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 $\varphi \in L^2(\mathbb{R})$.

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.
