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?
The inequality $\left|\sin x\right|\leq\min(1,|x|)$ is enough to prove that the integral is converging.
Integration by parts and the well-known $\int_{0}^{+\infty}\frac{\sin x}{x}\,dx=\frac{\pi}{2}$ then give: $$\begin{eqnarray*} I = \int_{0}^{+\infty}\frac{\sin^4 x}{x^2}\,dx &=& \int_{0}^{+\infty}\frac{4\cos x \sin^3 x}{x}\,dx\\&=&\int_{0}^{+\infty}\frac{\sin(2x)-\frac{1}{2}\sin(4x)}{x}\,dx\\&=&\left(1-\frac{1}{2}\right)\int_{0}^{+\infty}\frac{\sin x}{x}\,dx=\color{red}{\frac{\pi}{4}}.\end{eqnarray*}$$
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.
