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definiendum $ \zeta: \mathbb{C}\setminus\{???\} \to \mathbb C$
definiendum $ x\mapsto x^x$


$x^x={\mathrm e}^{x\log(x)}=\left({\mathrm e}^x\right)^{\log(x)}$
todo: write down the above with an expanded $\log$ to third order

Because of this, the local minimum of $x^x$ is that of $x\log(x)$, namely $\frac{1}{\mathrm e}\approx 0.37$, and then see Secretary problem (Wikipedia)


$x^x = x \sum_{n=0}^\infty \prod_{k=1}^n (1-x)\left(1-\frac{1+x}{k}\right) = x \sum_{n=0}^\infty \frac{1}{n!}(1-x)_n(1-x)^n $

with the Pochhammer symbol $(1-x)_n:=\prod_{k=0}^{n-1} (k-x)=\prod_{k=1}^n (k-(1+x))$.

From this

$x^x = \dfrac{x}{(1-x)^2} x \sum_{n=0}^\infty \dfrac{1}{\mathrm{Beta}(1-x,n)}(1-x)^n$

and all of this appears more naturally via

$x^x \equiv (1-t)^{1-t} = (1-t)\dfrac{1}{(1-t)^t}$


as Binomials relate to Gammas and Betas.

For moral talk regarding the rewriting $x=1-t$, see Zeta functions.

An integral

$P_n(x) := (1-x) \, x^n \prod_{k=0}^{n-1} \, (x+k)$

this must come from the Expression at the beginning, but I don't know how I arrived there.
Maybe there I switched from $x$ to $1-x$ or $\int_0^1$ to $\int_1^0$.

$\int_0^1 \, x^x \, dx = \sum_{n=0}^\infty \frac{1}{n!} \int_0^1 \, P_n(x) \, dx = \frac{1}{2} + \frac{1}{12} + \frac{1}{24} + \frac{131}{5040} + \frac{1093}{60480} + \dots$


Refinement of

Link to graph
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