Introduction:
I have proved the following:
Suppose that Poisson events are occuring at the constant rate of $\lambda$ per unit time. Let the random variable $Y$ denote the waiting time for the rth event, then $Y$ has pdf: $$f_Y(y) = \frac {\lambda^r} {(r-1)!} y^{r-1}e^{-\lambda y} $$ (1)
so $Y$ is defined for $r \in \mathbb N$.
However the gamma pdf:
$$f_Y(y) = \frac {\lambda^r} {\Gamma(r)} y^{r-1}e^{-\lambda y}$$
is a generalization of this function (right?), since $\Gamma(r) = (r-1)!$ for any positive integer $r$.
To justify that the gamma pdf is a well-defined function i must that the gamma function:
$$\Gamma(r) = \int_0^{\infty} y^{r-1}e^{-y} dy$$ is a well-defined function (I've already verified $$\int_0^{\infty}f_Y(y) dy = \int_0^{\infty} \frac {\lambda^r} {(r-1)!} y^{r-1}e^{-\lambda y} dy = 1$$ under the assumption that it is).
Question:
1.
How do I show that the gamma function is well-defined, that is the improper definite integral exists for any real number $r >0$ ? I know it exists for any positive integer $t > r$ (the integral with boundaries $0$ and $\infty$ of (1) is equal to $1$?) so $\Gamma(r) \le \Gamma(t)$ (right?) This implies that $\Gamma(r)$ is bounded, but how do I show the limit exists for $y \rightarrow \infty$ ? Also the functions of $y$ in $\Gamma(r)$ are both continuous, which implies the function is integrable ?