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?