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Given the integral: $$S=\int_{-a}^{a}\frac{\sin(ax)}{ax}$$ I get from MAPLE this result: $$\lim_{a\to \infty}S=0$$ My question is: how can I prove this result? Thanks.

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It's a sin to write sin instead of \sin. :-) – Asaf Karagila Sep 3 '12 at 11:07
up vote 5 down vote accepted

By a change of variables, you can rewrite your integral above as

$$S= \int_{-a^2}^{a^2} \frac{1}{a}\frac{\sin u}{u} du$$

which we can rewrite as $$\int_0^{a^2}\frac{2}{a} \frac{\sin u}{u} du = \frac{2}{a}\int_0^{a^2} \frac{\sin u}{u} du$$

Now the limit of the integral as $a \rightarrow \infty $ is $\pi/2$ (this can be proved using knowledge of the Dirichlet Kernel and Riemann-Lebesgue), while the limit as $a \rightarrow \infty$ of $2/a$ is zero. Therefore the limit of $S$ as $a \rightarrow \infty$ is zero.

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Just to note that all you need to do to prove the result requested is to show that $\int_0^{a^2} \frac{\sin u}{u} du$ is bounded, which can be done without evaluating the integral. – Mark Bennet Sep 3 '12 at 12:03
@MarkBennet This can be done by integration by parts I believe. – user38268 Sep 3 '12 at 12:04
I was thinking that the interval of integration could be divided into pieces of length $\pi$, which gives the sum of alternating series whose terms tend to zero. – Mark Bennet Sep 3 '12 at 12:14

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