$\int _{0}^{\infty }\! \left( {\it W} \left( -{{\rm e}^{-1 -\epsilon}} \right) +1+\epsilon \right) {{\rm e}^{-\epsilon}}{d\epsilon}={\rm e} - 1$ 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 
 A: 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
$$\int _{0}^{\infty }\! {\it W} \left( -{{\rm e}^{-1 -\epsilon}} \right) {{\rm e}^{-\epsilon}}{d\epsilon}=e\left[x\left(W(x)+\frac{1}{W(x)}-1\right)\right]_{x=-e{-1}}^{x=0}$$
$W(0)=0$
$W(e^{-1})=-1$
$Limit_{x=0}\left(\frac{x}{W(x)}\right)=1$
$$\int _{0}^{\infty }\! {\it W} \left( -{{\rm e}^{-1 -\epsilon}} \right) {{\rm e}^{-\epsilon}}{d\epsilon}=e(0+1+0)-(-1+\frac{1}{-1}-1)=2$$
$$\int _{0}^{\infty }\!\left( 1+\epsilon\right) {{\rm e}^{-\epsilon}}{d\epsilon}=\left[ -(\epsilon+2)e^{-\epsilon}\right]_0^\infty=e-3$$
$$\int _{0}^{\infty }\!\left( {\it W} \left( -{{\rm e}^{-1 -\epsilon}} \right)+1+\epsilon\right) {{\rm e}^{-\epsilon}}{d\epsilon}=(e-3)+2=e-1$$
A: 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 $t=ue^u$.
