How would we evaluate the definite integral $ \int_0^{e^{\frac{1}{e}}}\frac {-W(-\ln(x))}{\ln(x)}dx $

Question

How would we evaluate the definite integral: $$ \int_0^{e^{\frac{1}{e}}}\frac {-W(-\ln(x))}{\ln(x)}dx $$

Here $W$ is the Lambert W function. More information about this function can be found here.

Edit 1: Found a mistake in the limits so corrected it.

Edit 2: Here are some alternate forms that I derived that might help:

$$ \int_{-1/e}^{\infty} e^{(-u-W(u))}du $$ $$ -\int_{-1/e}^{\infty} \frac{W(x)}{Inv(W(x))}du $$ $$ \int_{-1/e}^{\infty} \frac { \sum_{n=0}^{ \infty } \frac {(-1)^{n-1}n^{n-1}x^{n}} {n!}} {x\sum_{n=0}^{ \infty }\frac{x^n}{n!} } dx$$

Edit 3: After a bit of more investigation, I have managed to further simplify the integral: $$ \int_0^{e^{\frac{1}{e}}}\frac {-W(-\ln(x))}{\ln(x)}dx = -\int_{-1/e}^{\infty} \frac{W(u)}{u(e^u)}du = -\int_{-1/e}^{\infty} \sum_{n=0}^{\infty} \frac {(-1)^nT(n)u^n }{n!}du = \sum_{n=0}^{\infty}\left[ \frac {(-1)^nT(n)u^{(n+1)} }{n!(n+1)} \right]_{-1/e}^{\infty} $$

where T(n) counts the number of forests of rooted labeled trees using labels in a subset of {1,⋯,n} , which is also the hyperbinomial transform of the constant sequence of 1's. (See A088957.)

Now the problem is that I don't know how to evaluate this summation due to T(n), and even wolfram alpha cant do it. Any help on evaluating this summation?

Answer 1

Not an answer

Change variables $y=-W(-\log x)$ to convert it to the integral $$ \int_{-\infty}^{1}(1-y)\exp(ye^{-y})\;dy $$ But that is still not elementary of course.

Answer 2

Not an answer but approximation.

If we consider the function $$f(x)=-\frac {W(-\ln(x))}{\ln(x)}$$ we can notice that $xf(x)$ looks like a power law.

Then, I thought that we could get some reasonable approximations writing $$xf(x)=\sum_{n=1}^p a_n x^{bn}$$ Using $p=4$, the results from a quick and dirty regression leads to $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a_1 & +0.49967 & 0.00119 & \{+0.49731,+0.50203\} \\ a_2 & +0.58096 & 0.00575 & \{+0.56954,+0.59239\} \\ a_3 & +1.27106 & 0.04029 & \{+1.19105,+1.35106\} \\ a_4 & -0.72606 & 0.02198 & \{-0.76971,-0.68242\} \\ b & +1.13077 & 0.00065 & \{+1.12948,+1.13206\} \end{array}$$ and computing $$I(t)=-\int_0^t \frac {W(-\ln(x))}{\ln(x)}\,dx$$ we should have the following results (not fantastic, I agree !) $$\left( \begin{array}{ccc} t & \text{approximation} & \text{exact} \\ 0.05 & 0.015218 & 0.015313 \\ 0.10 & 0.034027 & 0.034118 \\ 0.15 & 0.054945 & 0.055038 \\ 0.20 & 0.077617 & 0.077711 \\ 0.25 & 0.101866 & 0.101959 \\ 0.30 & 0.127595 & 0.127687 \\ 0.35 & 0.154749 & 0.154842 \end{array} \right)$$

For sure, this could be made better at the price of more parameters using for example $$xf(x)=\sum_{n=1}^p a_n x^{b_n}$$