Calculating $\int_0^\infty\, e^{x^2-x} \operatorname{erfc}(x)\;dx$ I am trying to find
$$I=\int_{0}^{\infty }{\,{{e}^{{{x}^{2}}-x}}\operatorname{erfc}\left( x \right)dx}$$
where $\operatorname{erfc}$ is the  complementary error function.
My Work:
$${{e}^{{{x}^{2}}-x}}=\sum\nolimits_{n=0}^{\infty }{\frac{{{\left( {{x}^{2}}-x \right)}^{n}}}{n!}}=\sum\nolimits_{n=0}^{\infty }{\left( \frac{1}{n!}\sum\nolimits_{k=0}^{n}{{{\left( -1 \right)}^{k}}\left( \begin{align}
  & n \\ 
 & k \\ 
\end{align} \right){{x}^{2n-k}}} \right)}=\sum\nolimits_{n=0}^{\infty }{\left( \sum\nolimits_{k=0}^{n}{\frac{{{\left( -1 \right)}^{k}}{{x}^{2n-k}}}{\left( n-k \right)!k!}} \right)}$$
then
$$I=\sum\nolimits_{n=0}^{\infty }{\left( \sum\nolimits_{k=0}^{n}{\frac{{{\left( -1 \right)}^{k}}}{\left( n-k \right)!k!}\int_{0}^{\infty }{{{x}^{2n-k}}\operatorname{erfc}\left( x \right)dx}} \right)}$$
I already know that
$$\int_{0}^{\infty }{{{x}^{2n-k}}\operatorname{erfc}\left( x \right)dx}=\frac{\Gamma \left( \frac{2n-k+2}{2} \right)}{\sqrt{\pi }\left( 2n-k+1 \right)}$$
Is it possible to continue???I appreciate other available methods or hints.  
 A: With the following identity (the proof is easy)
$$\mathrm{e}^{x^2}\mathrm{erfc}(x) = 
\frac{1}{\sqrt{\pi}}\int_0^\infty \mathrm{e}^{-t^2/4}\mathrm{e}^{-xt}\mathrm{d} t, \quad x \ge 0,$$
we have
\begin{align}
\int_0^\infty \mathrm{e}^{x^2-x} \mathrm{erfc}(x) \mathrm{d} x
&= \int_0^\infty \mathrm{e}^{-x} \frac{1}{\sqrt{\pi}}\int_0^\infty \mathrm{e}^{-t^2/4}\mathrm{e}^{-xt}\mathrm{d} t \mathrm{d} x\\
&= \frac{1}{\sqrt{\pi}}\int_0^\infty \mathrm{e}^{-t^2/4}\int_0^\infty \mathrm{e}^{-(1+t)x}\mathrm{d}x \, \mathrm{d} t\\
&= \frac{1}{\sqrt{\pi}}\int_0^\infty \mathrm{e}^{-t^2/4}\frac{1}{1+t} \mathrm{d} t.
\end{align}
It can be expressed in terms of special functions:
$$I = \frac{\pi \mathrm{erfi}(\tfrac{1}{2}) - \mathrm{Ei}(\tfrac{1}{4})}{2\sqrt{\pi}\sqrt[4]{\mathrm{e}}}.$$
Also, it can be expressed in terms of infinite series (for real $x$):
$$\mathrm{erfi}(x) = \frac{2}{\sqrt{\pi}}\int_0^x \mathrm{e}^{t^2}\mathrm{d} t
= \frac{2}{\sqrt{\pi}}\sum_{k=0}^\infty \frac{x^{2k+1}}{k!(2k+1)}$$
and 
$$\mathrm{Ei}(x) = \sum_{k=1}^\infty \frac{x^k}{k k!} + \ln x + \gamma.$$
