$\gamma$ is the lower incomplete gamma function. Is $\gamma(1, x) \ge \gamma(k, kx)$ when $k \in Z^+$, $x \in (0,1)$?
Concerning the lower incomplete gamma function
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special-functions
gamma-function
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0@J.M. yes I saw that when I was "googling for bounds" but I was not able to convert it into a solution... – 2012-08-11
1 Answers
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It doesn't appear to be the case.
Addendum: Notice that $\gamma(1,x) = 1-e^{-x}$, so $\gamma(1,x) < 1$ for $x\in(0,1)$. But $\gamma(k,k x) \sim \frac{(k x e^{-x})^k}{k(1-x)} \qquad (k\to\infty).$ (See DLMF 8.11.6.) Thus, for any $x\in(0,1)$ we can find a $k$ large enough so $\gamma(k,k x) > 1 > \gamma(1,x)$.