Evaluating $\int_{-\infty}^\infty e^{i(ax^2+bx+c)}\frac{\operatorname{sin}^2x}{x^2}dx$ Is there a closed form for the following integral?
$$\int_{-\infty}^\infty e^{i(ax^2+bx+c)}\frac{\operatorname{sin}^2x}{x^2}dx$$
Nothing of this form seems to appear in Gradshteyn and Ryzhik.
 A: There is a closed form expression in terms of the error function with complex argument.
I'll ignore the leading factor $\exp{(i\,c)}.$ The answer is found by setting $q=-i \,a$ in the following
$$ \quad \text{Define} \quad I(q,b):=\int_{-\infty}^\infty \exp{(-q\,x^2+i\,b\,x)}\frac{\sin^2(x)}{x^2} dx $$
$$F(q,b):=2\sqrt{q\pi} \exp{\big(-\frac{b^2}{4q}\big)}+b\,\pi\,\text{erf}\big(\frac{b}{2\sqrt{q}}\big) $$
Then
$$  I(q,b)=\frac{1}{4}\Big(F(q,b-2)-2F(q,b)+F(q,b+2) \Big) $$
To start the proof, assume $q>0.$  There is no problem with $q$ being purely imaginary because the rest of the integrand is without singularities.
Note that
$$ \frac{d}{dq} I(q,b)= -\int_{-\infty}^\infty \exp{(-q\,x^2+i\,b\,x)}\sin^2(x) dx $$
Use $\sin^2(x)=(1-\cos{2x})/2.$ We therefore have for the left-hand side an expression that is a linear combination (3 terms) of the well-known integrals 
$$ \int_{-\infty}^\infty \exp{(-q\,x^2+i\,c\,x)}dx = \sqrt{\frac{\pi}{q}} \exp{\big(-\frac{c^2}{4q}\big)}.$$
The $c$ will take values of $b-2, \, b, \text{ and } b+2.$
We need to integrate the previous and Mathematica finds
$$ \int \sqrt{\frac{\pi}{q}} \exp{\big(-\frac{c^2}{4q}\big)} dq = F(q,c)+\kappa$$
where $\kappa$ is a constant that could depend on $c$ but not $q$. The appropriate linear combination gives you the stated answer up to a constant $\kappa'.$  To prove  $\kappa'=0,$ it suffices to choose $q=0$ as in the following calculation:
$$\frac{1}{4}\Big(F(0,b-2)-2F(0,b)+F(0,b+2)\Big)+\kappa'=\int_{-\infty}^\infty \exp{(i\,b\,x)}\frac{\sin^2(x)}{x^2} dx =0$$
The integral is zero by symmetry and since $\text{erf}(1/0)=1$ we have
$$\frac{\pi}{4}\Big( (b-2) -2b + (b+2) \Big) + \kappa' = 0 \implies \kappa' = 0.$$
