Let be $x \in \mathbb{R}_{+}^{*}$, let be $f : x \mapsto \int_0^1 e^{-t} t^{x - 1} \textrm{d}t$ (also the lower gamma incomplete function evaluated as $\gamma(x, 1)$).
One can write $f$ as the following series expansion:
$\begin{equation*} f(x) = \displaystyle \sum_{n=0}^{+\infty} \dfrac{(-1)^n}{n!(x + n)} \end{equation*}$
And we can consider $f$ over $\mathbb{R} \setminus \mathbb{Z}^{-}$ now.
I try to find variations of $f$ and sign of $f''$ over $]-1, 0[$ and $]0, +\infty[$.
I showed $f$ is decreasing over $]-1, 0[ \cup ]0, +\infty[$ because $f'(x) \leq \dfrac{-1}{x^2}$ (correct me if I'm wrong, but the expansion is alternating so this inequality holds?).
Also, I can show that $f''$ is negative over $]-1, 0[$ but fails to do so over $]0, +\infty[$ (with the same type of inequality, i.e. $f''(x) \leq \dfrac{2}{x^3}$).
Finally, I'd like to determine limits as $x \to +\infty, 0^{+}, 0^{-}, -1^{-}$:
- For $x \to 0^{+}, 0^{-}$, I showed $g(x) \sim \dfrac{1}{x}$ as $x \to 0$, so that $g(x) \to \pm \infty$ as $x \to 0$.
- For $x \to +\infty$, by exchange of limits and integral (which is possible due to uniform convergence over $]0, +\infty[$), $g(x) \to 0$.
- For $x \to -1^{+}$, I imagine that $f(x) \to +\infty$, a way could be that, due $xg(x) - g(x + 1) = C \in \mathbb{R}$, then $g(x) \to -\infty$ as $x \to -1^{+}$ because $g(x) \to +\infty$ as $x \to 0^{+}$.
This proves that $f$ is not decreasing over $]-1, 0[$, though, I don't know how to show it.