Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $dv=e^{-x} \Rightarrow v=-e^{-x}$

Then

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

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 ?

Thanks

share|improve this question
6  
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

2 Answers 2

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

share|improve this answer
    
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 $\mu \to 0^{+}$: $$ \color{#0000ff}{\Large\int_{0}^{\infty}\ln\pars{x}\expo{-x}\,\dd x} = \overbrace{\ \Psi\pars{1}\ }^{\ds{-\gamma}} \quad \overbrace{\ \Gamma\pars{1}\ }^{\ds{1}} = \color{#0000ff}{\Large -\,\gamma} $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.