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I came across a quetion: Let $h$ go to zero. What is the asymptotic of $\Gamma(x+o_{p}(h))$ where $x\in(0,2)$? The difficulty is the limitation of x goes to zero.

Can I obtain $$\Gamma(x+o_{p}(h))\sim\Gamma(x)$$ Any comments are welcomed.

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    $\begingroup$ What is $o_p$? $ $\endgroup$ – Szeto Jul 21 '18 at 13:37
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The series expansion of the function $\Gamma(x)$ is : $$\Gamma(x+\epsilon)=\Gamma(x)+\Gamma(x)\left(\psi^{(0)}(x)\right)\:\epsilon+\Gamma(x)\left(\left(\psi^{(0)}(x)\right)^2+\psi^{(1)}(x)\right) \:\frac{\epsilon^2}{2}+O(\epsilon^3)$$ This formula is restricted at $x>0$ because $\Gamma(x\to 0)\to\infty$.

The coefficients of the above Taylor series are computed as usual in terms of successive derivatives of $\Gamma(x)$, which involves the polygamma functions $\psi^{(n)}(x)$. The usual digamma function is $\psi(x)=\psi^{(0)}(x)$.

For $x=0$ and $\epsilon\to 0$ we have to look for asymptotic series : $$\Gamma(\epsilon)\sim \frac{1}{\epsilon}-\gamma+\frac{1}{12}(6\gamma^2+\pi^2)\epsilon+O(\epsilon^3)$$ $\gamma$ is the Euler-Mascheroni constant. This is derived from the asymptotic series $(35)$ in http://mathworld.wolfram.com/GammaFunction.html

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  • $\begingroup$ Can you tell me the reference for the following result: $\Gamma(x+\epsilon)=\Gamma(x)+\Gamma(x)\left(\psi^{(0)}(x)\right)\:\epsilon+\Gamma(x)\left(\left(\psi^{(0)}(x)\right)^2+\psi^{(1)}(x)\right) \:\frac{\epsilon^2}{2}+O(\epsilon^3)$, what $\psi^{(0)}$ and $\psi^{(1)}$? $\endgroup$ – steven Jul 21 '18 at 16:40
  • $\begingroup$ One can find the successive derivatives of $\Gamma(x)$ from the definition of the polygamma functions $\psi^{(n)}(x)=\frac{d^{n+1}}{dx^{n+1}}\ln(\Gamma(x))$. mathworld.wolfram.com/PolygammaFunction.html . This allows to obtain the above Taylors series after some calculus. $\endgroup$ – JJacquelin Jul 21 '18 at 17:10
  • $\begingroup$ I got it. Thanks for your kindly help. $\endgroup$ – steven Jul 22 '18 at 1:22

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