Help on the Integration of $\int_0^{\infty} e^{-bx}\sin ax^2 \, \mathrm{d}x$. I have had the misfortune of coming across the following integral, for real $b$ and $a > 0$:
$$\int\limits_{0}^{\infty} e^{-bx} \sin\left(ax^{2}\right) \, \mathrm{d}x.\tag{1}$$
Naturally, I proceeded by taking (1) as equivalent to the imaginary part of the following integral:
$$\Im\int\limits_{0}^{\infty} e^{-bx} e^{iax^{2}}\,\mathrm{d}x.$$
Completing the square in the exponential yielded
$$
\Im \, {e}^{ib^{2}/4a}\int\limits_{0}^{\infty}
\exp\left(ia\left[\, x + {ib \over 2a} \, \right]^{2} \right) \, \mathrm{d}x
$$
whereupon I took out the term $e^{ib^{2}/4a}$ and factored the remaining polynomial. However, I am not sure if this is the right approach. I want to make it clear, though, that I am not looking for a complete evaluation, but rather advice on how to proceed.
Update:
Continuing from where I left off, I make the substitution $u = x + ib/2a$.
$$\Im \, e^{ib^2/4a} \int\limits_{c}^{\infty+c} e^{iau^2} \, \mathrm{d}u, \tag{2}$$
where $c=ib/2a$. Wolfram Alpha returns a mildy tame expression for this, namely
$$\int e^{iax^2} \, \mathrm{d}x = -\frac{e^{i3\pi/4}}{2}\sqrt{\frac{\pi}{a}} \; \mathrm{erfi}\left(e^{i\pi/4}\sqrt{a} \, x\right)+K,$$
for some constant $K$, and so we seek
$$-\frac{e^{i3\pi/4}}{2}\sqrt{\frac{\pi}{a}} \; \mathrm{erfi}\left(e^{i\pi/4}\sqrt{a} \, x\right) \Bigg|_{c}^{\infty+c},$$
in order to compute the remaining integral in (2). Evaluating this at the respective limits, however, does not seem very easy.

If you are still looking for the answer to this, Mathematica 11.1 gives the answer in terms of FresnelC and FresnelS functions, however I managed to also tease a hypergeometric solution out.
$$
I(a,b) = \int\limits_{0}^{\infty} e^{-bx} \sin\left(ax^{2}\right) \; dx
$$
Take the Mellin transform of the integrand with respect to $b$,
$$
\int_0^\infty b^{s-1}e^{-b x}\sin(a x^2) \; db = x^{-s}\Gamma(s)\sin(a x^2)
$$
then integrate over $x$ assuming $a>0$
$$
\int_0^\infty x^{-s}\Gamma(s)\sin(a x^2) \; dx = \frac{1}{2}a^{\frac{1}{2}(s-1)}\cos\left(\frac{\pi(s+1)}{4}\right)\Gamma\left(\frac{1-s}{2}\right)\Gamma(s)
$$
then perform the inverse Mellin transform (using Mathematica)
$$
\frac{1}{2 \pi i} \int_{c-i \infty}^{c+i \infty} b^{-s}\frac{1}{2}a^{\frac{1}{2}(s-1)}\cos\left(\frac{\pi(s+1)}{4}\right)\Gamma\left(\frac{1-s}{2}\right)\Gamma(s) \; ds= I(a,b)
$$
this gives 
$$
I(a,b) = \frac{-2b\;_1F_2\left(1\bigg|\frac{3}{4},\frac{5}{4}\bigg|-\frac{c^2}{4}\right)+\sqrt{2\pi a}\left(\cos(c)+\sin(c)\right)}{4a}
$$
where $c=b^2/(4a)$. This seems to check out for a few numerical values.
