Γ(ϵ)∼1/ϵ

hbghlyj

Integrate[Exp[-1/x]/x^1.0001,{x,0,Infinity}]
证明:$$0<ϵ\ll1,\quad\int_0^∞{\exp(-\frac 1x)\over x^{1+ϵ}}dx∼\frac1ϵ$$

hbghlyj

$\int_0^∞ {\exp(-1/x)\over x^{1 + ϵ}} dx = \Gamma(ϵ)$
证明:(换元积分)
$\Gamma(ϵ)=\int_0^\infty t^{ϵ-1}e^{-t}dt$
令$t=1/x,dt=-dx/x^2$
$\Gamma(ϵ)=-\int_\infty^0 x^{1-ϵ}e^{-1/x}dx/x^2=\int_0^\infty x^{-1-ϵ}e^{-1/x}dx$

hbghlyj

如何证明 Γ(ϵ)∼1/ϵ
很简单 $ϵ\Gamma(ϵ)=\Gamma(ϵ+1)→\Gamma(1)=1$

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Laurent series $\Gamma(x)=x^{-1} - \gamma + \frac1{12}(6\gamma^2 + \pi^2)x + \frac16x^2\left(-\gamma^3 - \frac{\gamma \pi^2}2 + \psi^{(2)}(1)\right) + O(x^3)$
where $\gamma$ is the Euler-Mascheroni constant and $\psi^{(2)}(x)$ is the 2nd derivative of digamma function

hbghlyj

Limit[n!^(1/n), n -> 0]
$$\lim_{n\to0} \root n\of {n!} = e^{-\gamma}$$
这个怎么证明

战巡

hbghlyj 发表于 2024-3-27 02:00
Limit[n!^(1/n), n -> 0]
$$\lim_{n\to0} \root n\of {n!} = e^{-\gamma}$$
这个怎么证明

\[\lim_{n\to 0}\ln(\sqrt[n]{n!})=\lim_{n\to 0}\frac{\ln(\Gamma(n+1))}{n}\]
\[=\lim_{n\to 0}\frac{\frac{d}{dn}\ln(\Gamma(n+1))}{\frac{d}{dn}n}=\psi(1)=-\gamma\]