Evaluate $\int\limits_{0}^{\infty}\frac{n \sin x}{1+n^2x^2}dx$ Notice that $\frac{n \sin x}{1+n^2x^2}\to 0$ pointwise.
And we have,$$\int\limits_{0}^{\infty}\frac{n \sin x}{1+n^2x^2}dx=\int\limits_{0}^{1}\frac{n \sin x}{1+n^2x^2}dx+\int\limits_{1}^{\infty}\frac{n \sin x}{1+n^2x^2}dx$$
Then for $\int\limits_{1}^{\infty}\frac{n \sin x}{1+n^2x^2}dx$ portion we have,
$$\left|\frac{n \sin x}{1+n^2x^2}\right|\leq\left|\frac{n}{n^2x^2}\right|=\left|\frac{1}{nx^2}\right|\leq\frac{1}{x^2}$$
And $\frac{1}{x^2}$ is integrable on $(1,\infty)$
So by the dominated convergence theorem:
$$\int\limits_{1}^{\infty}\frac{n \sin x}{1+n^2x^2}dx\to\int\limits_0^{\infty}0=0$$
But how should I do the $\int\limits_{0}^{1}\frac{n \sin x}{1+n^2x^2}dx$?
Appreciate your help
 A: The change fo variable $u=nx$ gives
$$
\int^\infty_0\frac{\sin(u/n)}{1+u^2}\,du
$$
The integrand is dominated by $\frac{1}{1+u^2}$ which is integrable. Then, by dominated convergence
$$
\lim_{n\rightarrow\infty}\int^\infty_0\frac{n\sin x}{1+n^2x^2} =\lim_{n\rightarrow\infty}\int^\infty_0\frac{\sin(u/n)}{1+u^2}\,du=
\int^\infty_0\lim_{n\rightarrow\infty}\frac{\sin(u/n)}{1+u^2}\,du=0
$$
for $\sin(u/n)\xrightarrow{n\rightarrow\infty}0$ for all $u$.
A: And idea: substitute
$$x=\frac1u\implies dx=-\frac1{u^2}du\implies\int_0^1\frac{n\sin x}{1+n^2x^2}dx=\int_\infty^1-\frac{du}{u^2}\cdot\frac{n\sin\frac1u}{1+\frac{n^2}{u^2}}=$$
$$=\int_1^\infty\frac{n\sin\frac1u}{u^2+n^2}du$$
and now estimate
$$\left|\frac{n\sin\frac1u}{u^2+n^2}\right|\le\frac n{u^2+n^2}\,,\,\,\text{and}\;\int_1^\infty\frac n{n^2+u^2}du=\int_1^\infty\frac{d\left(\frac un\right)}{1+\left(\frac un\right)^2}=$$
$$=\left.\arctan\frac un\right|_1^\infty=\frac\pi2-\arctan\frac1n\xrightarrow[n\to\infty]{}\frac\pi2$$
Which means your integral is bounded above...
