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...
 A: 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.
A: The derivative of the integrand is
$$\frac{d}{dx}\left(\frac{1-e^{-tx}}{t^{3/2}}\right) = \frac{e^{-tx}}{\sqrt{t}}$$
We claim that
$$F'(x) = \int_0^\infty \frac{e^{-tx}}{\sqrt{t}}\,dt$$
for all $x > 0$.
Fix $a > 0$ and notice that for $x \ge a$ we have
$$\frac{e^{-tx}}{\sqrt{t}} \le \frac{e^{-ta}}{\sqrt{t}} \le \frac1{\sqrt{t}}\chi_{(0,1]}(t) + e^{-ta}\chi_{[1,\infty)}(t)$$
the latter being an integrable function on $(0,\infty)$.
For any such $x$ we have
$$F'(x) = \lim_{h\to 0} \frac{F(x+h) - F(x)}h = \lim_{h\to 0} \int_0^\infty \frac1h\left[\frac{1-e^{-t(x+h)}}{t^{3/2}} - \frac{1-e^{-tx}}{t^{3/2}}\right]\,dt$$
The mean value theorem implies that $\exists \theta(h) \in (x,x+h)$ such that $$\frac1h\left[\frac{1-e^{-t(x+h)}}{t^{3/2}} - \frac{1-e^{-tx}}{t^{3/2}}\right] = \frac{e^{-t\theta}}{\sqrt{t}}$$
so $$F'(x) = \lim_{h\to 0} \int_0^\infty \frac{e^{-t\theta(h)}}{\sqrt{t}} \,dt =  \int_0^\infty \lim_{h\to 0}\frac{e^{-t\theta(h)}}{\sqrt{t}} \,dt = \int_0^\infty \frac{e^{-tx}}{\sqrt{t}}\,dt$$
We were able to swap the integral and the limit according to the Lebesgue dominated convergence theorem since 
$$\left|\frac{e^{-t\theta(h)}}{\sqrt{t}} \right| \le \frac1{\sqrt{t}}\chi_{(0,1]}(t) + e^{-ta}\chi_{[1,\infty)}(t), \quad\forall t > 0$$
Notice that it is important that the dominating function does not depend on $h$ in any way.
Since $a > 0$ is arbitrary, we conclude $F'(x) = \int_0^\infty \frac{e^{-tx}}{\sqrt{t}}\,dt$ for all $x > 0$.

For surjectivity it suffices to show that $\lim_{x\to 0} F(x) = 0$ and $\lim_{x\to\infty} F(x) = \infty$. This together with continuity yields the desired result via the intermediate value theorem.
For $x \le 1$ we have
$$\frac{1-e^{-tx}}{t^{3/2}} \le \frac1{\sqrt{t}}\chi_{(0,1]}(t) + \frac1{t^{3/2}}\chi_{[1,\infty)}(t)$$
having used $e^{-tx} \ge 1-tx$. The latter function is integrable on $(0,\infty)$ so Lebesgue dominated convergence theorem implies
$$\lim_{x\to 0} F(x) = \lim_{x\to 0} \int_0^\infty \frac{1-e^{-tx}}{t^{3/2}}\,dt = \int_0^\infty \lim_{x\to 0}\frac{1-e^{-tx}}{t^{3/2}}\,dt = \int_0^\infty 0 \,dt = 0$$
You already know that for $x_1,x_2 \in (0,\infty)$ with $x_1 < x_2$ we have $$\frac{1-e^{-tx_1}}{t^{3/2}} < \frac{1-e^{-tx_2}}{t^{3/2}}$$ 
Lebesgue monotone convergence theorem thus implies
$$\lim_{x\to \infty} F(x) = \lim_{x\to \infty} \int_0^\infty \frac{1-e^{-tx}}{t^{3/2}}\,dt = \int_0^\infty \lim_{x\to \infty}\frac{1-e^{-tx}}{t^{3/2}}\,dt = \int_0^\infty \frac1{t^{3/2}} \,dt = \infty$$
