We have to evaluate the limit of a sequence $\{f_n\}$ which is $$\lim_{n\rightarrow \infty }\int \frac{n\sin(x/n)}{x(1+x^2)}\,dx$$ using Lebesgue Dominated Convergence Theorem. The hint is: $\int \frac{1}{1+x^2}\,dx=\pi$.
My solution so far: I was looking for a dominating function $g$ and searched this forum as well to find $$|f_n|=\Big|\frac{n\sin(x/n)}{x(1+x^2)}\Big|\leq \Big|\frac{1}{1+x^2}\Big|.$$ But now I don't know how to take the next step, I would like to show the conclusion of LDCT i.e $\lim_{n\rightarrow \infty}\int f_n = \int f$.
Best Answer
$|f_{n}|\leq \frac{1}{1+x^{2}}$ and $$\int_{\Bbb{R}}\frac{1}{1+x^{2}}\,d\lambda=\pi<\infty$$ (evaluate using $\arctan$).
Thus all the conditions for DCT are satisfied and hence you can switch the order of limit and integration .
$$\lim_{n\to\infty}\int_{\Bbb{R}}f_{n}\,d\lambda =\int_{\Bbb{R}}\lim_{n\to\infty}\bigg(\frac{\sin(\frac{x}{n})}{\frac{x}{n}}\cdot\frac{1}{(1+x^{2})}\bigg)\,d\lambda=\int_{\Bbb{R}}\frac{1}{1+x^{2}}\,d\lambda=\pi$$