For which values of x is the Gamma function Riemann Integrable?

Question

For which values of $x$ is the Gamma function $\Gamma (x) = \int\limits_0^{\infty}t^{x-1}e^{-t} dt$ Riemann Integrable?

I have read from elsewhere that the Gamma function converges for $x>0$, but:

1. What is the proof for $x>0$?
2. Does this itself (convergence) show that it is Riemann Integrable?
3. What about for $x\leq 0$?

Thanks!

Answer 1

You have

$$t^{x-1}e^{-t}\underset{t\to 0}{\sim} t^{x-1}$$

which is integrable on $(0,1]$ if and only if $x>0$.

And

$$t^{x-1}e^{-t}=\underset{t\to\infty}{o}(e^{-t/2})$$

which is always integrable on $[1,+\infty)$.

So, the integral defining $\Gamma(x)$ exists (i.e., converges) if, and only if, $x>0$.