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.
