A question about interchanging an improper integral and a limit

Question

I was reading this article https://brilliant.org/wiki/integration-tricks/#differentiation-under-the-integral-sign and I didn't understand what allows them in the second example (the one where they compute $\int\limits_0^{\infty} \frac{\sin x}{x}$) to put $\lim\limits_{a\to \infty}$ under the integral sign. What theorem is this?
EDIT : The specific part I have trouble understanding is why $\lim \limits_{a \to \infty} \int_0^{\infty} e^{-ax} \frac{\sin x} {x} dx= \int_0^{\infty} \lim \limits_{a \to \infty} e^{-ax} \frac{\sin x} {x} dx $

Answer 1

Note that as $a \to \infty$,

$$0 \leqslant \left|\int_0^\infty e^{-ax} \frac{\sin x}{x} \, dx \right| \leqslant \int_0^\infty e^{-ax} \left|\frac{\sin x}{x}\right| \, dx \leqslant\int_0^\infty e^{-ax} = \frac{1}{a} \to 0$$

We can also justify switching the limit and integral by the dominated convergence theorem or simply by the uniform convergence of the improper integral.