Show if a function is an $L^1$ function

Question

Is $\frac{\sin x}{x^2}$ an $L^1$ function on $[\pi,\infty)$? Does knowing $\frac{\sin x}{x}$ and $\frac{1}{x}$ are not in $L^1 ([\pi,\infty))$ help?

Answer 1

$$ \left|\frac{\sin x}{x^2}\right|\leqslant\frac{1}{x^2} $$ Since $\frac{1}{x^2}\in L^1([\pi,\infty))$, $\frac{\sin x}{x^2}$ is $L^1$ on $[\pi,\infty)$.