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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 $

ChemistryGeek
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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.

RRL
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  • Thank you for the elementary approach! How do you actually apply the theorems you mentioned? I know the DCT, but doesn't it work only for sequences of functions? – ChemistryGeek Feb 19 '20 at 06:49
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    @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 $a \to \infty$. – RRL Feb 19 '20 at 17:00