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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!

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    $\begingroup$ 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 $\mu \in (1,2)$. $\endgroup$ – Daniel Fischer Jan 7 '16 at 16:52
  • $\begingroup$ @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! $\endgroup$ – A.Ward.2016 Jan 7 '16 at 16:58
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    $\begingroup$ 1. The integrand is obviously positive in $(0, \infty)$. 2. We have $\Gamma(1) = 1 = \Gamma(2)$. $\endgroup$ – Paul K Jan 7 '16 at 17:00
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    $\begingroup$ 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. $\endgroup$ – Daniel Fischer Jan 7 '16 at 17:01
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    $\begingroup$ 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. $\endgroup$ – Daniel Fischer Jan 7 '16 at 17:11

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