Integral of Sinc times Exponent of Squared variable I would like to integrate this in my research:
$$\int\limits_{-\infty}^\infty{\frac{e^{i b x^2}\sin{(a x)}}{x}}dx$$
where a and b are both real and greater than zero. If possible, I would like to take this a step further and integrate
$$\int\limits_{-\infty}^\infty{\frac{e^{i b (x-c)^2 }\sin{(a x)}}{x}}dx$$
where c is complex.
The topic is turbulence, and you can determine an exact answer on Mathematica when letting a and b be, say, 3 and 5 (with c=0). Fresnel integrals appear.
The routes which I have attempted:


*

*I have written $1/x$ as $\int_\infty^\infty{e^{-s x}}dx$. 

*Following (1), I have expanded $e^{i b (x-const)^2}$ (given the Laplace transform of $x^m$), but the resulting series included terms with with $(2n)!$ in the numerator such that the series diverged when summing from $n=0$ to $n=\infty$.

*I have represented $\frac{\sin(a x)}{x}$ as $\int_0^a {\cos(\alpha x)}d\alpha$, which of course yields diverging terms since the $1/x$ is what allows for convergence.

*Following (1), I have written the $e^{i b x^2}$ term as the derivative of the sum of Fresnel integral functions. Taking a contour integral yields a function with Fresnel integrals. This result can be verified on Mathematica. The problem is that, once I have arrived here (i.e. taking the Laplace transform of $e^{i b x^2}$), the integral over s becomes complicated (to the point that Mathematica can't solve it analytically even when a,b,c are specified). This appears to not be the way that Mathematica is solving the original integral.

 A: For first, get rid of the extra parameter by setting $c=\frac{b}{a^2}$:
$$I= \int_{\mathbb{R}}e^{ibx^2}\frac{\sin(ax)}{x}\,dx = \int_{\mathbb{R}}e^{icx^2}\frac{\sin x}{x}\,dx=\text{Im PV}\int_{\mathbb{R}}e^{icx^2+ix}\frac{dx}{x} $$
then translate the $x$ variable in order to get:
$$ I = \text{Im}\left(e^{-\frac{i}{4c}}\,\text{PV}\int_{\mathbb{R}}e^{icx^2}\frac{dx}{x-\frac{1}{2c}}\right)$$
Now the inner integral can be evaluated in terms of the $\text{Erfi}$ function. By taking the imaginary part of the integral multiplied by $e^{-\frac{i}{4c}}$, the Fresnel integrals make their appearance:

$$ \int_{-\infty}^{+\infty}e^{icx^2}\frac{\sin x}{x}\,dx = \pi(1+i)\left(C\left(\frac{1}{\sqrt{2\pi c}}\right)-S\left(\frac{1}{\sqrt{2\pi c}}\right)\right) \tag{1}$$

where:
$$ S(x) = \int_{0}^{x}\sin\left(\frac{\pi t^2}{2}\right)\,dt,\qquad C(x) = \int_{0}^{x}\cos\left(\frac{\pi t^2}{2}\right)\,dt.$$
Now $(1)$ can be checked also by differentiating both sides with respect to $c$, then by considering the limit of both sides as $c\to 0^+$. Another chance is to take the Fourier transform of $\frac{\sin x}{x}$, that is just a multiple of the indicator function of $(-1,1)$, and integrate it against the inverse Fourier transform of $e^{icx^2}$, that is given by the same function multiplied by a real constant times $\sqrt{2ic}$.
