Help finishing this exercise! Given the following functions:

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


Show that $F$ is well-defined in $(0,+\infty)$ and it is differentiable in that same interval. Compute $F'$ and $\lim_{t \to \infty} F(t)$. Finally, compute $F(t)$ and $\lim_{t \to 0^+} F(t)$
Compute $F_s'$ in $(0, +\infty)$. Show that $F_s$ is a equicontinous family of functions
Show that $\lim_{s \to \infty}F_s(t) = F(t)$ uniformly for $t \in (0,+\infty)$
Show that $\int_0^\infty \dfrac{\sin{x}}{x}\,dx = \dfrac{\pi}{2}$


I have already asked you about (3) (Show that $\lim_{s \to \infty}F_s(t) = F(t)$ uniformly for $t \in (0,+\infty)$). Now I have 1,2 and 3 solved, but I don't quite know how to finish up things in (4)

Using the definition I have

$$ \int_0^\infty \dfrac{\sin{x}}{x}\,dx := \lim_{L \to \infty} \int_0^L \dfrac{\sin{x}}{x}\,dx$$

and I am guessing I should connect this somehow with $\lim_{t \to 0^+} F(t)$ (which I have already computed). Also, I have noticed that I have not used that $\{ F_s \}$ is equicontinous, maybe that is necessary?. Could you give me any hints about how to complete this?
Given as a solution from this post:

This can also be solved using double integrals.

Using the double integral:

$$\int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dy \Bigg)\, dx = \int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dx \Bigg)\,dy$$
Notice that
$$\int_{0}^{\infty} e^{-xy} \sin x\,dy = \frac{\sin x}{x}$$

We can then substitute this in and say:

$$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dx \Bigg)\,dy$$
Now the right hand side can be found easily, using integration by parts.

$$\begin{align*}
I &amp;= \int e^{-xy} \sin x \,dx = -e^{-xy}{\cos x} - y \int e^{-xy} \cos x \, dx\\
&amp;= -e^{-xy}{\cos x} - y \Big(e^{-xy}\sin x + y \int e^{-xy} \sin x \,dx \Big)\\
&amp;= \frac{-ye^{-xy}\sin x - e^{-xy}\cos x}{1+y^2}.
\end{align*}$$
Thus $$\int_{0}^{\infty} e^{-xy} \sin x \,dx = \frac{1}{1+y^2}$$
Thus $$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty}\frac{1}{1+y^2}\,dy = \frac{\pi}{2}.$$
Given as a solution from this post:

This can also be solved using double integrals.

Using the double integral:

$$\int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dy \Bigg)\, dx = \int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dx \Bigg)\,dy$$
Notice that
$$\int_{0}^{\infty} e^{-xy} \sin x\,dy = \frac{\sin x}{x}$$

We can then substitute this in and say:

$$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dx \Bigg)\,dy$$
Now the right hand side can be found easily, using integration by parts.

$$\begin{align*}
I &amp;= \int e^{-xy} \sin x \,dx = -e^{-xy}{\cos x} - y \int e^{-xy} \cos x \, dx\\
&amp;= -e^{-xy}{\cos x} - y \Big(e^{-xy}\sin x + y \int e^{-xy} \sin x \,dx \Big)\\
&amp;= \frac{-ye^{-xy}\sin x - e^{-xy}\cos x}{1+y^2}.
\end{align*}$$
Thus $$\int_{0}^{\infty} e^{-xy} \sin x \,dx = \frac{1}{1+y^2}$$
Thus $$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty}\frac{1}{1+y^2}\,dy = \frac{\pi}{2}.$$