Okay, so I'm looking now at the absolute integrability, over the real line, of
$$f(x) = \operatorname{sinc}^2(2 \pi x) = \left\{\begin{matrix} \frac{\sin^2(2 \pi x)}{x^2} & x \neq 0\\ 1 & x = 0 \end{matrix}\right.$$
Since the integral we're ultimately concerned with is $\int_{-\infty}^\infty | f(x) |dx$, I first proved that $| f | = f$, but I'm not sure where to go from there.
Direct computation of the integral isn't impossible - it evaluates to $2\pi^2$ - but I feel it's a bit beyond the coursework since it involves the sine integral ($\operatorname{Si}(x)$) and what not. So while direct computation works, I feel there's a more elegant solution to this.
My gut guess is that there's some sort of way I can utilize the comparison test. Since I know the integral works out to a finite value, I need some function $g$ where $|g| \geq |f|$ and where $g$ is absolutely integrable on $\mathbb{R}$. My initial guess was $1/x^2$, since $\sin^2(2\pi x) \in [0,1]$, but that isn't absolutely integrable, so that's out.
Any ideas?
$\int_0^{1}\frac {\sin ^{2} (2\pi x)} {x^{2}} <\infty$ because the integrand is a bounded measurable function. $\int_1^{\infty}\frac {\sin ^{2} (2\pi x)} {x^{2}}\leq \int_1^{\infty}\frac 1 {x^{2}} <\infty$. (Since the integrand is an even function the integral on the negative line is same as the one on the positive real line).