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To prove that the $\Gamma (s) > 0$ for all $s > 0$ we consider the following, $\Gamma(s) \geq \int\limits_0^1 e^{-t} t^{s-1} dt \geq e^{-1} \int\limits_0^1 t^{s-1} dt$. I do not know how to show the second inequality.

Zoey
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  • $t \mapsto e^{-t}$ is strictly monotonically decreasing on $[0,1]$. Therefore it is bounded below by the constant $e^{-1}$, which, by the linearity of integration, "comes out". – Eman Yalpsid Mar 22 '17 at 10:08

1 Answers1

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If $f \geq g$ in $[0,1]$, then $\int\limits_0^1 f(t) dt \geq \int\limits_0^1 g(t) dt$. In the second inequality $f(t) = e^{-t}t^{s-1}$, and $g(t) = e^{-1} t^{s-1}$. The latter is smaller, because the function $e^{-t}$ decreases in $[0,1]$ and its smallest value there is $e^{-1}$.

ajr
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  • Just what I was typing! Nice and clearly put, not just saying $\int_0^1 f \ge f_{\max}$ but showing for general $f$ and $g$. Perhaps leave it as an exercise for the OP to show your first statement? (+1) – Sam OT Mar 22 '17 at 10:08