Stuck with integral $\int_{-\infty}^\infty \left( \frac{\sin(a t+b)}{at+b} \right)^2 \, dt$ I am stuck with the following integral:
$$\int_{-\infty}^\infty \left( \frac{\sin(a t+b)}{at+b} \right)^2 \, dt$$
I would like to show that $\varphi(t)=\frac{\sin(at+b)}{at+b}$ belongs to $L^2(\mathbb{R})$ and/or $L^1(\mathbb{R})$, i.e. $\int_{-\infty}^\infty | \varphi |^2 \,dt < \infty$   and/or $\int_{-\infty}^\infty | \varphi | \,dt < \infty $.
So far, I know that it is $|\frac{\sin(at+b)}{at+b}| \leq |\frac{1}{at+b}|$, but as $\int_{-\infty}^\infty |\frac{1}{at+b}|^2 \, dt$ does not converge, I cannot be conclusive. Looking at the plot it can be stated that it converges and then $\varphi \in L^2(\mathbb{R})$.
 A: Under the obvious substitution,
$$\int_{-\infty}^\infty\frac{\sin^2(at+b)}{(at+b)^2}\,dt
=\frac1{|a|}\int_{-\infty}^\infty\frac{\sin^2 t}{t^2}\,dt.$$
The integrand is bounded near zero, and $O(t^{-2})$ at infinity, so
the integral converges.
By one of those amazing coincidences,
$$\int_{-\infty}^\infty\frac{\sin^2 t}{t^2}\,dt
=\int_{-\infty}^\infty\frac{\sin t}{t}\,dt=\cdots$$
a very well-known integral.
A: Yet another strategy: once the original integral has been reduced to $\frac{2}{|a|}\int_{0}^{+\infty}\frac{\sin^2 x}{x^2}\,dx$, one may invoke $\mathcal{L}(\sin^2 x)(s)=\frac{2}{s(4+s^2)}$ and $\mathcal{L}^{-1}\left(\frac{1}{x^2}\right)(s)=s$ to further reduce it to
$$ \frac{4}{|a|}\int_{0}^{+\infty}\frac{ds}{s^2+4} = \frac{\pi}{|a|}.$$
A: One standard strategy is$$\int_{-\infty}^\infty\frac{\sin^2 y}{y^2}dy=\int_{-\infty}^\infty\frac{e^{2iy}+e^{-2iy}-2}{-4}\int_0^\infty ze^{-zy}dzdy=\int_{-\infty}^\infty\frac{\frac{z}{z-2i}+\frac{z}{z+2i}-2}{-4}dz.$$I'll leave the rest to you.
A: I finally came out with this solution:
I computed the Fourier transform of $\varphi(t)$, which is:
$$\hat{\varphi}(w) = \sqrt{\frac{\pi}{2} } \frac{e^{iwb/a}}{a}\chi_{[-a,a]}(w)$$
Then I learned that it is bandlimited, and as $\varphi \in L^2(\mathbb{R})$, then $\varphi$ lives in the Paley-Wiener space $PW_{a}$. Therefore, as in $L^2[-a,a]$ the Fourier transform is a unitary operator, it holds that:
$$\langle x,y \rangle = \langle \mathcal{F}x, \mathcal{F}y \rangle$$
so,
$$I = \int_{-\infty}^{+\infty}{\left( \frac{sen(at+b)}{at+b} \right)^2 dt}$$ 
and:
$$I = \langle \varphi, \varphi \rangle_{L^2({\mathbb{R}})} = \langle \hat{\varphi}, \hat{\varphi}\rangle_{L^2[-a,a]} = \frac{\pi}{2} \frac{1}{a^2} ||\chi_{[-a,a]}||^2 = \frac{\pi}{2} \frac{2a}{a^2} = \frac{\pi}{a}$$
