Prove improper integral exists and continuously differentiable For $x\in(0,\infty)$, let $F(x)=\int_0^\infty\frac{1-e^{-tx}}{t^{\frac{3}{2}}}\ dt$. 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 $$f(t) = \frac{1-e^{-xt}}{t^{3/2}}$$
For $t \to 0^+$ you have
$$\frac{1-e^{-xt}}{t^{3/2}} \sim \frac{xt}{t^{3/2}} = \frac{x}{t^{1/2}}$$
since $\int_0^1 \frac{x}{t^{1/2}} dt$ converges, you have convergence of $\int_0^1 f(t) dt$.
For $t \to + \infty$ you have
$$\frac{1-e^{-xt}}{t^{3/2}} \sim \frac{1}{t^{3/2}}$$
since $\int_1^{+\infty} \frac{1}{t^{3/2}} dt$ 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
$$\frac{1-e^{-x_1t}}{t^{3/2}}<\frac{1-e^{-x_2t}}{t^{3/2}}$$
thus integrating you have
$$\int_0^{+ \infty} \frac{1-e^{-x_1t}}{t^{3/2}} \mathrm dt <\int_0^{+ \infty} \frac{1-e^{-x_2t}}{t^{3/2}} \mathrm dt$$
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:
$$F(x)=\sqrt{x} \int_{0}^{+ \infty} \frac{1-e^{-u}}{u^{3/2}}  \mathrm du = F(1) \sqrt x$$
which is a well-known function: continuous, strictly increasing, differentiable and bijective.
