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It is required to calculate $\lim_{n\to\infty}\int_{[0,\infty)}\frac{n\sin\frac{x}{n}}{x(1+x^2)}dx$. The following is my attempt.

Let $f_n(x)=\frac{n\sin\frac{x}{n}}{x(1+x^2)}$ for each $n\in\mathbb{N},x>0$. Then $f_n(x)\leq \frac{n\frac{x}{n}}{x(1+x^2)}=\frac{1}{1+x^2}$ for each $n\in\mathbb{N}, x>0$. Define $g:(0,\infty)\to\mathbb{R}$ as $g(x)=\frac{1}{1+x^2}$. Then $f_n\leq g$ a.e. on $[0,\infty)$, and $\int_{[0,\infty)}g$ exists finitely. Also $(f_n)$ converges to $g$ pointwise a.e. on $[0,\infty)$. Therefore by Lebesgue Dominated Convergence theorem, $\lim_{n\to\infty}\int_{[0,\infty)}\frac{n\sin\frac{x}{n}}{x(1+x^2)}dx=\int_{[0,\infty)}g$.

Is my solution alright? Thank you.

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    $\begingroup$ I'd be careful to say that $\lvert f_n\rvert\leq g$ as opposed to $f_n\leq g$ -- remember, the fact that your pointwise limit is nonnegative doesn't mean that your individual functions are nonnegative. $\endgroup$ – Nick Peterson May 24 '17 at 16:44
  • $\begingroup$ I agree. Is that the only error though? $\endgroup$ – Janitha357 May 24 '17 at 16:48
  • $\begingroup$ the result should be $$\frac{\pi}{2}$$ $\endgroup$ – Dr. Sonnhard Graubner May 24 '17 at 17:19
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It is enough to apply the dominated convergence theorem. The function $$ f_n(x) = \frac{\sin\frac{x}{n}}{\frac{x}{n}}\cdot\frac{1}{1+x^2} $$ belongs to $L^1(\mathbb{R}^+)$ and it is dominated in absolute value by the integrable function $f(x)=\frac{1}{1+x^2}$.
The pointwise limit is precisely given by $f(x)$: $$ \forall x\in\mathbb{R}^+,\qquad \lim_{n\to +\infty} f_n(x) = f(x) $$ hence it follows that $$ \lim_{n\to +\infty}\int_{0}^{+\infty}f_n(x)\,dx = \int_{0}^{+\infty}f(x)\,dx = \frac{\pi}{2}.$$

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