# Integral form of $\Gamma (x)$

While trying to represent the poles and analytic continuation of $\Gamma (x)$ , the author uses the following equality:

$$\int\nolimits_{0}^{1}t^{x-1}e^{-t}dt=\sum_{n=0}^{+\infty}\frac{{(-1)}^{n}}{(n+x)n!}.$$ Any help in figuring out how?

In order to expand my comment: put for $0\leq t\leq 1$: $\displaystyle S_n(t)=\sum_{k=0}^n\frac{(-t)^k}{k!}$. We have $$\sup_{0\leq t\leq 1}|\exp(-t)-S_n(t)| = \sup_{0\leq t\leq 1}\sum_{k=n+1}^{+\infty}\frac{(-t)^k}{k!}\leq \sum_{k=n+1}^{+\infty}\frac 1{k!},$$ and since the series $\displaystyle\sum_{k=0}^{+\infty}\frac 1{k!}$ is convergent, the sequence $\{S_n\}$ converges uniformly to $t\mapsto \exp (-t)$ on $\left[0,1\right]$. Therefore, we can switch the series and the integral for $x\geq 1$:
\begin{align*} \int_0^1t^{x-1}e^{-t}dt&=\int_0^1t^{x-1}\left(\sum_{k=0}^{+\infty}\frac{(-1)^kt^k}{k!}\right)dt\\ &=\sum_{k=0}^{+\infty}\frac{(-1)^k}{k!}\int_0^1 t^{x+k-1}dt\\ &=\sum_{k=0}^{+\infty}\frac{(-1)^k}{k!(x+k)}. \end{align*}
For $0<x<1$, we can use the dominated convergence theorem since $|t^{x-1}S_n(t)|\leq e\cdot t^{x-1}$ which is integrable on $\left[0,1\right]$ (we can use a more simple argument in the case $x\geq 1$).