EDIT: I originally did the integral over $(-\infty,\infty)$ and got an elementary solution, instead of over the interval $[0,\infty)$ where we get special functions. This has been amended.
Define $I(a,b) = \int_0^{+\infty} \exp(-a t^2-bt)dt$ for $a\in \mathbb{R}_+$ and $b\in\mathbb{C}$. If we complete the square, we get
$I(a,b) = \int_0^{+\infty} \exp\left[ -a\left(t+\frac{b}{2a}\right)^2+\frac{b^2}{4a} \right]dt$
$=\exp\left(\frac{b^2}{4a}\right) \int_0^{+\infty} \exp\left[ -a\left(t+\frac{b}{2a}\right)^2\right]dt$
$=\exp\left(\frac{b^2}{4a}\right)\int_{b/2a}^\infty e^{-au^2}du$
$=\frac{1}{2} \sqrt{\frac{\pi}{a}}\exp\left(\frac{b^2}{4a}\right) \mathrm{erfc}\left(\frac{b}{2\sqrt{a}}\right).$
In the above, $\mathrm{erfc}(\cdot)$ denotes the complementary error function (which is defined for complex numbers). Now notice that the sine in the original integral can be split into complex exponentials, giving
$(*)=\frac{I(a^2,b-ci)-I(a^2,b+ci)}{2i}.$