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?
Best Answer
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 $z>0$.