The Question
What do we know about $$\xi(z)=\frac{1}{\Gamma(z)}\sum_{n=1}^\infty \frac{\Gamma(n+z-1)}{n^n}= \int_0^1 \frac{dx}{(1+x\ln x)^z} $$
Specifically, what can be said about $\xi$ regarding it's rationality at integer values? Obviously, they're surely all irrational(because a number plucked from the reals should be irrational) but of course we are interested in demonstrably irrational.
- Is $\xi(z)$ irrational for (some) integer $z$?
- Are infinitely many $\xi(z)$ irrational? Is atleast one?
Exposition I'm not sure that these thoughts will prove helpful in making any claims but here goes.
Generally we may write $$\int_{0}^{1}\frac{\left(-x\ln x\right)^{k}}{\left(1+x\ln x\right)^{z}}dx=\int_{0}^{1}\frac{1}{\left(1+x\ln x\right)^{z}}dx-\sum_{n=1}^{k}\int_{0}^{1}\frac{\left(-x\ln x\right)^{k-n}}{\left(1+x\ln x\right)^{z-1}}dx$$ But in the case $z=1$ this is simply
$$\int_0^1 \frac{(-x \ln x)^k}{1+x\ln x}dx=\sum_{n=k+1}^\infty \frac{\Gamma(n)}{n^n}$$
And $(-x \ln x)$ is maximally $1/e$ so we have some of the conditions which resemble Beuker's proof of the irrationality of $\zeta(s)$ for $s \le 3$. That is we have an integral that we can manipulate for bounds on the tail of our series. But this type of argument works for small values of $s$ because the growth rate of $\operatorname{lcm}(1,2,\dots n)^s$ can be overcome but of course $\operatorname{lcm}(1,2, \dots n^n)$ grows even more rapidly.
On the other hand, maybe $\xi$ is better behaved than $\zeta$ in the sense that $ \xi$ is an entire function.
$\xi(2)$ is asked about on MSE and crossposted to MO There is a pretty continued fraction form there $$\xi(2)=\frac{1}{2+\underset{n=2}{\overset{\infty}{K} }\frac{-n^{2n-1}}{n^n+(n+1)^n}}=1+\frac{1}{2+\frac{-8}{13+\frac{-243}{91+\frac{-16384}{881+\frac{-1953125}{10901+\frac{-362797056}{164305+\frac{...}{...}}}}}}}$$ I'm hopeful that this continued fraction can be generalized to $\xi(z)$.
