Can $\int_0^{\infty} e^{-a^2 t^2 - b t} \sin c t \mathrm dt$ be done? I'm a physicist and I've been encountering integrals like
$\int_0^{\infty} e^{-a^2 t^2 - b t} \sin c t \;\mathrm dt$, where everything is real.
Mathematica could not solve it and I could not find it in any references. Any idea if this can be solved? How?
 A: 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
$$(*)=\frac{I(a^2,b-ci)-I(a^2,b+ci)}{2i}.$$
A: This is the imaginary part of $J(a,z)=\displaystyle\int_0^{+\infty}\exp(-a^2t^2+zt)\mathrm{d}t$ for $z=-b+\mathrm{i}c$ hence let us compute $J(a,z)$ for every nonzero real number $a$ and every complex number $z$. 
Expanding $\mathrm{e}^{zt}$ as a series in $t$, one gets
$$
J(a,z)=\sum_{n\ge0}\frac{z^n}{n!}\int_0^{+\infty}\exp(-a^2t^2)t^n\mathrm{d}t.
$$
The change of variable $u=a^2t^2$ shows that each integral in the RHS is a multiple of the function $\Gamma$ evaluated at $\frac12(n+1)$, more precisely,
$$
J(a,z)=\frac1{2a}\sum_{n\ge0}j_n\left(\frac{z}a\right)^n,\qquad j_n=\frac1{n!}\Gamma\left(\frac{n+1}2\right).
$$
For every nonnegative integer $k$,
$$
j_{2k}=\frac{\sqrt\pi}{2^{2k}}\frac1{k!},\quad
j_{2k+1}=\frac{k!}{(2k+1)!}.
$$
Finally, the integral of interest is
$$
\frac1{4\mathrm{i}}\sum_{n\ge0}(-1)^n\frac{j_n}{a^{n+1}}((b-\mathrm{i}c)^n-(b+\mathrm{i}c)^n).
$$
