Finding whether improper integrals exist

Question

For which $p>0$ does the following improper integral exist? $$\int^{\infty}_{1} x^{-p}\sin{x} \ dx$$ how do I find the value of p?

Answer 1

The integral is convergent for any $ p > 0 $; just integrate by parts to get $$ \int_{1}^{+\infty}{\frac{\sin(x)}{x^p}} = -\frac{\cos(x)}{x^p}\Big|_{1}^{+\infty} - p\int_{1}^{+\infty}{\frac{\cos(x)}{x^{p+1}}}, $$ and the integral at the rhs obviously converges since its modulus is no greater than $ 1/x^{p+1} $ (which is summable on $ (1,+\infty) $).

If $ p \le 0 $ the integral is apparently non convergent, since it oscillates without being "tamed".