How to calculate $\int^{\infty}_0 \frac{e^{-x^2}}{(x+\frac{1}{2})^2} dx$? I am supposed to use the Euler-Possion-Integral $\int^{\infty}_0 e^{-x^2} dx = \frac{\pi}{2}.$
Here is what I've done so far:
$$\begin{array}\ \int^{\infty}_0 \frac{e^{-x^2}}{(x+\frac{1}{2})^2} dx &= -\int^{\infty}_0 e^{-x^2}d(\frac{1}{x+\frac{1}{2}})\\ &= 2-2\int^{\infty}_0 \frac{xe^{-x^2}}{x+\frac{1}{2}} dx\\ &= 2-2[ \int^{\infty}_0 e^{-x^2} dx - \frac{1}{2}\int^{\infty}_0 \frac{e^{-x^2}}{x+\frac{1}{2}} dx]\\ &= 2-\pi + \int^{\infty}_0 \frac{e^{-x^2}}{x+\frac{1}{2}} dx \end{array}$$
I don't know what to do next. I've tried integration by part, and it makes the problem more complicated.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
\int_{0}^{\infty}{\expo{-x^{2}} \over \pars{x + 1/2}^{\, 2}}\,\dd x &
\,\,\,\stackrel{\mrm{IBP}}{=}\,\,\,
2 + \int_{0}^{\infty}{\expo{-x^{2}}\pars{-2x} \over x + 1/2}\,\dd x \\[5mm] & =
2 - 2 \int_{0}^{\infty}\pars{1 - {1/2 \over x + 1/2}}\expo{-x^{2}}\,\dd x
\\[5mm] & =
2 - 2\
\underbrace{\int_{0}^{\infty}\expo{-x^{2}}\,\dd x}
_{\ds{\root{\pi} \over 2}}\ +\
\underbrace{\int_{0}^{\infty}{\expo{-x^{2}} \over x + 1/2}\,\dd x}
_{\ds{\mrm{G}\pars{1 \over 2}}}
\\[5mm] & =
\bbx{\large 2 - \root{\pi} + \mrm{G}\pars{1 \over 2}} \\ &
\end{align}
$\ds{\mrm{G}}$ is the
Goodwin-Staton Integral.
