# A question about interchanging an improper integral and a limit

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 $$\lim \limits_{a \to \infty} \int_0^{\infty} e^{-ax} \frac{\sin x} {x} dx= \int_0^{\infty} \lim \limits_{a \to \infty} e^{-ax} \frac{\sin x} {x} dx$$

• @ChemstryGeek Maybe this is related topic: math.stackexchange.com/questions/781398/… Feb 18, 2020 at 21:10
• @Sunghee Yun I don't think it is. I edited my post so that it is more clear what I am asking about. Feb 18, 2020 at 22:30
• What is the downvote for? Feb 18, 2020 at 22:34
• The link is behind a log-in wall Feb 18, 2020 at 22:36
• @user27182 For me it works and I am not logged in. Anyway, with my edit this should be intelligible without the whole article Feb 18, 2020 at 22:37

Note that as $$a \to \infty$$,
$$0 \leqslant \left|\int_0^\infty e^{-ax} \frac{\sin x}{x} \, dx \right| \leqslant \int_0^\infty e^{-ax} \left|\frac{\sin x}{x}\right| \, dx \leqslant\int_0^\infty e^{-ax} = \frac{1}{a} \to 0$$
• @ChemistryGeek: We can apply the DCT for any sequence $a_n$ such that $a_n \to \infty$ as $n \to \infty$. Here we have $e^{-x}$ as an integrable dominating function. Hence $\lim_{n \to \infty}\int_0^\infty f(x,a_n) \, dx = \int_0^\infty \lim_{n \to \infty} f(x,a_n) \, dx$. Since this is true for any sequence it allows us to switch integral and limit as $a \to \infty$.