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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.

Bloodpolyhydron
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1 Answers1

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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.

Felix Marin
  • 89,464