# Gamma function is strictly positive and increasing

Rudin defined Gamma function for $s>0$ as integral $\Gamma(s)=\int \limits_{0}^{\infty}x^{s-1}e^{-x}dx$. How to prove that $\Gamma(s)$ strictly positive function and monotone increasing? Can anyone show please a rigorous proof!

• Positivity is fairly straightforward. Monotonicity on the other hand isn't, since $\Gamma(s)$ is decreasing on $(0,\mu]$ and increasing on $[\mu,+\infty)$ for some value $\mu \in (1,2)$. – Daniel Fischer Jan 7 '16 at 16:52
• @DanielFischer, 1) Why strict positivity is obvious? 2) Why $\Gamma(s)$ is decreasing on $(0,\mu]$ for some $\mu \in (1,2)$? Can you reveal you answer please. I would be greatful for you! – A.Ward.2016 Jan 7 '16 at 16:58
• 1. The integrand is obviously positive in $(0, \infty)$. 2. We have $\Gamma(1) = 1 = \Gamma(2)$. – Paul K Jan 7 '16 at 17:00
• 1) because the integrand is strictly positive for every $s\in (0,+\infty)$. 2) We have $\lim\limits_{s\searrow 0} \Gamma(s) = +\infty = \lim\limits_{s\to +\infty} \Gamma(s)$, further $\Gamma$ is convex. – Daniel Fischer Jan 7 '16 at 17:01
• For every $s\in (0,+\infty)$, we have $x^{s-1} e^{-x} > 0$ for all $x \in (0,+\infty)$, therefore the integral is strictly positive. – Daniel Fischer Jan 7 '16 at 17:11