# These calculations are correct ? About $\int\frac{e^{-x}}{x}dx$

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

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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

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

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$$\int_{0}^{\infty}x^{\mu}\,\expo{-x}\,\dd x = \Gamma\pars{\mu + 1}$$ 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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