Gamma function is $\Gamma(x)=\int_0^\infty e^{-t}t^{x-1}dt$. This is an integral that depends on a parameter, $x$. I want to show that this function is well defined. In order to show it, I could alredy prove thet $t\to f(t,x)=e^{-t}t^{x-1}$ is measurable, and that for all $t\in\mathbb{R}_0^+$ and some $x_0\in\mathbb{R}^+$ the limit of the function exists.
However, my third condition to finally prove it is well defined is the following:
There exists a function $g(t)$ integrable over $\mathbb{R}_0^+$ such that $|f(t,x)|\leq g(t)~\forall t\in\mathbb{R}_0^+~\forall x\in\mathbb{R}^+$
I can't find an upper bond that depends only on the variable $x$. I've tried with $x^{x-1}$ but it won't work, and I can always find a way to break that bound. Maybe I have to take a different approach to show it's well defined (doing the integral, using series criteria, etc) but I need to use the theorem about integrals that depend on the parameter and the only condition I can't show is the one I quoted.
An upper bound that depends only on the variable $x$ is $$ \Gamma(x) < \frac{\sqrt{2\pi} \cdot e^{1/12x} \cdot x^x}{\sqrt{x}e^x} $$
You can prove that the gamma function is well defined by proving that $$ \int_0^\infty t^{x-1} e^{-t} \; \mathrm d t $$ is convergent.
The proof can be found here beginning from page 8.