$$\int_0^{\infty} e^{-x} \prod_{m=0}^{\infty} ( 1-e^{-2^m x}) \space dx$$
Is the amount more than Bob that Alice gets when dividing objects according to the Thue-Morse fair sharing sequence (ABBABAAB…), when the values of the objects are $1, \frac12, \frac13, \dots$ (of course, the amount each gets would be infinite, but we can take the limit of differences).
That is shown in this answer to a closely related question, which uses the Mellin transform (something I do not know about) to find an alternate form of the sum
$$Z_{TM}(s) = \sum_{n=0}^\infty \frac{(-1)^{t_n}}{(n+1)^s}
= \frac{1}{\Gamma(s)} \int_0^\infty x^{s-1} e^{-x} T(e^{-x}) \, dx$$
Where $t_n$ is the $n$th term in the Thue-Morse sequence and
$$T(z) = (1-z) (1-z^2) (1-z^4) (1-z^8) \cdots
= \prod_{m=0}^\infty \bigl( 1 – z^{2^m} \bigr)$$
In the case $s=1$ this gives
$$Z_{TM}(1) = \int_0^\infty e^{-x} T(e^{-x}) \space dx$$
And $Z_{TM}(1)$ is precisely the quantity I'm interested in. The case $s=1$ was excluded from the question I linked, as a result of how their $\zeta_{TM}$ differs from $Z_{TM}$, but the case $s=1$ seems particularly interesting, and the integral it gives is a bit simpler.
So does $Z_{TM}(1)$ have any closed form solution?
Best Answer
I doubt it: It's sufficiently complicated.
It looks like your value is approximately $0.39876108810841881240743054440027306033680$ ; if one discovers a closed form they might consider adding it to the oeis page: oeis.org/A351404