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I got a clock as a gift recently. It has a very novel face in that the hour positions are given by a complex formula. For the most part, I have been able to verify the calculations presented as accurate, but the two o'clock identity has me stumped.

$$ \frac{\gamma}{\displaystyle{\int_0^\infty\frac{e^{-x^2}-e^{-x}}{x}\,dx}} $$

I know that $\gamma$ is the Euler-Mascheroni constant. And WolframAlpha tells me that $\int_0^\infty\frac{e^{-x^2}-e^{-x}}{x}\,dx=\frac{\gamma}{2}$ which makes sense because

$$ \frac{\gamma}{\displaystyle{\int_0^\infty\frac{e^{-x^2}-e^{-x}}{x}\,dx}}=\frac{\gamma}{\displaystyle{\frac{\gamma}{2}}}=2 $$

It is not clear how I would show $\int_0^\infty\frac{e^{-x^2}-e^{-x}}{x}\,dx=\frac{\gamma}{2}$. Could anyone shed some light on this or point me to a source (book, article, etc...) where I can read up on this. The usual internet (listed above) resources have not been helpful to me.

Laars Helenius
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2 Answers2

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$$\frac{e^{-x^2}-e^{-x}}{x} = \int_x^1 dt \, e^{-x t} = \int_{x^2}^x dy \, e^{-y}$$

Therefore, let's integrate and reverse the order of integration as follows.

$$\begin{align} \int_0^{\infty} dx\, \frac{e^{-x^2}-e^{-x}}{x} &= \int_0^{\infty} \frac{dx}{x} \, \int_{x^2}^x dy \, e^{-y} \\ &= \int_0^{\infty} dy \, e^{-y} \int_y^{\sqrt{y}} \frac{dx}{x} \\ &= \int_0^{\infty} dy \, e^{-y} \left [\log{\sqrt{y}}-\log{y} \right ] \\ &=-\frac12 \int_0^{\infty} dy \, e^{-y} \, \log{y} \\ &= \frac{\gamma}{2} \end{align}$$

Ron Gordon
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  • Thanks! Is there a particular reason you wrote most of your integrals with the differential in front of the integrand, outside of notational convenience? – Laars Helenius Jan 27 '15 at 07:48
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    @LaarsHelenius: It's how I've been writing integrals since my days as an optical physicist, where we regularly used multiple integrals. The differential in front treats the integral as an operator and makes it easier to make sense of multiple integrals. I just got used to writing things this way. You can see this throughout my work here. – Ron Gordon Jan 27 '15 at 07:51
  • Cool. I never thought of the differential I that manner before. Thanks for sharing! – Laars Helenius Jan 27 '15 at 07:53
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ The first step is 'integration by parts': \begin{align}&\color{#66f}{\large\int_{0}^{\infty}\frac{\expo{-x^{2}} - \expo{-x}}{x}\,\dd x} =-\ \overbrace{\int_{0}^{\infty}\ln\pars{x}\expo{-x^{2}}\pars{-2x}\,\dd x} ^{\ds{\dsc{x}\ \mapsto\ \dsc{x^{1/2}}}}\ +\ \int_{0}^{\infty}\ln\pars{x}\expo{-x}\pars{-1}\,\dd x \\[5mm]&=-\,\half\int_{0}^{\infty}\ln\pars{x}\expo{-x}\pars{-1}\,\dd x =-\,\half\,\lim_{\mu \to 0}\,\totald{}{\mu} \int_{0}^{\infty}x^{\mu}\expo{-x}\,\dd x =-\,\half\,\lim_{\mu \to 0}\,\totald{\Gamma\pars{\mu + 1}}{\mu} \\[5mm]&=-\,\half\,\Gamma'\pars{1} =-\,\half\,\Gamma\pars{1}\Psi\pars{1}=\color{#66f}{\large\half\,\gamma} \end{align}

Felix Marin
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