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.

  • 1
    $\begingroup$ 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

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 $$ \lim_{n\to +\infty}\int_{0}^{+\infty}f_n(x)\,dx = \int_{0}^{+\infty}f(x)\,dx = \frac{\pi}{2}.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.