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.
 A: 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.
