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I need to prove that

$\lim_{n\rightarrow \infty} \int_0^{n^2} n\sin(x/n) e^{-x^2}dx=1/2$

I tried substituting $t=\frac{x}{n^2}$, but it was not very useful because I can't find a bounding function for the integrand which does not depend on $n$ in order to apply the Dominated convergence Theorem. It would not help anyway as the limit of the integrand is zero which would give the wrong result.

I think I need a good substitution, if anyone could give a hint about that it would be very appreciated!

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1 Answer 1

up vote 1 down vote accepted

First recall $$ \lim_{y\rightarrow 0}\frac{\sin y}{y}=1. $$

It follows easily that the integrands converge pointwise to $xe^{-x^2}$ on $[0,+\infty)$..

Now recall $|\sin y|\leq |y|$ for all $y$.

So $$ |n\sin (x/n)e^{-x^2}|\leq xe^{-x^2} $$ for all $x\geq 0$.

The latter is integrable on $[0,+\infty)$.

So Lebesgue domninated convergence theorem applies and yields that the limit is equal to $$ \int_0^{+\infty}xe^{-x^2}dx=\frac{1}{2} $$ by an easy change of variable.

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