I am trying to find a close form representation for the following integral:
$$ A(x;a,b,c)= \int_{0}^{x}\frac{\sin\left(a k+b k^{2}\right)+\sin\left(c k-b k^{2}\right)}{k}dk $$ for $0<x \ll \infty$. I can prove that the integral exists and it is asymptotically bounded in $x$. Looking at its asymptotic properties, I also have the hunch that it is related with Sine integral function, error function as well as Fresnel integrals.
If a close form is hard, I would also appreciate an analytic approximation or close form expression of the Taylor series coefficients with respect to $x$ or a recurrence relation that can help to compute it.
I don't believe there is a closed form for the integral. But the Maclaurin series wrt $x$ can be obtained from the Maclaurin series for the integrand. Expand the two sines. I'll look at just one of the terms. Note that
$$\eqalign{ \dfrac{\sin(ak) \cos(bk^2)}{k} &= \sum_{i=0}^\infty \sum_{j=0}^\infty (-1)^{i+j} \dfrac{a^{2i+1} b^{2j}}{(2i+1)!(2j)!} k^{2i+4j}\cr &= \sum_{m=0}^\infty \sum_{j=0}^{\lfloor m/2 \rfloor} (-1)^{m-j} \dfrac{a^{2m-4j+1} b^{2j}}{(2m-4j+1)! (2j)!} k^{2m}} $$ so that the integral is
$$ \int_0^x \dfrac{\sin(ak) \cos(bk^2)}{k}\; dk = \sum_{m=0}^\infty \sum_{j=0}^{\lfloor m/2 \rfloor} (-1)^{m-j} \dfrac{a^{2m-4j+1} b^{2j}}{(2m-4j+1)! (2j)! (2m+1)} x^{2m+1}$$
The coefficient of $x^{2m+1}$ can be expressed using a hypergeometric function: $$(-1)^m {\frac {{a}^{2\,m+1}}{\left( 2\,m+1 \right)! } {\mbox{$_4$F$_1$}(-\frac{m}2,-\frac{m}2-\frac14,-\frac{m}2+\frac14,-\frac{m}2+\frac12;\,\frac12;\,-64\,{\frac {{b}^{2}}{{a}^{4}}})} } $$