$\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\erf{\operatorname{erf}}\def\Ei{\operatorname{Ei}}$
For $x\in\mathbb R,\ x\ne-1$
\begin{align}
I(x)&=
\int_0^1
(1-\Wm(-\tfrac t\e))^x
–
(1-\Wp(-\tfrac t\e))^x
\, dt
\tag{1}\label{1}
\\
&=
\frac{\e\,(2+\e\,(x-1)\,\Gamma(x+2,1))}{x+1}
=f(x)
\tag{2}\label{2}
,
\end{align}
where $\Wp,\ \Wm$ are the real branches
of the Lambert $\W$ function,
and $\Gamma$ is
the incomplete gamma function.
For integer values $x=n$, $I(n)$ follows the pattern
of $\e\cdot a_n$ fromA093964.
\begin{align}
I(1)&=f(1)=\e
,\\
I(2)&=f(2)=6\,\e
,\\
I(3)&=f(3)=33\,\e
,\\
&\dots
\end{align}
Some other special cases of $x$:
\begin{align}
I(\tfrac12)&=
\int_0^1 \sqrt{1-\Wm(-\tfrac t\e)} \, dt
–
\int_0^1 \sqrt{1-\Wp(-\tfrac t\e)} \, dt
\\
&=
(\tfrac32\,\sqrt2+\tfrac14\,\e^2\,\sqrt\pi\,(\erf(\sqrt2)-1))
\\
&-(\tfrac32\,\sqrt2-\tfrac12\,\e
+\tfrac14\,\e^2\,\sqrt\pi\,(\erf(\sqrt2)-\erf(1)))
\\
&=
\tfrac12\,\e+\tfrac14\,\e^2\,\sqrt\pi\,(\erf(1)-1)
=f(\tfrac12)
\tag{3}\label{3}
\\
&\approx 0.844113386646
,\\
I(-2)&=
\int_0^1 \frac1{(1-\Wm(-\tfrac t\e))^2}
–
\frac1{(1-\Wp(-\tfrac t\e))^2} \, dt
\\
&\approx -.57344306156
\\
&=
\e\,(3\,\e\,\Ei(1,1)-2)
=f(-2)
\tag{4}\label{4}
,
\end{align}
where $\Ei(a,x) = x^{a-1} \Gamma(1-a,x)$.
\begin{align}
I(-\Omega)
&\approx -0.4015641473638446
\approx f(-\Omega)
,\quad \Omega=\W(1)\approx .56714329
.
\end{align}
Also,
\begin{align}
I(-1)&\approx -0.523798568446
\\
&\approx \e\,(1-2\,\e\,\Ei(1,1))=\lim_{x\to -1}f(x)
.
\end{align}
Questions:
1) Is this correct/known? Any reference/confirmation?
2) Is it possible to transform \eqref{2}
in order to cure the nasty case of $x=-1$?
$\endgroup$
Best Answer
By simple change of variables, we have $$ - \int_0^1 {\left( {1 - W_0 \left( { - \frac{t}{e}} \right)} \right)^x dt} = \int_0^1 {(1 + s)^x e^{1 - s} (s - 1)ds} $$ and $$ \int_0^1 {\left( {1 - W_{ - 1} \left( { - \frac{t}{e}} \right)} \right)^x dt} = \int_1^{ + \infty } {(1 + s)^x e^{1 - s} (s - 1)ds} . $$ Consequently, $$ I(x) = \int_0^{ + \infty } {(1 + s)^x e^{1 - s} (s - 1)ds} = e\int_0^{ + \infty } {\frac{{e^{ - s} s}}{{(1 + s)^{ - x} }}ds} - e\int_0^{ + \infty } {\frac{{e^{ - s} }}{{(1 + s)^{ - x} }}ds} \\ = e\int_0^{ + \infty } {\frac{{e^{ - s} }}{{(1 + s)^{ - x - 1} }}ds} - 2e\int_0^{ + \infty } {\frac{{e^{ - s} }}{{(1 + s)^{ - x} }}ds} = e^2 \Gamma (x + 2,1) - 2e^2 \Gamma (x + 1,1) \\ = e^2 \Gamma (x + 2,1) - 2e^2 \frac{{\Gamma (x + 2,1) - e^{ - 1} }}{{x + 1}} = e\frac{{2 + e(x - 1)\Gamma (x + 2,1)}}{{x + 1}}. $$ Regarding your second question, you may write $$ I(x) = e^2 \Gamma (x + 2,1) - 2e^2 \Gamma (x + 1,1) = e(e(x - 1)\Gamma (x + 1,1) + 1). $$