$\Gamma ( \alpha)$ function

Question

Gamma function of $\alpha$ is defined as

$\Gamma \left( \alpha \right) = \int\limits_0^\infty {y^{\alpha - 1} e^{ - y} dy}$

Gamma function exist for $ \alpha > 0 $ why???

I think the reason is $\Gamma(\alpha) $ diverges for $ \alpha < 0$ but how to prove divergence..

Answer 1

We have: $$y^{\alpha-1}e^{-y}\sim_0y^{\alpha-1},$$ so the integral $\int_0^1 y^{\alpha-1}e^{-y}dy$ is convergent if and only if $\alpha-1>-1\iff \alpha>0$.Moreover, we have $$y^{\alpha-1}e^{-y}=_\infty o(\frac{1}{y^2}),$$ so the integral $\int_1^\infty y^{\alpha-1}e^{-y}dy$ is convergent for all $\alpha\in\mathbb{R}$.

Finally we conclude that $\Gamma(\alpha)=\int_0^\infty y^{\alpha-1}e^{-y}dy$ is convergent if and only if $\alpha>0$.

Answer 2

On $[0,1]$, $e^{-y}\geq 1/e$, and $y^{\alpha-1}e^{-y}\geq 0$ for all $y>0$, so we have

$$\int_0^{\infty} y^{\alpha-1}e^{-y} dy > \int_0^{1} y^{\alpha-1}e^{-y} dy > \frac{1}{e}\int_0^{1} y^{\alpha-1} dy.$$ This last integral is divergent when $\alpha\leq 0$.