Showing that the gamma function is integrable for $x \in \mathbb{C}_{Re>0}$

Question

Let $\mathbb{C}_{Re>0}=\{x \in \mathbb{C} \space | \space Re(x)>0 \}$.

For all $x \in \mathbb{C}_{Re>0}$ the function $\gamma_x: \mathbb{R}_+ \to \mathbb{C}, \gamma_x(t)=t^{x-1}\exp(-t)$ is integrable.

How do I show this?

Answer 1

Note that $\forall x,a,b\in\mathbb{R}:x^{a+ib}=x^ae^{ib\ln(x)}$. Hence, for any $x=a+bi$ with $a>0$ and $b\in\mathbb{R}$: $$\int_{\mathbb{R}^+}|t^{x-1}e^{-t}|dt = \int_{\mathbb{R}^+}|t^{a-1}e^{-t}|dt=\int_{\mathbb{R}^+}t^{a-1}e^{-t}dt.$$

We know that $\lim_{t\to 0}t^{a-1}e^{-t} = 0$, so there are no problems around zero. Let's look at the asymptotic behavior of $t^{a-1}e^{-t}$ at infinity. First of all, note that $t^{a-1}=O(t^{a-1})$ as $t\to \infty$. Furthermore, it is true that $$e^{-t} = o\left(\frac{1}{t^b}\right)\hspace{0.4cm}\text{as}\hspace{0.4cm}t\to\infty,$$ for all $b>0$. Take $b=2a$, then $$t^{a-1}e^{-t} = O\left(\dfrac{1}{t^{a+1}}\right)\hspace{0.4cm}\text{as}\hspace{0.4cm}t\to\infty.$$ Since $a>0$, this means that $t^{a-1}e^{-t}$ is integrable on $\mathbb{R}^+$.