Examine the convergence of $\int_0^1 \frac{\sin(1/x)}{x^p} \space dx $
$$\bigg| \frac{\sin(1/x)}{x^p}\bigg| \le \frac 1{x^p} $$
Now $$ \int _0^1 \frac {dx}{x^p} $$ is convergent for $p\ < 1$
So by comparisons test given integral is convergent for $p<1$
When $ 1\le p <2$
By the transformation $\frac 1x=u $
$$ \int_0^1 \frac{\sin(1/x)}{x^p}\space dx=\int_1^{\infty} u^{p-2}\sin u \space du$$
Again the right integral is $\int_1^{\infty} \frac{\sin u}{u^{2-p}} \space du$
Let $\phi(u)=\frac 1{u^{2-p}} $. Then $\phi$ is monotonic and bounded on $[1, \infty]$ and $\displaystyle{\lim_{u \to \infty}}\phi(u)=0 $
The integral $\int_0^X \sin u \space du $ is bounded on $[1,X] $ for all $X> 1$
Then the given integral is convergent for is convergent by Dirichlet's test when $ 1\le p <2$
When $p=2$ then it is divergent . But when $p>2$ , I have attempted to show the integral is divergent but without success (Like I tried to break the interval in $[2k\pi ,(2k+1)\pi]$ and get some inequality )
Please. Any hint regarding the case $p>2$ ?
Let $p>2$. For every $n \in \mathbb{N}$, let $\displaystyle u_n = \int_{\pi}^{n\pi} u^{p-2} \sin(u) du$.
Then for every $n \in \mathbb{N}$, one has $$|u_{n+1}-u_n| = \left|\int_{n\pi}^{(n+1)\pi} u^{p-2} \sin(u) du \right| = \int_{n\pi}^{(n+1)\pi} u^{p-2}\left| \sin(u)\right| du $$
(since $\sin$ keeps the same sign over $[n \pi, (n+1)\pi]$). So $$|u_{n+1}-u_n| \geq (n\pi)^{p-2} \int_{n\pi}^{(n+1)\pi} \left| \sin(u)\right| du =2(n\pi)^{p-2}$$
In particular, $(|u_{n+1}-u_n|)_{n \in \mathbb{N}}$ does not tend to $0$, and hence $(u_n)_{n \in \mathbb{N}}$ cannot converge.
Hence, the integral $\displaystyle \int_1^{+\infty} u^{p-2} \sin(u) du$ diverges, and by substitution, $$\int_0^1 \dfrac{\sin(1/x)}{x^p} dx \text{ diverges}$$