Is $\frac {\sin^2(x)}{x^2}$ Lebesgue integrable on $[0,\infty)$? How would someone show $$x\mapsto\frac{\sin^2(x)}{x^2}$$ is Lebesgue integrable on $[0, \infty)$? Also that  

$$\int_{[0,\infty)} \frac {\sin^2(x)}{x^2} d\lambda = \int_{[0,\infty)} \frac {\sin(x)}{x} d \lambda$$ 

My thoughts so far:

$$\frac{\sin^2(x)}{x^2} \leq \frac{1}{x^2}$$
Since $1/x^2$ is Lebesgue integrable, then $\sin^2x/x^2$ must be as well.

But I don't know how to proceed with showing the other part.
Since
$$
\frac{\sin x}{x} \to 1\quad\text{as $x\to 0$},
$$
there is some $\delta &gt; 0$ and some $M &gt; 0$ such that
$$
\bigg|\frac{\sin^2 x}{x^2}\bigg|&lt; M\quad \text{if $0\le x &lt; \delta$}.
$$
Since everything in sight is non-negative, by additivity,
$$
\int_0^\infty\frac{\sin^2 x}{x^2}\,dx = \int_0^\delta\frac{\sin^2 x}{x^2}\,dx + \int_\delta^\infty\frac{\sin^2 x}{x^2}\,dx,
$$
and the first integral is convergent because it is bounded above by $M\delta$. The second integrand is bounded above by $x\mapsto \frac{1}{x^2}$, which belongs to $L^1\big([\delta,\infty)\big)$, by the $p$-test if you like. Hence $x\mapsto \frac{\sin^2 x}{x^2}$ is Lebesgue integrable on $[0,\infty)$.
Show that $f(x)=\frac{\sin^2\!x}{x^2}$ is continuous in $[0,\infty)$ and you are done.