Does anyone know how can I evaluate this integral
\begin{equation} \int\limits_0^\infty dx\;e^{-x^2} \operatorname{Ci}(ax) \end{equation} where $\operatorname{Ci}(x)$ is the cosine integral function \begin{equation} \operatorname{Ci}(x)=\int\limits_\infty^x dt\;\frac{\cos(t)} t \end{equation}
I'm assuming $\text{Re}(a) > 0$ here.
Let $$J(a) = \int_0^\infty e^{-x^2} \operatorname{Ci}(a x)\; dx $$
Differentiating, we get
$$ J'(a) = \int_0^\infty e^{-x^2} \frac{\cos(ax)}{a}\; dx = \frac{\sqrt{\pi}}{2a} e^{-a^2} $$
Now we should expect $J(a) \to 0$ as $a \to \infty$, so
$$ J(a) = - \int_a^\infty J'(s)\; ds = -\frac{\sqrt{\pi}}{4} \operatorname{Ei}(1,a^2/4) $$ according to Maple. This can also be written using the incomplete Gamma function: $$ J(a) = -\frac{\sqrt{\pi}}{4} \Gamma(0,a^2/4)$$