Why do we split the gamma function at x=1?

Question

I am trying to prove that the gamma function converges.

Namely i'm trying to prove, $\displaystyle{\Gamma(t)=\int_{0}^{\infty} x^{t-1}e^{-x}} dx$ converges.

I understand how but most places such as here (Page 8) rewrites $\displaystyle{\Gamma(t)=\int_{0}^{\infty} x^{t-1}e^{-x}} dx$ as $\displaystyle{\Gamma(t)=\int_{0}^{1} x^{t-1}e^{-x}} dx + \displaystyle{\int_{1}^{\infty} x^{t-1}e^{-x}} dx $ first. Why is that. Why do we split the integral at $1$?

I'm a little confused by that.

Answer 1

The integral is split in this proof because the reasons behind its convergence between 0 and 1 and between 1 and $+\infty$ are not the same. For instance, the integral converges between 0 and 1 because the $x$ term converges "quicker" than the exponential term diverges whereas between 1 and $+\infty$ it is because of the exponential term that the integral converges. Splitting the integral here allows the author to use a Riemann integral to show this result in an clear and simple way

Answer 2

The assertion “the integral $\int_0^\infty x^{t-1}e^{-x}\,\mathrm dx$ converges” is equivalent to “for some $a\in(0,\infty)$, both integrals $\int_0^a x^{t-1}e^{-x}\,\mathrm dx$ and $\int_a^\infty x^{t-1}e^{-x}\,\mathrm dx$ converge”. A natural choice for $a$ is to take $a=1$, but any other number greater than $0$ will do.