$\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Ei{\operatorname{Ei}}$
Is there a known closed form for the integral
\begin{align}
I&=\int_0^1
\frac{\Wp(-\tfrac t\e)}{\Wm(-\tfrac t\e)}
\,dt
\approx 0.151216902884937
\tag{1}\label{1}
,
\end{align}
where $\Wp,\Wm$ are two real branches of the Lambert $\W$ function?
An alternative form of \eqref{1} is
\begin{align}
I&=\e\cdot\!\!\int_0^1
\frac{\sqrt[1-t]{t}(1-t+t\,\ln t)(t-1-\ln t)}{(1-t)^3}
\, dt
\tag{2}\label{2}
.
\end{align}
Using series expansion of $\Wp$
it can be expressed in terms of the infinite sum:
\begin{align}
I&=\e-2-
\e\cdot\sum_{n=1}^\infty
\frac{\Gamma(n+2,n+1)}{\Gamma(n+2)\,n^3\,(1+\tfrac1n)^{n+1}}
\tag{3}\label{3}
.
\end{align}
Also, the closed form of \eqref{1}
can be found, using closed form of either
\begin{align}
I_2&=\int_0^1 \left(-\Wp(-\tfrac t\e)-\frac1{\Wm(-\tfrac t\e)}\right)^2\, dt
\approx 0.62200121658
\\
\text{or }\quad
I_3&=\int_0^1 \left(-\Wp(-\tfrac t\e)+\frac1{\Wm(-\tfrac t\e)}\right)^2\, dt
\approx 0.01713360504
,
\end{align}
or both, since
\begin{align}
I_2+I_3&=
20+4\,\e\,(\Ei(1,1)-2)
\approx 0.639134821620414414482
,
\end{align}
where
\begin{align}
\Ei(1,1)&=\int_1^\infty \frac{\exp(-t)}t \, dt
\approx 0.21938393439552
.
\end{align}
Any ideas?
$\endgroup$
Best Answer
I managed to get to the equivalent representation $$1+e\int_{0}^{1}t^{\frac{1}{1-t}}\left(1-\frac{1}{t}\right)dt = 1+e\int_{0}^{1}t^{\frac{1}{1-t}}dt-e\int_{0}^{1}t^{\frac{t}{1-t}}dt$$ but wasn't able to find a closed form.
I will start off from $(2)$: $$I=e\int_0^1\frac{\sqrt[1-t]{t}(1-t+t\,\ln t)(t-1-\ln t)}{(1-t)^3}\, dt$$
This is equivalent to $$e\int_{0}^{1}t^{\frac{1}{1-t}}\left(\frac{1-t+t\ln t}{\left(1-t\right)^{2}t}\right)\frac{t\left(t-1-\ln t\right)}{1-t}dt$$
Then integrating by parts yields $$e\int_{0}^{1}t^{\frac{1}{1-t}}\left(\frac{\ln t}{\left(t-1\right)^{2}}+\frac{t-2}{t-1}\right)dt$$
This can be split as $$e\int_{0}^{1}t^{\frac{1}{1-t}}\left(\frac{\ln t}{\left(1-t\right)^{2}}+\frac{1}{t\left(1-t\right)}\right)dt+e\int_{0}^{1}t^{\frac{1}{1-t}}\left(\frac{\ln t}{\left(1-t\right)^{2}}+\frac{t-2}{t-1}-\left(\frac{\ln t}{\left(1-t\right)^{2}}+\frac{1}{t\left(1-t\right)}\right)\right)dt$$
Which is then equal to $$1+e\int_{0}^{1}t^{\frac{1}{1-t}}\left(\frac{t-2}{t-1}-\frac{1}{t\left(1-t\right)}\right)dt$$
Using partial fraction decomposition, this simplifies to $$1+e\int_{0}^{1}t^{\frac{1}{1-t}}\left(1-\frac{1}{t}\right)dt = 1+e\int_{0}^{1}t^{\frac{1}{1-t}}dt-e\int_{0}^{1}t^{\frac{t}{1-t}}dt$$