Limit of Lebesgue Integrals $\lim_{n\to \infty} \int_{[0,\infty]} \frac{n\sin(\frac{x}{n})}{x(1+x^2)}dm$ Can you assist me in solving this limit:
$$\lim_{n\to \infty} \int_{[0,\infty]} \frac{n\sin(\frac{x}{n})}{x(1+x^2)}\,dm$$
where $m$ is the Lebesgue Measure on $\mathbb{R}$?
I thought I should try to use the dominated convergence theorem, but didn't succeed in bounding that integrand, through substitution either.
 A: One may use that
$$
|\sin x | \le |x|,\qquad x \in \mathbb{R},
$$ giving, as $n \to \infty$,
$$
\left|\sin\Big(\frac{x}{n}\Big)\right|\le \left|\frac{x}{n}\right| \implies n\left|\sin\Big(\frac{x}{n}\Big)\right|\le |x|,\qquad x \in \mathbb{R},
$$ then
$$
\left|\int_{[0,\infty)} \frac{n\sin(\frac{x}{n})}{x(1+x^2)}\:dm\right|\le \int_{[0,\infty)} \frac{ |x|}{x(1+x^2)}\:dm=\int_{[0,\infty)} \frac{ 1}{1+x^2}\:dm=\frac{\pi}2.
$$ Can you take it from here?
A: Hint: The integrand equals $[\sin(x/n)/(x/n)]\cdot 1/(1+x^2).$ How big can $(\sin u)/u$ be for real $u?$
A: Using the Laplace transform, since
$$ \mathcal{L}\left(\sin\frac{x}{n}\right) = \frac{n}{1+n^2 s^2},\qquad \mathcal{L}^{-1}\left(\frac{1}{x(1+x^2)}\right) = 1-\cos(s) $$
we have that:
$$ I(n) = \int_{0}^{+\infty}\frac{n\sin\frac{x}{n}}{x(1+x^2)}\,dx = \int_{0}^{+\infty}\frac{1-\cos(s)}{s^2+\frac{1}{n^2}}\,ds $$
so, by the dominated convergence theorem:
$$ \lim_{n\to +\infty} I(n) = \int_{0}^{+\infty}\frac{1-\cos(s)}{s^2}\,ds \stackrel{\text{IBP}}{=}\int_{0}^{+\infty}\frac{\sin s}{s}\,ds = \color{blue}{\frac{\pi}{2}}. $$
