Looking at some straightforward generalizations of the Sophomore's Dream identity I derived the following identity which works for $\Re(b) > 0$: $$ \int_0^1 x^{a x^b} dx = \sum_{n=0}^{\infty} \frac{(-a)^n}{(bn + 1)^{n+1}} $$ In particular, if you let $-a = b = z$, you find the following two functions are equal for $\Re(z)>0$: $$ f(z) = \int_0^1 x^{-z x^z} dx $$ $$ g(z) = \sum_{n=0}^{\infty}\frac{z^n}{(z n + 1)^{n+1}} = \frac{1}{z}\sum_{n=0}^{\infty} \frac{1}{(n + \frac{1}{z})^{n+1}} $$ The sum for $g(z)$ is clearly an analytic function on $\mathbb{C}\setminus\{0,-1,-\frac1 2,-\frac1 3,...\}$. Since they agree for $\Re(z) > 0$, this means $f$ is also analytic in the right half-plane.
The surprising thing is that $f(z)$ exists for all $z \in \mathbb{R}$, but $f(z) \ne g(z)$ for $z < 0$. Is there some simple reason why $f(z)$ for negative $z$ is not the analytic continuation of $f$?
EDIT: I am pretty sure $f$ is smooth as a real-valued function, even at $0$. That leads me to wonder if it is real-analytic for negative $z$. I think so, but I am not sure.