# 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.

• Try using contour integration with a semi-circular contour. – aleden Sep 26 '18 at 23:01

## 1 Answer

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 $$\frac{\pi}{4}\Big( (b-2) -2b + (b+2) \Big) + \kappa' = 0 \implies \kappa' = 0.$$