The reason for finiteness of the integral definition of the $\Gamma$-function in $\operatorname{Re}z>1$ is,
$$
\left|\int\limits_0^{\infty}t^{z-1}e^{-t}\mathrm{d}t\right|\leq \int_0^{\infty}\left|t^{z-1}e^{-t}\right|\mathrm{d}t=\int_0^{\infty}t^{\alpha}e^{-t}dt,
$$ where $\alpha>0$. Now, for lower value of $t$ close to zero, $t^{\alpha}$ is small, and as $t$ increases, we can see by L' Hospital rule that $\lim_{t\rightarrow \infty} t^{\alpha}/e^t=0$. Hence, intuitively the integral converges. But can anyone give a more formal explanation of this?
Best Answer
Splitting the integrals up $$\int\limits_0^\infty t^{\alpha-1}e^{-t}dt=\int\limits_0^1 t^{\alpha-1}e^{-t}dt+\int\limits_1^\infty t^{\alpha-1}e^{-t}dt$$
Now, we have that
$$0\le t\le 1\;\implies\;\; t^{\alpha-1}e^{-t}\le t^{\alpha-1}\;,\;\;\text{and}\;\;\int\limits_0^1t^{\alpha-1}dt=\left.\frac{t^\alpha}\alpha\right|_0^1=\frac1\alpha$$
and we also have that for sufficiently big $N$, with $N\leq t$
$$t^{\alpha-1}e^{-t}\stackrel{\text{}}\le e^{-t/2}\;,\;\;\text{and}\;\;\int\limits_1^\infty e^{-t/2}dt=\left.-2e^{-t/2}\right|_1^\infty=2 e^{-1/2}.$$