# Assymptotics of incomplete gamma function $\gamma(x /\ln 2,x)$

I am trying to prove that $$\frac{\gamma(x/ \ln 2 ,x)}{\Gamma(x / \ln 2)}<ae^{-bx}$$ for some positive $a$ and $b$, where $\gamma(x,s) = \int_0^s t^{x-1}e^{-t}dt$ is lower incomplete gamma function.

Asymptotics of $\Gamma(x)$ is well known but I struggle to come with good bounds for the incomplete gamma.

• Do you want your inequality to hold for any $x>0$ or just for any $x$ large enough? – Jack D'Aurizio Sep 18 '17 at 18:53
• Anyway, my previous remark still stands. Since $t^{x\log 2}e^{-t}$ is concentrated around $t=x\log 2 \ll x$, I would expect the LHS to converge to $1$ as $x\to +\infty$, violating your inequality for any $b>0$. – Jack D'Aurizio Sep 18 '17 at 18:55
• I think that if one shows it for large enough $x$ than it should be possible to adjust the coefficients to account for smaller $x$ – Boris K. Sep 18 '17 at 18:57
• Thank you, instead of $\ln 2$ it should have been $\log_2 e=(\ln 2)^{-1}$. Apparently that should hold or for any constant greater than 1. – Boris K. Sep 18 '17 at 19:19

The inequality from http://dlmf.nist.gov/8.10.E11 might help $$(1-e^{-\alpha_{a}x})^{a}\leq P\left(a,x\right)\leq(1-e^{-\beta_{a}x})^{a},$$ with $$P(a,x) = \frac{\gamma(a,x)}{\Gamma(a)}$$ and $\alpha_{a}, \beta_{a}$ given by http://dlmf.nist.gov/8.10.E12