Let$\ x$ be a real number between$\ 0$ and$\ 1$. Is it possible to write$\ e^{x}$ as a function of$\ \Gamma \left(x+1\right)$?

Question

In particular, I'm looking for a relation between$\ e^x$ and$\ \frac{1}{ \Gamma \left(x+1\right) }$, which would be of help for a proof.

Answer 1

There is the relation: $$\frac{1}{\Gamma(1+z)}=e^{y(z)},$$ where $$y(z)=\gamma z-\sum_{k=2}^\infty \frac{(-1)^k \zeta(k)z^k}{k},$$ and $\gamma$ is Euler's gamma constant and $\zeta(x)$ is the Riemann zeta function.

Answer 2

Sure, let $y=e^x$ and $z=1/\Gamma(x+1)$, then $$y=\exp(\Gamma^{-1}(1/z)-1),$$ provided $x$ is restricted to some domain where $x\mapsto\Gamma(x+1)$ is injective with inverse $\Gamma^{-1}$, say $$x\geqslant0.47.$$