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.


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 $\sqrt{2ic}$.
