Solving an integral with parameter I try to calculate the derivative of the following function:
$$f: (0, \infty) \to \mathbb R,\ x \mapsto \int_1^\infty \frac{e^{-xt^2}}{t} dt$$
I managed to show that the function is well-defined. Further I tried to calculate the derivative by using a theorem for integrals with parameters. If I set $g(x, t) = \frac{e^{-xt^2}}{t}$, I know that
$$ \frac{\partial g}{\partial x}(x,t) = -t e^{-xt^2}$$
is continuous and hence $g(x, \cdot) \in C^1(1, \infty)$ for every $x \in (0, \infty)$. Setting $h(t) = t$ I further have that
$$\left\vert \frac{\partial g}{\partial x}(x,t) \right\vert = t e^{-xt^2} \leq h(t) \qquad \text{for all } t \in (1, n)$$ 
for an arbritary $n \in \mathbb N$ and all $x \in (0, \infty)$. Notice that $h$ is an integrable on $(1, n)$.
Hence I know
$$\frac{d}{d x} \int_1^n \frac{e^{-xt^2}}{t} dt= \int_1^n \frac{d}{d x} \left(\frac{e^{-xt^2}}{t} \right)dt = \int_1^n -t e^{-xt^2} dt = \left[  \frac{1}{2x} e^{-xt^2} \right]_1^n = \frac{e^{-n^2x} - e^{-x}}{2x}.$$
But now I'm struggeling to give an argument so that I have
\begin{align*} \frac{df}{d x}(x) &= \frac{d}{d x} \int_1^\infty \frac{e^{-xt^2}}{t} dt = \frac{d}{d x} \lim_{n \to \infty} \int_1^n \frac{e^{-xt^2}}{t} dt \\
&\overset{?}{=} \lim_{n \to \infty} \frac{d}{d x} \int_1^n \frac{e^{-xt^2}}{t} dt = \lim_{n \to \infty} \frac{e^{-n^2x} - e^{-x}}{2x} = - \frac{e^{-x}}{2x}.
\end{align*}
How can I deduce that $\overset{?}{=}$ holds? Is this even the right way to prove it? Or do I need a "better" majorant $h$ to solve this question? I would appreciate some hints on the topic :)
 A: Let
$$ f(x) = \int_{1}^{+\infty}e^{-xt^2}\frac{dt}{t}.\tag{1}$$
Then for any $x>0$ and $h>0$:
$$ \frac{f(x+h)-f(x)}{h} = \int_{1}^{+\infty}e^{-xt^2}\left(\frac{e^{-ht^2}-1}{h}\right)\frac{dt}{t} \tag{2}$$
so, in order to prove that $f'(x)\stackrel{?}{=}\int_{1}^{+\infty}\frac{\partial}{\partial x}e^{-xt^2}\frac{dt}{t}=\int_{1}^{+\infty}-t e^{-xt^2}\,dt = -\frac{e^{-x}}{2x}$ as expected, it is enough to show that
$$\forall x>0,\qquad \lim_{h\to 0^+}\int_{1}^{+\infty}e^{-xt^2}\left(\frac{e^{-ht^2}-1}{h}+t^2\right)\frac{dt}{t}=0.\tag{3}$$
Since $e^{-ht^2}=\frac{1}{e^{ht^2}}\leq\frac{1}{1+ht^2}$, we have $\left(\frac{e^{-ht^2}-1}{h}+t^2\right)\leq\frac{h t^4}{1+h t^2}\leq h t^4$, that is enough to prove $(3)$. Since the dominated convergence theorem applies and $\lim_{x\to +\infty}f(x)=0$,
$$ f(x) = \int_{1}^{+\infty}e^{-xt^2}\frac{dt}{t} = \color{blue}{\frac{1}{2}\int_{x}^{+\infty}\frac{du}{u e^u}} \tag{4}$$
that also follows from the substitution $t=\frac{\sqrt{u}}{x}$. Additionally, by the Cauchy-Schwarz inequality,
$$ \forall x>0,\qquad 0\leq f(x) \leq \frac{1}{ 2e^x\sqrt{2x}}.\tag{5}$$
A: For all $x\ge x'>0$ and for $t\ge 0$, we have
$$\left|\frac{\partial g(x,t)}{\partial x}\right|\le te^{-x't^2}$$
Since $\int_1^\infty te^{-x't^2}\,dt=\frac{e^{-x'}}{2x'}$ converges, then for $x\ge x'>0$, $\int_1^\infty \frac{\partial g(x,t)}{\partial x}\,dt$  converges uniformly and therefore
$$\frac{d}{dx}\int_1^\infty \frac{e^{-xt^2}}{t}\,dt=\int_1^\infty \frac{\partial }{\partial x}\left(\frac{e^{-xt^2}}{t}\right)\,dt = -\frac{e^{-x}}{2x}\tag 1$$
Since $(1)$ is valid for an arbitrary $x\ge x'>0$, we conclude that for $x\in (0,\infty)$
$$\frac{d}{dx}\int_1^\infty \frac{e^{-xt^2}}{t}\,dt=-\frac{e^{-x}}{2x}$$
