Improper integral: $\int_0^\infty \frac{\sin^4x}{x^2}dx$ I have been trying to determine whether the following improper integral converges or diverges:
$$\int_0^\infty \frac{\sin^4x}{x^2}dx$$
I have parted it into two terms. The first term: $$\int_1^\infty \frac{\sin^4x}{x^2}dx$$ converges (proven easily using the comparison test), but the second term:
$$\int_0^1 \frac{\sin^4x}{x^2}dx$$ troubles me a lot. What could I do?

For the second term, you can re-write your function as $\frac{\sin x}{x} \cdot \frac{\sin x}{x} \cdot \sin^2(x)$.  Note that on $[0,1]$, this function is continuous (by the Fundamental Trig Limit you can extend the function to be defined as $f(0)=1$ at $x=0$).  But then any continuous function on a closed interval is integrable, so $\int_0^1 \frac{\sin^4(x)}{x^2}$ converges.  Hence the whole integral converges.
