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I'd like some help to evaluate this integral :

$$I=\int^\infty_0 \frac{x-1}{\ln(x)}\,e^{-x} \,dx$$

I tried to use parameter then I've got an integral of gamma function which I don't know how to integrate it .

Any help will be greatly appreciate .

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Have you tried integrating by parts? – Neal Sep 19 '11 at 16:17
"Solve" is the wrong word. One may evaluate an expression; one does not solve an expression. One can solve an equation or solve a problem. – Michael Hardy Sep 19 '11 at 17:53
Surly:yes,I have. @Michael Hardy : English is not my mother language, so some times I can't choose the right words. I'm very sorry and thanks for the alert. – Lina Sep 19 '11 at 18:39
@Lina: I very frequently see people using the word "solve" in this context. It seems to be one of those words that students use whenever they don't know which word to use. – Michael Hardy Sep 21 '11 at 12:28
up vote 7 down vote accepted

First let $x = \mathrm{e}^t$, changing variables $\mathrm{d} x = \mathrm{e}^t \mathrm{d} t$

$$ I = \int_{-\infty}^\infty \exp( t - \mathrm{e}^t ) \frac{\mathrm{e}^t - 1}{t} \mathrm{d} t $$ In order to evaluate the integral, first evaluate

$$ \mathcal{I}_s = \int_{-\infty}^\infty \exp( t - \mathrm{e}^t ) \mathrm{e}^{s t} \mathrm{d} t = \int_{-\infty}^\infty \exp( (s+1)t - \mathrm{e}^t ) \mathrm{d} t $$ Changing variables back to $x$: $$ \mathcal{I}_s = \int_0^\infty x^{s} \mathrm{e}^{-x} \mathrm{d} x = \Gamma(s+1) $$ Now use $\int_0^1 \mathrm{e}^{s t} \mathrm{d} s = \frac{\mathrm{e}^t - 1}{t}$ to get $$ I = \int_0^1 \mathcal{I}_s \, \mathrm{d}s = \int_0^1 \Gamma(1+s) \, \mathrm{d} s $$ The integral $I$, thus, hardly has a closed form. Its approximate numerical value: $$ I = 0.9227459506806306051438805 $$

Added: Rereading the answer, we can forgo changes of variables, observing $\int_0^1 x^s \mathrm{d} s = \frac{x-1}{\log x}$. Then

$$ I = \int_0^\infty \frac{x-1}{\log x} \mathrm{e}^{-x} \mathrm{d} x = \int_0^\infty \int_0^1 x^s \mathrm{e}^{-x} \mathrm{d} s \mathrm{d} x = \int_0^1 \int_0^\infty x^s \mathrm{e}^{-x} \mathrm{d} x \mathrm{d} s = \int_0^1 \Gamma(1+s) \mathrm{d} s $$

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Thank you very much . – Lina Sep 19 '11 at 18:52

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