Can one deduce a general form for the integral:
$$\mathcal{J}(a)=\int_0^\infty \frac{\sin (x^2+ax)}{x}\, {\rm d}x$$
I guess that the Feymann trick should do here. Differentiating $J$ with respect to $a$ one gets:
$$J'(a)= \int_0^{\infty} \cos (x^2 +ax)\, {\rm d}x$$
Using the formula $\cos (\theta+\phi)=\cos \theta \cos \phi - \sin \theta \sin \phi $ and letting $\theta=x^2, \; \phi=ax$ one gets that the latter integral should equal to:
$$J'(a)=\int_0^\infty \cos x^2 \cos ax \, {\rm d}x - \int_0^\infty \sin x^2 \sin ax \, {\rm d}x $$
I can do the latter integrals for some specific values of $a$. For example if $a=1$ the first one is known,
$$\int_0^\infty \cos x^2 \cos x \, {\rm dx} = \frac{1}{2}\sqrt{\frac{\pi}{2}} \left( \sin \frac{1}{4} + \cos \frac{1}{4} \right)$$
while the second one evaluates to something that involves the Fresnel cosine and sine integrals.
So, perhaps this is not best approach for this specific integral. Do you see another pattern?