Alternative representations for $\int_1^e e^{e^{t W_{-1}\left(\frac{-\ln(t)}{t}\right)}}dt$ and $\int_{e^{-e}}^{e^{\frac1e}}e^{t+W_0(-\ln(t))}dt$?

Question

Some time ago I posted an answer in this post. The original question was about evaluating $$ \lim_{n\to\infty} \int_{0}^{\infty} \frac{e^x}{^{2n}x} \mathrm{d}x $$ where $^{2n}x$ is the power tower of height $2n$. In Knuth arrow notation $^{2n}x = x \mathbin{\uparrow}\mathbin{\uparrow} 2n$.

In my answer, I posted an integral representation of the limiting value, given by $$ e^{1+e^{-e}}-1 - \int_{1}^{e} e^{ e^{y W_{-1}\left( \frac{-\ln(y)}{y}\right)}} \mathrm{d}y+\int_{e^{-e}}^{e^{\frac1e}} e^{x +W_0\left( -\ln(x)\right)} \mathrm{d}x $$

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Afterward, I found that in some cases the Lambert-W function in an integral can lead to nice simplifications like in this, this, or this example. I tried replicating some of the ideas used in these questions but mostly ended up with integrals involving things like $e^{e^u}$ which didn't seem any more straightforward.

Nevertheless, just because I couldn't seem to find some interesting alternate representations for the integral doesn't mean they don't exist. So my question is

Are there some other interesting representations for $e^{1+e^{-e}}-1 - \int_{1}^{e} e^{ e^{y W_{-1}\left( \frac{-\ln(y)}{y}\right)}} \mathrm{d}y+\int_{e^{-e}}^{e^{\frac1e}} e^{x +W_0\left( -\ln(x)\right)} \mathrm{d}x $? Either in integral, infinte sum or special function form.

Answer 1

Let’s do the $$\int_{e^{-e}}^{\sqrt[e]e} e^{x+\text W(-\ln(x))}dx $$ part:

$$\int_{e^{-e}}^{\sqrt[e]e} e^{x+\text W(-\ln(x))}dx = -\int_e^{-\frac1e} \frac{e^{e^{-x}} e^{-x} x}{\text W(x)}dx =\int_{-\frac1e}^e\frac{x}{\text W(x)}\sum_{m=0}^\infty \frac{e^{-x(m+1)}}{m!}dx$$

which is valid. Integration by parts on $\int xf(x) dx$ would complicate the problem, so another sum is okay:

$$\int_{-\frac1e}^e\frac{x}{\text W(x)}\sum_{m=0}^\infty \frac{e^{-x(m+1)}}{m!}dx= \int_{-\frac1e}^e\frac{x}{\text W(x)}\sum_{m=0}^\infty \frac1{m!}\sum_{n=0}^\infty \frac{(-x(m+1))^n}{n!} dx$$

The last series expansion is also valid with this demo, so:

$$\int_{-\frac1e}^e\frac{x}{\text W(x)}\sum_{m=0}^\infty \frac1{m!}\sum_{n=0}^\infty \frac{(-x(m+1))^n}{n!} dx=\sum_{m=0}^\infty \frac1{m!}\sum_{n=0}^\infty \frac{(-(m+1))^n}{n!} \int_{-\frac1e}^e\frac{x^{n+1}}{\text W(x)} dx $$

A goal is to use this formula

If one is careful with the summation bounds, one finds this double sum of an En function which can be rewritten in terms of incomplete gamma functions.

Therefore, when $\sum\limits_{m,n\ge0}=\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty$. Note the En function and the Ei function:

$$\int_{e^{-e}}^{\sqrt[e]e} e^{x+\text W(-\ln(x))}dx =2\text{Ei}(1)-\text{Ei}\left(\sqrt[e]e\right)-\text{Ei}\left(e^{-e}\right)+e^{\sqrt[e]e-1}-e^{e^{-e}-1}-\sum_{m,n\ge0}\frac{(m+1)^n}{(n+2)m!n!}\left(\text E_{-n}(n+2)+(-1)^n\text E_{-n}(-(n+2))\right)$$

Is it possible to simplify this double sum?

For the second part,

$$\int _1^e e^{e^{x\text W_{-1}\left(-\frac{\ln(x)}x\right)}}dx=\int_{-\frac1e}^0 \frac{e^{e^{ \frac{\text W(x)\text W_{-1}(x)}x}}}{x^2}\left(\text W(x)+\frac1{\text W(x)+1}-1\right)dx$$

which looks much harder than the first integral above