Interval of p where $\int_0^\infty{\sqrt{x}\sin(\frac{1}{x^p})}dx$ converges $$\int_0^\infty{\sqrt{x}\sin{\left(\frac{1}{x^p}\right)}}dx$$
In my attempt I used the small angle approximation of $\sin{x}$. I stated that  as x approaches infinity $\frac{1}{x^p}$ approaches 0, so $\sin{\left(\frac{1}{x^p}\right)} = \frac{1}{x^p}$ and simplified the equation to $$\int_0^\infty{\sqrt{x}\cdot\frac{1}{x^p}}dx$$ getting $$\int_0^\infty{\frac{1}{x^{p-\frac{1}{2}}}}dx$$. I stated that for the integral to converge, $p-\frac{1}{2}>1$ because of how the sum for p-series needs to meet this requirement.
Am I doing this correctly?

Your answer is correct but there is a flaw in  the argument. $\int_0^{\infty} \frac   1{x^{p-\frac 1  2}}dx=\infty$ for all $p$. What you have to do is to split the integral into integrals from $0$ to $1$ and $1$ to $\infty$. The first integral converges for all $p$. The second integral converges iff $p >\frac  1 2$ by your argument. 
