generalization of sophomore’s dream integral/series identities

This post grew out of one of my MO questions and Sivaram Ambikasaran’s questions on M.SE. We’re currently poking at M(z)=1+g(z) to see how deformed exponential and trigonometric functions would operate, but the following piece of reasoning occurred to me tonight and is too good not to share:
Define g(z)=\sum_{n=1}^{\infty} \frac{z^{n}}{n^n}

g(z) can be used to evaluate integrals of the “Sophomore’s dream” form:

\int_{0}^{1} x^{ax} dx

So:

\int_{0}^{1} x^{ax} dx = \sum_{n=0}^{\infty} \int_{0}^{1} \frac{(ax)^{n}(\log x)^{n}}{n!} dx

We can easily factor that a^{n} out:

\int_{0}^{1} x^{ax} dx = \sum_{n=0}^{\infty} a^{n} \int_{0}^{1} \frac{x^{n}(\log x)^{n}}{n!} dx

 
Wikipedia’s page on the Sophomore’s dream gives an explicit evaluation
of the integral inside as:
\int_{0}^{1} \frac{x^{n}(\log x)^{n}}{n!} dx = \frac{(-1)^{n}}{(n+1)^{n+1}}

Therefore, making the appropriate substitution:
\int_{0}^{1} x^{ax} dx = \sum_{n=0}^{\infty} a^{n} \frac{(-1)^{n}}{(n+1)^{n+1}} = -\frac{1}{a}\sum_{n+1=1}^{\infty}  \frac{(-a)^{n+1}}{(n+1)^{n+1}} =  -\frac{1}{a}\sum_{m=1}^{\infty}  \frac{(-a)^{m}}{(m)^{m}} = -\frac{1}{a}g(-a)

Let’s do a numerical test:
\int_{0}^{1} x^{x/2} dx \approx 0.883790326730436614380642\ldots
\sum_{n=0} \frac{\left(-\frac{1}{2}\right)^{n}}{(n+1)^{n+1}}\approx 0.8837903267304366143806\ldots

(Additional: This document discusses another direction to go in terms of generalizing the sophomore’s dream integrals)

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