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$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Ei{\operatorname{Ei}}$

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?

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    $\begingroup$ I managed to simplify $(2)$ into $1+e\int_{0}^{1}\frac{\sqrt[1-t]{t}\ln\left(t\right)}{1-t}dt$. I think they are equal, but I'm not 100% sure. $\endgroup$ Sep 1 '20 at 7:32
  • $\begingroup$ @Varun Vejalla: WolframAlpha agrees. Consider posting an answer. $\endgroup$
    – g.kov
    Sep 1 '20 at 11:22
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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 $$1+e\int_{0}^{1}t^{\frac{1}{1-t}}\left(1-\frac{1}{t}\right)dt = 1+e\int_{0}^{1}t^{\frac{1}{1-t}}dt-e\int_{0}^{1}t^{\frac{t}{1-t}}dt$$

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