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
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Michael durable
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asked Mar 6 "16 at 16:28
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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
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IanIan
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To evaluateeginequationint fracxe^x-1 inter-base.netrmdx ag1labeleq:161007-1endequation

begin witheginequationI_1 = int frac1e^x-1 inter-base.netrmdx = int frace^-x1-e^-x inter-base.netrmdx = ln(1-e^-x) ag2labeleq:161007-2endequation

and rewrite our result aseginalignln(1-e^-x) &= ln<-e^-x(-e^-x+1)> \&= ln(-1) + ln(e^-x) + ln(1-e^x) \&= ipi ,- x + ln(1-e^x)endalignand thuseginequationI_1 = ipi ,- x + ln(1-e^x)endequation

Now us evaluate equation eqrefeq:161007-1, our initial integral, via integration by parts. Because that $int ainter-base.netrmdb = abdominal muscle - ,int binter-base.netrmda$ we have actually $a = x$, $inter-base.netrmdb = (e^x-1)^-1$, and also $b = I_1$eginequationint fracxe^x-1 inter-base.netrmdx = x I_1 ,- int I_1 inter-base.netrmdxendequation

eginequationint I_1 inter-base.netrmdx = int inter-base.netrmdx = ipi x ,- frac12x^2 + int ln(1-e^x) inter-base.netrmdxendequationFor the integral top top the appropriate hand side, us make the substitution $y = e^x$,eginequationint ln(1-e^x) inter-base.netrmdx = int fracln(1-y)y inter-base.netrmdy = -inter-base.netrmLi_2(y) = -inter-base.netrmLi_2(e^x)endequation

Putting every one of the pieces together, us haveeginalignint fracxe^x-1 inter-base.netrmdx &= ipi x ,- x^2 + xln(1-e^x) - left(ipi x ,- frac12x^2 -inter-base.netrmLi_2(e^x) ight) \&= inter-base.netrmLi_2(e^x) + xln(1-e^x) ,- frac12x^2 + inter-base.netrmconstendalign