How to prove $\int_{0}^{1} \frac{\sin(x)^2}{x^{5/2}}$ is divergent

Question

The original problem is to test the convergence of $\int_{0}^{\infty} \frac{\sin(x)^2}{x^{5/2}}$. It is easy to prove that $\int_{1}^{\infty} \frac{\sin(x)^2}{x^{5/2}}$ is convergent, but I have checked in https://www.integral-calculator.com that $\int_{0}^{1} \frac{\sin(x)^2}{x^{5/2}}$ is not. How could I prove it? Thanks in advance.

Answer 1

Since $\lim_{x\to0}\frac{\sin x}x=1$, you know that $\lim_{x\to0}\frac{\sin^2x}{x^2}=1$. So, near $0$ your function behaves as $\frac1{x^{1/2}}$ and therefore it converges.

Answer 2

Actually, it does converge. (I'm not sure where you went wrong in using WA.) For a proof, note the integrand approximates $x^{-1/2}$ at small $x$, with antiderivative $\approx 2\sqrt{x}$.