Trying to find the following improper integral's divergence, which I've split into two integrals:
$\int_{0}^{\infty} \frac{\sin{x}}{x^2} dx = \int_{0}^{1} \frac{\sin{x}}{x^2} dx + \int_{1}^{\infty} \frac{\sin{x}}{x^2} dx$.
For the second integral, I've shown that it's convergent by the comparison test with $ \frac{\sin{x}}{x^2} \leq \frac{1}{x^2}$. For the first, I'm not sure how to show it's divergent.
I've given that $\sin{x} \approx x $ when $x$ is small, which means you can compare the integral to $\int_{0}^{1} \frac{1}{x}$ which is always smaller in the interval $0<x<1$, but I've not shown that the integral of $\frac{1}{x}$ is divergent (I know it is as I've seen the p-test online, but it's not something covered yet).
Is there another way to approach this question?