This can sound an extremely basic, however I do the efforts looking on integral tables and also on wolphram inter-base.net, yet I couldn"t get the method for addressing it.The procedures I went through were:

Part integration with \$u=x\$.That lead to the allude I got stuck, \$int ln(1-e^x),dx\$. The calculation was \$-operatornameLi2(e^x)\$.Final an outcome was \$operatornameLi2(e^x) - x^2/2 + xlog(1-e^x)\$

My questions are:

Is there any type of shortcut because that this integral? possibly a way to obtain a cleaner result (i.e. There is no the Li2 term).How deserve to I do it because that the expression \$0

Here"s the an outcome from wolphramhttp://www.wolframalpha.com/input/?i=%E2%88%AB+x%2F(e%5Ex+-1)

Thanks

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edited Mar 6 "16 in ~ 16:35 Michael durable
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asked Mar 6 "16 at 16:28 PatrickPatrick
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If i didn"t know anything about special functions, I would probably shot this:

\$\$fracxe^x-1=fracx e^-x1-e^-x=x sum_n=1^infty e^-nx.\$\$

This equality hold (and the collection converges) because that \$x>0\$; close to \$x=0\$ girlfriend may inspect that the integrand is bounded, for this reason neglecting that point is no loss.

Then you have reason come hope the you can combine term by term; this is simple application that integration by parts, giving

\$\$int x e^-nx dx = x e^-nx/(-n) + int e^-nx/n dx = -x e^-nx/n - e^-nx/n^2.\$\$

So one antiderivative is

\$\$sum_n=1^infty -x e^-nx/n - e^-nx/n^2\$\$

and in particular

\$\$int_0^infty fracxe^x-1 dx = sum_n=1^infty 1/n^2 = fracpi^26.\$\$

There are various convergence theorems that can be offered to justification this interchange of limitless summation and integration.

But this is certainly a "special function", so you won"t find response in terms of a finite mix of elementary school functions.

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edited Mar 6 "16 at 16:40
reply Mar 6 "16 at 16:33 IanIan
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