Prove that $f(x)=\int_0^{\infty} \frac{\exp(-xt)}{1+t} dt$ is convergent for all $x>0$ and differentiable

Question

Prove that $f(x)=\int_0^{\infty} \frac{\exp(-xt)}{1+t} dt$ is convergent for all $x>0$ and differentiable.

Would you have any hint, please?

Answer 1

Hint:

Recall that if $f_n(x)$ converges at at least one point and $f_n'(x) \to g(x)$ uniformly, then $f_n(x) \to f(x)$ uniformly with $f'(x) = g(x)$.

Consider

$$f_n(x) = \int_0^n \frac{e^{-xt}}{1+t} \, dt \\ f'_n(x) = -\int_0^n \frac{te^{-xt}}{1+t} \, dt, $$

and for $x \in [\alpha, \infty)$

$$\frac{te^{-xt}}{1+t} \leqslant e^{-\alpha t}$$