Asymptotic of gamma function

Question

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.

Answer 1

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