# Calculating $\lim_{n \rightarrow \infty} \int_{[1, \infty)} \frac{n \sin(\frac{x}{n}) }{x^3}\,dx$

I want to compute $$\lim_{n \rightarrow \infty} \int_{[1, \infty)} \frac{n \sin(\frac{x}{n}) }{x^3}\,dx$$ I tried to use Lebesgue's dominated convergence theorem. It holds: $$|f_n(x)|= | \frac{n \sin(\frac{x}{n}) }{x^3}\ | \leq \frac{n x/n}{x^3} =\frac{1}{x^2}$$ The last function is lebesgue integrable.You can compute $$f_n \rightarrow \frac{x}{x^3} = \frac{1}{x^2}$$ So $$f_n$$ converges pointwise. Therefore the theorem says I can swap limit and integral. So I get $$\int_{[1, \infty)} \frac{1}{x^2} dx$$ Now I think there is a theorem, that says: If the improper-Riemann-integral over f converges, f is also lebesgue integrable. This is the case. I only don't know, wether the values of these integrals are the same?

I would say: $$\int_{[1, \infty)} \frac{1}{x^2} dx = \int_{1}^ {\infty} \frac{1}{x^2} dx = 1$$ Is this right?

• Yes. It is right. (Any absolutley convergent Riemann-integrable function is Lebesge integrable and vice-versa) – Tito Eliatron Jun 3 at 18:28
• @TitoEliatron: The values are the same? – Steven33 Jun 3 at 18:30
• If the improper-Riemann-integral over f converges, f is also lebesgue integrable This is false. See $\sin(x)/x$ en.wikipedia.org/wiki/Improper_integral . However, it is true for positive functions. In that cases both integrals agree. It follows from the monotone convergence theorem. – mlainz Jun 3 at 18:36
• @mlainz Ok in this case it is true, because $\frac{1}{x^2}$ is a positive function. But I cannot say something about the values of these integrals. – Steven33 Jun 3 at 18:42
• @TitoEliatron: Thank you very much: I appreciate your help:) – Steven33 Jun 3 at 19:11