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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?

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    $\begingroup$ Yes. It is right. (Any absolutley convergent Riemann-integrable function is Lebesge integrable and vice-versa) $\endgroup$ – Tito Eliatron Jun 3 at 18:28
  • $\begingroup$ @TitoEliatron: The values are the same? $\endgroup$ – Steven33 Jun 3 at 18:30
  • $\begingroup$ 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. $\endgroup$ – mlainz Jun 3 at 18:36
  • $\begingroup$ @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. $\endgroup$ – Steven33 Jun 3 at 18:42
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    $\begingroup$ @TitoEliatron: Thank you very much: I appreciate your help:) $\endgroup$ – Steven33 Jun 3 at 19:11

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