If $f(x)=\sin^2(x)/x^6$, then does the integral for $f(x)$ from $0$ to $\infty$ converge or diverge?

Question

So, I was able to find that for $\frac{\sin(x)}{x}$ it converges to $\pi/2$, but for an expression more like the one I wrote, with the $x$ being raised to a higher exponent than the $\sin(x)$, I haven't been able to find anything.

Thanks!

Answer 1

The integral $$ \int_0^\infty\frac{\sin^n(x)}{x^m}dx $$ with positive integer $m,n$ converges only if $n\ge m$. To see that the condition is necessary observe that the integrand behaves like $x^{n-m}$ in the vicinity of $0$.

Answer 2

Note that, for $0<x<\pi/3$, $$ \sin x>\frac{x}{2} $$ so $$ \int_{\varepsilon}^{\pi/3}\frac{\sin^2x}{x^6}\,dx\ge\int_{\varepsilon}^{\pi/3}\frac{1}{4x^4}\,dx $$ This decides for nonconvergence at 0.