Define $F:[0,\infty)\to\mathbb{R}$ by $$F(t)=\int_{0}^{\infty}e^{-tx}\dfrac{\sin x}{x}\,dx.$$ Prove that $F$ is continuous at $t=0$.
I already proved that $$F'(t)=-\dfrac{1}{1+t^2}, \quad t>0.$$ I don't know if this result could be useful for the proof of the continuity at 0.
Recall that for a sequence of continuous functions, if we have uniform convergence $F_n(t) \to F(t)$ as $n \to \infty$ for $t \in [a,b]$ then $F$ is continuous.
This improper integral is convergent and we have for each $t$,
$$\lim_{n\to \infty}F_n(t) = \lim_{n\to \infty}\int_{1/n}^{n}e^{-tx}\dfrac{\sin x}{x}\,dx = \int_{0}^{\infty}e^{-tx}\dfrac{\sin x}{x}\,dx = F(t)$$
Since the integrand is continuous for $(x,t) \in [1/n,n]\times[0,1]$ we have for each $n$ that $F_n$ continuous for $t \in [0,1]$. If we can show that the convergence is uniform, then it follows that $F$ is continuous on $[0,1]$ and, in particular, at $t= 0$.
In this case, the uniform convergence of the improper integral follows from Abel's test. See here and here.
The conditions are met here since $x \mapsto e^{-tx}$ is bounded and monotone for $(x,t) \in [0,\infty) \times [0,1]$ and the improper integral,
$$\int_0^ \infty \frac{\sin x}{x} \, dx, $$
is convergent and, thus, uniformly convergent for all $t \in[0,1]$ trivially since there is no dependence on $t$.
Proof that $F_n$ is continuous on $[0,1]$
The integrand $f(x,t) = e^{-tx}\dfrac{\sin x}{x}$ is uniformly continuous on the compact set $[1/n,n]\times [0,1]$. Take any $t_0 \in [0,1]$. For any $\epsilon > 0$ there exists $\delta > 0$ independent of $x,t$ such that if $|t- t_0| < \delta$, we have
$$|f(x,t) - f(x,t_0)| < \epsilon/(n-1/n)$$
Thus,
$$|F(t) - F(t_0)| \leqslant \int_{1/n}^n|f(x,t) - f(x,t_0)| \, dx < (n-1/n)\epsilon/(n - 1/n) = \epsilon$$