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Was trying to calculate $$\int_{0}^{\infty}e^{-x}\ln x dx=-\gamma$$ and I found this question:

I want to analyze $$\int\frac{e^{-x}}{x}dx$$

With $u=\displaystyle\frac{1}{x} \Rightarrow du = \displaystyle\frac{-1}{x^{2}} dx $, and $dv=e^{-x} \Rightarrow v=-e^{-x}$


$$\int\frac{e^{-x}}{x}dx = \displaystyle\frac{1}{x}\cdot-e^{-x}-\int-e^{-x}\cdot\displaystyle\frac{-1}{x^{2}} dx = -\displaystyle\frac{e^{-x}}{x}-\int \displaystyle\frac{e^{-x}}{x^{2}} dx$$

Integrating from the same form gives:

$$\int\frac{e^{-x}}{x}dx = -\displaystyle\frac{e^{-x}}{x} + \displaystyle\frac{e^{-x}}{x^{2}} + 2\int\frac{e^{-x}}{x^{3}}dx$$

Are these calculations are correct?, and more is valid say :

$$\int\frac{e^{-x}}{x}dx = \displaystyle\sum\limits_{n=0}^\infty (-1)^{n+1}n!\frac{e^{-x}}{x^{n+1}}\ ?$$

$\bf{EDIT}$: This series helps me to calculate it ? : $$\int_{0}^{\infty}e^{-x}\ln xdx=-\gamma$$ I don't know how to turn this series in something harmonic. If not, is this the way to calculate that this integral converges to $-\gamma$, which is the form ?


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You have just derived an asymptotic series for the exponential integral. – J. M. Nov 27 '10 at 6:42
Note that $\mathrm{Ei}(x)=-\mathrm{PV}\int_{-x}^{\infty}\frac{\exp(-u)}{u}\mathrm du=\gamma+\ln(x)+\int_0^x \frac{\exp(u)-1}{u}\mathrm du$ – J. M. Nov 28 '10 at 6:44
up vote 6 down vote accepted

The series diverges, but converges to your integral asymptotically: If you add up the first $n$ terms the ratio of the error to the $n$th term goes to zero as $x$ goes to infinity

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Thanks, then i can to continue trying with $-\gamma$ – Bryan Yocks Nov 27 '10 at 6:41

$\newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{{\rm e}^{#1}}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\pp}{{\cal P}}% \newcommand{\sgn}{{\rm sgn}}% \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}$

$$ \int_{0}^{\infty}x^{\mu}\,\expo{-x}\,\dd x = \Gamma\pars{\mu + 1} $$ We take the derivative respect $\mu$: $$ \int_{0}^{\infty}x^{\mu}\ln\pars{x}\,\expo{-x}\,\dd x = \Gamma\,'\pars{\mu + 1} = \Psi\pars{\mu + 1}\Gamma\pars{\mu + 1} $$ We take the limit $\mu \to 0^{+}$: $$ \color{#0000ff}{\Large\int_{0}^{\infty}\ln\pars{x}\expo{-x}\,\dd x} = \overbrace{\ \Psi\pars{1}\ }^{\ds{-\gamma}} \quad \overbrace{\ \Gamma\pars{1}\ }^{\ds{1}} = \color{#0000ff}{\Large -\,\gamma} $$

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