt le frac1sqrttchi_(0,1](t) + e^-tachi_[1,infty)(t)$$
the latter being an integrable function on $(0,infty)$.
For any such $x$ we have
$$F'(x) = lim_hto 0 fracF(x+h) - F(x)h = lim_hto 0 int_0^infty frac1hleft[frac1-e^-t(x+h)t^3/2 - frac1-e^-txt^3/2right],dt$$
The mean value theorem implies that $exists theta(h) in (x,x+h)$ such that $$frac1hleft[frac1-e^-t(x+h)t^3/2 - frac1-e^-txt^3/2right] = frace^-tthetasqrtt$$
so $$F'(x) = lim_hto 0 int_0^infty frace^-ttheta(h)sqrtt ,dt = int_0^infty lim_hto 0frace^-ttheta(h)sqrtt ,dt = int_0^infty frace^-txsqrtt,dt$$
We were able to swap the integral and the limit according to the Lebesgue dominated convergence theorem since
$$left|frace^-ttheta(h)sqrtt right| le frac1sqrttchi_(0,1](t) + e^-tachi_[1,infty)(t), quadforall t > 0$$
Notice that it is important that the dominating function does not depend on $h$ in any way.

Since $a > 0$ is arbitrary, we conclude $F'(x) = int_0^infty frace^-txsqrtt,dt$ for all $x > 0$.

For surjectivity it suffices to show that $lim_xto 0 F(x) = 0$ and $lim_xtoinfty F(x) = infty$. This together with continuity yields the desired result via the intermediate value theorem.

For $x le 1$ we have
$$frac1-e^-txt^3/2 le frac1sqrttchi_(0,1](t) + frac1t^3/2chi_[1,infty)(t)$$
having used $e^-tx ge 1-tx$. The latter function is integrable on $(0,infty)$ so Lebesgue dominated convergence theorem implies
$$lim_xto 0 F(x) = lim_xto 0 int_0^infty frac1-e^-txt^3/2,dt = int_0^infty lim_xto 0frac1-e^-txt^3/2,dt = int_0^infty 0 ,dt = 0$$

You already know that for $x_1,x_2 in (0,infty)$ with $x_1 < x_2$ we have $$frac1-e^-tx_1t^3/2 < frac1-e^-tx_2t^3/2$$

Lebesgue monotone convergence theorem thus implies
$$lim_xto infty F(x) = lim_xto infty int_0^infty frac1-e^-txt^3/2,dt = int_0^infty lim_xto inftyfrac1-e^-txt^3/2,dt = int_0^infty frac1t^3/2 ,dt = infty$$