Prove that $f(x)$ is defined and differentiable for $x>0$ Let $$f(x)=\int_0^{\infty} \frac{e^{-xt}}{1+t}dt$$
Prove that $f$ is defined and differentiable for $x>0$.
I have no idea how to deal with this...
 A: Hint
$f$ is well defined : Since $|e^{-xt}|\leq \frac{1}{xt}$, if $x>0$ and $t>1$,
$$\left|\frac{e^{-xt}}{1+t}\right|\leq \frac{1}{xt(t+1)}\leq \frac{1}{xt^2}.$$
$f$ is differentiable : Use dominated convergence to compute $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.$$
It can be usefull to suppose $|h|<1$.
A: Let $f(x)$ be defined as 
$$f(x)=\int_0^\infty\frac{e^{-xt}}{1+t}\,dt$$
for $x>0$.  Forming the difference quotient $\frac{f(x+h)-f(x)}{h}$ reveals 
$$\begin{align}
\frac{f(x+h)-f(x)}{h}&=\int_0^\infty \frac{e^{-xt}}{1+t}\left(\frac{e^{-ht}-1}{h}\right)\,dt\tag 1\\\\
&=\int_0^\infty \frac{e^{-(x+h)t}}{1+t}\left(\frac{1-e^{ht}}{h}\right)\,dt \tag 2
\end{align}$$
Noting that for $h>0$ $$\left|\frac{e^{-ht}-1}{h}\right|\le t$$
while for $h<0$  $$\left|\frac{1-e^{ht}}{h}\right|\le t$$
we see that for $h>0$ $$\left|\frac{e^{-xt}}{1+t}\left(\frac{e^{-ht}-1}{h}\right)\right|\le \frac{te^{-xt}}{1+t}$$
while for $-x/2<h <0$ $$\left|\frac{e^{-(x+h)t}}{1+t}\left(\frac{1-e^{ht}}{h}\right)\right|\le \frac{te^{-xt/2}}{1+t}$$
Inasmuch as $\frac{te^{-xt}}{1+t}$ and $\frac{te^{-xt/2}}{1+t}$ are integrable for all $x>0$, then we can apply the Dominated Convergence Theorem to assert that
$$\begin{align}
f'(x)&=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\\\\
&=\lim_{h\to 0}\int_0^\infty \frac{e^{-xt}}{1+t}\left(\frac{e^{-ht}-1}{h}\right)\,dt\\\\
&=\int_0^\infty \frac{e^{-xt}}{1+t}\lim_{h\to 0}\left(\frac{e^{-ht}-1}{h}\right)\,dt\\\\
&=\int_0^\infty\frac{te^{-xt}}{1+t}\,dt
\end{align}$$
