I would really appreciate feedback/guidance. This is from a past year exam and my professor didn't cover much on improper integrals. I'm trying to prove that $\int_0^{\infty} \frac{\sin x}{x^p}\, dx$ converges for $0<p<2$ using the comparison test for improper integrals. I know that from the test, if $|f(x)| \leq g(x) \forall x\geq a$ then $\int_a^{\infty} g(x)\, dx$ converges.
EDIT: Here's my second attempt at the proof after much discussion in the comments below.
We know that $\frac{\sin x}{x^p}$ will be continuous on 1 because $\lim_{x\rightarrow 1} \frac{\sin x}{x^p} = lim_{x\rightarrow 1} x^{-p} \sin(x) = 1^{-p}\sin(1) = \sin(1)$.
However, for 0 we have that:
$\lim_{x \rightarrow 0} \frac{\sin x}{x^p} = \frac{\sin 0} {0^{p}} = \frac{0}{0^{-p}} = \frac{1}{0^{p-1}}$. Since we can't have 0 in the denominator, we only have continuity on 0 if p-1 < 1, which implies p<2.
And so, by a theorem, we know that $\frac{\sin(x)}{x^p}$ is Riemann integrable on [0,1] for p<2. Furthermore, by the improper limit comparison test, if we prove that $\int_1^{\infty} f(x)\, dx$, then we can show that $\int_0^{\infty} f(x)\, dx$ converges. Hence we have that
$\lim_{b \rightarrow \infty} \int_1^{b} \frac{\sin(x)}{x^p}\,dx = \lim_{b \rightarrow \infty} \int_1^{b} \frac{\sin(x)}{x^p}\,dx =\lim_{b\rightarrow \infty} -\frac{cos x}{{p}} |_p^{b} – \int_1^{b} \frac{\cos x}{(p-1)x^{p+1}}\,dx$
And so I know that since $\int \frac{cos x}{x^{p+1}} < \int \frac{1}{x^{p+1}}$, and since the RHS will converge if and only if p+1 > 1, we have that p>0.
And so, we conclude that the integral converges for $0<p<2$.
Best Answer
Integrate by parts with $u=x^{-p}$ and $v=-\cos(x)$. Then, we can write
$$\int_1^L \frac{\sin(x)}{x^p}\,dx=\left.\left(-\frac{\cos(x)}{x^p}\right)\right|_{1}^{L}-p\int_1^L \frac{\cos(x)}{x^{p+1}}\,dx$$