This question is about the gamma function defined only for $z \in \mathbb{R}, z \gt 0$.
So the definition is as follows
$$\Gamma(z) = \int_{0}^{\infty} x^{z-1} \cdot e^{-x}\ dx$$
(again: there are no complex numbers/variables here).
I am reading this integral as if it's improper only at its upper bound (but maybe I am wrong here).
My question is what happens when $z=1$?
Is the integral improper also at its lower bound (the point zero)?
I mean, when $z=1, x=0$ we have $x^{z-1} = 0^0$ which is not really well defined. Right?
Or maybe... when $z=1$, should we just understand/read up-front the integral as being equal to
$$\int_{0}^{\infty} e^{-x}\ dx$$
I guess the same confusion applies when $0 \lt z \lt 1$
In that case ($z \lt 1$), is the integral improper also at its lower bound (the point zero)?
Maybe I should just (from the very start) read this integral as being improper at both its ends. Is that how I should read it?
The integral is improper in both the lower and upper limits of integration.
The best way to study its convergence is to splite the integral in two:
$$\int_{0}^{\infty} e^{-t}t^{z-1} dt = \lim_{\epsilon \to 0 } \int_{\epsilon}^{1} e^{-t}t^{z-1} dt + \lim_{\delta \to \infty } \int_{1}^{\delta} e^{-t}t^{z-1} dt$$
Note that for $t>0\quad 0<e^{-t}<1$ and $z\in \mathbb{R}$
we have
$$ e^{-t}t^{z-1}<t^{z-1}$$
then if $\epsilon>0$
$$\int_{\epsilon}^{1} e^{-t}t^{z-1} dt < \int_{\epsilon}^{1} t^{z-1}dt = \frac{1}{z} -\frac{\epsilon^{z}}{z}$$
So if $z>0$ $$\int_{\epsilon}^{1} e^{-t}t^{z-1} dt < \int_{\epsilon}^{1} t^{z-1}dt < \frac{1}{z} $$ Then $$ \int_{0}^{1} e^{-t}t^{z-1} dt = \lim_{\epsilon \to 0 } \int_{\epsilon}^{1} e^{-t}t^{z-1} dt \leq \frac{1}{z} \quad z>0 $$
Now for any $n\in \mathbb{N}$
$$e^{t}> \frac{t^n}{n!} \Longrightarrow e^{-t}< \frac{n!}{t^n} $$
if $z>0$ we can always find $n\in \mathbb{N}$ such that $n>z+1$ and
$$ \int_{1}^{\delta} e^{-t}t^{z-1}dt < \int_{1}^{\delta} \frac{n!}{t^{n+1-z}} = n!\frac{1-\delta^{z-n}}{n-z} < \frac{n!}{n-z}, $$
Hence
$$ \int_{1}^{\infty} e^{-t}t^{z-1}dt = \lim_{\delta \to \infty} \int_{1}^{\delta} e^{-t}t^{z-1}dt \leq \frac{n!}{n-z}$$
So both improper integrals exist if $z>0$.