Prove the improper integral of the Gamma function $\Gamma(t)$ converges for $z \in \mathbb C$ with $Re(z) > 0$:

Prove the improper integral of the Gamma function $\Gamma(t)$ converges for $z \in \mathbb C$ with $Re(z) > 0$: The gamma function $\Gamma(t)$ is defined by the following improper integral $$\Gamma(t) = \int^{\infty}_0 x^{t-1}e^{-x}dx$$

It is said that the integral converges for $z \in \mathbb C$ with $Re(z) >0$, and I've been trying to find a way of proving this.

Using integration by parts we know $\Gamma(t+1) = t\Gamma(t)$. However this doesn't really prove anything does it ? Because here we assume the improper definite integral converges (correct me if I'm wrong). Also if $t$ is not a positive integer, but some positive real number, we must evaluate $\Gamma(t)$ for some $t \in (0,1)$.

Could someone tell me how to determine that this integral actually converges ?

Solution 1:

Hints using the comment of Daniel:

$$\int\limits_0^\infty x^{t-1}e^{-x}dx=\int\limits_0^1 x^{t-1}e^{-x}dx+\int\limits_1^\infty x^{t-1}e^{-x}dx$$

Now, we have that

$$0\le x\le 1\;\implies\;\; x^{t-1}e^{-x}\le x^{t-1}\;,\;\;\text{and}\;\;\int\limits_0^1x^{t-1}dx=\left.\frac{x^t}t\right|_0^1=\frac1t$$

and we also have that

$$1\le x\;\implies\;\;x^{t-1}e^{-x}\stackrel{\text{Prove this!}}\le e^{-x/2}\;,\;\;\text{and}\;\;\int\limits_1^\infty e^{-x/2}dx=\left.-2e^{-x/2}\right|_1^\infty=2 e^{-1/2}$$