# Asymptotic of gamma function

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.

• What is $o_p$? $– Szeto Jul 21 '18 at 13:37 ## 1 Answer The series expansion of the function$\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)$in http://mathworld.wolfram.com/GammaFunction.html • Can you tell me the reference for the following result:$\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)}$? – steven Jul 21 '18 at 16:40 • – JJacquelin Jul 21 '18 at 16:54 • One can find the successive derivatives of$\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. – JJacquelin Jul 21 '18 at 17:10
• I got it. Thanks for your kindly help. – steven Jul 22 '18 at 1:22