Show that $\lim_{s \to \infty}F_s(t) = F(t)$ uniformly for $t \in (0,+\infty)$

Question

Given the following functions:

$$ F(t)= \int_0^\infty e^{-tx}\dfrac{\sin{x}}{x}\,dx, \quad t>0$$ $$ F_s(t)= \int_0^s e^{-tx}\dfrac{\sin{x}}{x}\,dx, \quad t \geq 0, s>0$$

Show that $\lim_{s \to \infty}F_s(t) = F(t)$ uniformly for $t \in (0,+\infty)$

This is what I've tried:

$$\lim_{s \to \infty}{F_s(t)} = F(t) \, \textrm{uniformly} \iff \lim_{s \to \infty} || F_s(t) -F(t) || _\infty = 0$$

$$\begin{array}{rcl} \lim_{s \to \infty}|| F_s(t) -F(t) || _\infty &=& \lim_{s \to \infty} \sup \, \{ | \int_0^se^{-tx}\dfrac{\sin{x}}{x}\,dx - \int_o^\infty e^{-tx}\dfrac{\sin{x}}{x}\,dx|, t>0\} \\ &=& \lim_{s \to \infty} \sup \, \{ | \int_0^se^{-tx}\dfrac{\sin{x}}{x}\,dx - \lim_{L \to \infty} \int_o^L e^{-tx}\dfrac{\sin{x}}{x}\,dx|, t>0\}\\ &\overset{(*)}{=}& \lim_{s \to \infty} \lim_{L \to \infty} \sup \, \{ | \int_0^se^{-tx}\dfrac{\sin{x}}{x}\,dx - \int_o^L e^{-tx}\dfrac{\sin{x}}{x}\,dx|, t>0\} \\ &= & 0\end{array}$$

Im not sure about the step marked $(*)$. Is this right? If it is wrong, how could I show that?

Answer 1

Hint: For the step $(*)$ use

$$\left|\int_0^sf(x,t)dx - \lim_{L\to\infty}\int_0^L f(x,t)dx\right| = \left|\int_s^\infty e^{-xt}{\sin(x)\over x} dx\right| < \int_s^\infty e^{-xt}{dx \over x} < \frac{e^{-st}}{st}$$

instead. Now take limsup.

Edit: The estimate above is only good enough to prove uniform convergence on $[\epsilon,\infty)$ for any $\epsilon > 0$. To extend to $(0,\infty)$ a better estimate is needed.