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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.

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    $\begingroup$ You can get the proper font and spacing for $\operatorname{erfc}$ using \operatorname{erfc}. $\endgroup$ – joriki Mar 26 at 3:18
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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 $$\mathrm{Ei}(x) = \sum_{k=1}^\infty \frac{x^k}{k k!} + \ln x + \gamma.$$

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