Is $\frac {\sin^2(x)}{x^2}$ Lebesgue integrable on $[0,\infty)$?

Question

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.

Answer 1

Show that $f(x)=\frac{\sin^2\!x}{x^2}$ is continuous in $[0,\infty)$ and you are done.

Answer 2

You can check that $\int^1_0\frac{\sin^2(x)}{x^2}dx$ is Riemann integrable, hence Lebesgue integrable. On the other hand, the Dominated Convergence Theorem shows that $\int^\infty_1\frac{\sin^2(x)}{x^2}dx$ is Lebegue integrable:

consider the sequence of functions $g_n(x)=\frac{\sin^2(x)}{x^2}\chi_{[1,n]}(x)$ and note that $g_n(x)\le f(x)=1/x^2$. Since $f\in L^1(\mathbb R)$, we may apply the DCT to conclude that $\lim g_n(x)=\frac{\sin^2(x)}{x^2}$ is Lebesgue integrable.

Now, combining these two integrals, using the additivity of the Lebegue integral, we have that $\int^{\infty}_0\frac{\sin^2(x)}{x^2}dx\in L^1(\mathbb R).$

Answer 3

Since $$ \frac{\sin x}{x} \to 1\quad\text{as $x\to 0$}, $$ there is some $\delta > 0$ and some $M > 0$ such that $$ \bigg|\frac{\sin^2 x}{x^2}\bigg|< M\quad \text{if $0\le x < \delta$}. $$ 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 $[0,\infty)$.

Answer 4

AOrtiz has shown that the integral is convergent. To complete the answer, I will show that $$ \int_0^\infty \frac {\sin^2x}{x^2}\,dx= \int_0^\infty \frac {\sin x}{x}\,dx. $$ Take $\epsilon>0$. Then, integrating by parts and using that $2\sin^2x=1-\cos(2\,x)$, we have \begin{align} \int_\epsilon^\infty \frac {\sin x}{x}\,dx&=-\frac{\cos x}{x}\Bigr|_\epsilon^\infty-\int_\epsilon^\infty \frac {\cos x}{x^2}\,dx\\ &=\frac{\cos\epsilon}{\epsilon}-\int_{\epsilon/2}^\infty \frac {\cos(2\,t)}{2\,t^2}\,dt\\ &=\frac{\cos\epsilon}{\epsilon}+\int_{\epsilon/2}^\infty \frac {1-\cos(2\,t)}{2\,t^2}\,dt-\int_{\epsilon/2}^\infty \frac {dt}{2\,t^2}\\ &=\frac{\cos\epsilon-1}{\epsilon}+\int_{\epsilon/2}^\infty \frac {\sin^2t}{t^2}\,dt. \end{align} Let $\epsilon\to0$ and you are done.