Integrals invariant to the choice of the real branch of the Lambert W function used in the integrand

Question

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

Consider for example, this integral:

\begin{align} \int_0^1 \W(-\tfrac t\e) \, dt \tag{1}\label{1} . \end{align}

Its value depends on the choice of the real branch used:

\begin{align} \int_0^1 \Wp(-\tfrac t\e) \, dt &=\e-3 \tag{2}\label{2} ,\\ \int_0^1 \Wm(-\tfrac t\e) \, dt &=-3 \tag{3}\label{3} . \end{align}

On the other hand, the value of the integral

\begin{align} \int_0^1 \frac{-\W(-\tfrac t\e)\,(2+\W(-\tfrac t\e))}{1+\W(-\tfrac t\e)}\, dt \tag{4}\label{4} \end{align}

is invariant to the choice of the real branch used, and despite that the integrands represent completely different curves, in both cases the integral value is the same:

\begin{align} \int_0^1 \frac{-\Wp(-\tfrac t\e)\,(2+\Wp(-\tfrac t\e))}{1+\Wp(-\tfrac t\e)}\, dt &=1 \tag{5}\label{5} ,\\ \int_0^1 \frac{-\Wm(-\tfrac t\e)\,(2+\Wm(-\tfrac t\e))}{1+\Wm(-\tfrac t\e)}\, dt &=1 \tag{6}\label{6} . \end{align}

Question: are there more known interesting examples with this kind of invariance?

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Answer 1

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

Two more examples:

\begin{align} \int_0^1 \sin(-\W(-\tfrac t\e))\, dt &= \int_0^1 \sin(-\Wp(-\tfrac t\e))\, dt \\ &= \int_0^1 \sin(-\Wm(-\tfrac t\e))\, dt \\ &=\tfrac12\,\cos(1) \approx .270151152934 . \end{align}

\begin{align} \int_0^1 \frac{1+\ln(-\W(-\tfrac t\e))(1+\W(-\tfrac t\e))} {1+\W(-\tfrac t\e)}\, dt &= \int_0^1 \frac{1+\ln(-\Wp(-\tfrac t\e))(1+\Wp(-\tfrac t\e))} {1+\Wp(-\tfrac t\e)}\, dt \\ &= \int_0^1 \frac{1+\ln(-\Wm(-\tfrac t\e))(1+\Wm(-\tfrac t\e))} {1+\Wm(-\tfrac t\e)}\, dt \\ &=0 . \end{align}

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Update:

This one showed up in a-little-game-around-lamberts-function-and-simple-and-beautiful-integral

\begin{align} \int_0^1 \left(2\,\sqrt{-\W(-\tfrac t\e)}+\frac 1{\sqrt{-\W(-\tfrac t\e)}} \right)\, dt &=\int_0^1 \left(2\,\sqrt{-\Wp(-\tfrac t\e)}+\frac 1{\sqrt{-\Wp(-\tfrac t\e)}} \right)\, dt \\ &=\int_0^1 \left(2\,\sqrt{-\Wm(-\tfrac t\e)}+\frac 1{\sqrt{-\Wm(-\tfrac t\e)}} \right)\, dt \\ &=4 . \end{align}

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Answer 2

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

One more non-trivial real-branch-invariant function:

\begin{align} f_{\W}(x)&=25\,\cos(2\,\W(-\tfrac x\e))-16\,\cos(\W(-\tfrac x\e)) . \end{align}

Again, the graphs of $f_{\Wp}$ and $f_{\Wm}$ are essentially different,

but the integral is invariant to the choice of the branch used:

\begin{align} \int_0^1 25\,\cos(2\,\W(-\tfrac x\e))&-16\,\cos(\W(-\tfrac x\e)) \, dx \\ &= \int_0^1 25\,\cos(2\,\Wp(-\tfrac x\e))-16\,\cos(\Wp(-\tfrac x\e)) \, dx \\ &= \int_0^1 25\,\cos(2\,\Wm(-\tfrac x\e))-16\,\cos(\Wm(-\tfrac x\e)) \, dx \\ &= 11+\sin(1)\,(4+3\,\sin(1)-4\,\cos(1)) \\ &\approx 14.6715 . \end{align}

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