Integrals with the special functions $Ci(x)$ and $erf(x)$ I'm looking for the solutions of the following two integrals:
$$I_1=\int\limits_0^\infty dx\, e^{-x^2}\text{Ci}(ax)$$
and
$$I_2=\int\limits_0^\infty dx\, e^{-ax}\text{erf}(x)$$
with
$$\text{Ci}(x)=\int\limits_\infty^x\frac{\cos(t)}{t}dt$$
and
$$\text{erf}(x)=\frac{2}{\sqrt{\pi}}\int\limits_0^xe^{-t^2}dt$$
Now I'm not 100% sure what is meant by "the solution of the integrals" since these will probably be not-evaluable. But I'm guessing that the question is to reduce the expression to one single special function in stead of the integral of a special function with an elementary function.
Mathematica yields me the answers:
$$I_1=-\frac{\sqrt{\pi}}{4}\Gamma\left(0,\frac{a^2}{4}\right)$$
and
$$I_2=\exp\left(\frac{a^2}{4}\right)\frac{1-\text{erf}(a/2)}{a}$$
A good first step for evaluating these integrals $I_1$ and $I_2$ seemed to fill in the integral representations of these special functions and try to switch te integrals over $x$ and $t$. However this has not (yet) been a success. I also tried to find a differential equation for these integrals, but also this was not so easy to do. Are there any tips/tricks to evaluate these integrals ?

For $I_2$, please note that this integral is just the Laplace transform of $\text{Erf}(x)$ w.r.t. to the variable $a$ 
Furthermore the Laplace transform of an integral $\mathcal{L}(\int f(x) dx)$ is just $\mathcal{L}(f(x))/a$ 
Using this we find that 
$$
I_2=\frac{2}{a\sqrt{\pi}}\int_0^{\infty}e^{- a x}e^{-x^2}dx
$$
Completing the square in the exponent  and changing variables $y=\frac{a}{2}+x$
This turns into
$$
I_2=2\frac{ e^{\frac{a^2}{4}}}{a\sqrt{\pi}}\int_{\frac{a}{2}}^{\infty}e^{-y^2}dy
$$
Now using straightforwardly the definition of the Errorfunction we obtain
$$
I_2=\frac{e^{\frac{a^2}{4}}}{a}\left(1-\text{Erf}(a/2)\right)
$$
As promised!
I will have a look on $I_1$ later
