$\int _{0}^{\infty }\! \left( {\it W} \left( -{{\rm e}^{-1 -\epsilon}} \right) +1+\epsilon \right) {{\rm e}^{-\epsilon}}{d\epsilon}={\rm e} - 1$

Question

How to prove $\int _{0}^{\infty }\! \left( {\it W} \left( -{{\rm e}^{-1 -\epsilon}} \right) +1+\epsilon \right) {{\rm e}^{-\epsilon}}{d\epsilon}={\rm e} - 1$ where W is the Lambert W function?

Maple can calculate it.. I was hoping to generalize it so that

$\int _{0}^{\infty }\! \left( {\it W} \left( -{{\rm e}^{- \left( {{\rm e}^{-{\frac {{\it \alpha}\,s}{{\it \beta}}}}}{{\it \beta}}^ {2}+\epsilon\,{\it \alpha} \right) \left( {{\rm e}^{-{\frac {{\it \alpha}\,s}{{\it \beta}}}}} \right) ^{-1}{{\it \beta}}^{-2}}} \right) { {\rm e}^{-{\frac {{\it \alpha}\,s}{{\it \beta}}}}}{{\it \beta}}^{2}+{ {\rm e}^{-{\frac {{\it \alpha}\,s}{{\it \beta}}}}}{\it \alpha}\,{\it \beta }\,s+{{\rm e}^{-{\frac {{\it \alpha}\,s}{{\it \beta}}}}}{{\it \beta}}^{2} +\epsilon\,{\it \alpha} \right) {{\rm e}^{-\epsilon}} \left( {{\rm e}^{ -{\frac {{\it \alpha}\,s}{{\it \beta}}}}} \right) ^{-1}{{\it \alpha}}^{-1 }{{\it \beta}}^{-1}{d\epsilon}$ could be expressed

Answer 1

Let $x=-e^{-1-\epsilon}$

$e^{-\epsilon}=-ex$ and $d\epsilon=-\frac{dx}{x}$ $$\int _{0}^{\infty }\! {\it W} \left( -{{\rm e}^{-1 -\epsilon}} \right) {{\rm e}^{-\epsilon}}{d\epsilon}=e\int _{-e^{-1}}^0 W(x)(-ex)(-\frac{dx}{x})=e\int _{-e^{-1}}^0 W(x)dx$$ $\int W(x)dx = x\left(W(x)+\frac{1}{W(x)}-1\right)$+constant $$\int _{0}^{\infty }\! {\it W} \left( -{{\rm e}^{-1 -\epsilon}} \right) {{\rm e}^{-\epsilon}}{d\epsilon}=e\left[x\left(W(x)+\frac{1}{W(x)}-1\right)\right]_{x=-e{-1}}^{x=0}$$ $W(0)=0$

$W(e^{-1})=-1$

$Limit_{x=0}\left(\frac{x}{W(x)}\right)=1$ $$\int _{0}^{\infty }\! {\it W} \left( -{{\rm e}^{-1 -\epsilon}} \right) {{\rm e}^{-\epsilon}}{d\epsilon}=e(0+1+0)-(-1+\frac{1}{-1}-1)=2$$

$$\int _{0}^{\infty }\!\left( 1+\epsilon\right) {{\rm e}^{-\epsilon}}{d\epsilon}=\left[ -(\epsilon+2)e^{-\epsilon}\right]_0^\infty=e-3$$

$$\int _{0}^{\infty }\!\left( {\it W} \left( -{{\rm e}^{-1 -\epsilon}} \right)+1+\epsilon\right) {{\rm e}^{-\epsilon}}{d\epsilon}=(e-3)+2=e-1$$

Answer 2

First of all, notice that your integral can be broken up into a sum of three integrals, of which the last two are trivial. Then let $t=-e^{-(x+1)}$ in the former. You will ultimately get an expression in terms of $\displaystyle\int W(t)~dt$, which, believe it or not, does possess an antiderivative, that can be evaluated by substituting $t=ue^u$.