Determine whether the integral $$\int_0^\infty \frac{\sin^2x}{x^2}~\mathrm dx$$ converges.
I know it converges, since in general we can use complex analysis, but I'd like to know if there is a simpler method that doesn't involve complex numbers. But I cannot come up with a function that I could compare the integral with.
Best Answer
Hint:$$x>1\implies0\le\frac{\sin^2(x)}{x^2}\le\frac1{x^2}\\0<x<1\implies1\le\frac{\sin^2(x)}{x^2}\le\frac1{\cos^2(x)}\le\frac1{\cos^2(1)}$$The second of the two coming from the proof of the derivative of $\sin$ using squeeze theorem.