Gamma function expansion

Question

I am trying to find the expansion of the $\Gamma(1+\epsilon)$ up to the quadratic order. The simplest way is (Can be found some other answers for almost the same question). $$ \Gamma(1+\epsilon)=\Gamma(1)+\epsilon \Gamma '(1)+\epsilon^2\Gamma''(1)+... $$ Now since $\Gamma'(1)=-\gamma_E$ and $\Gamma''(1)=\zeta(2)/2$ we get $$ \Gamma(1+\epsilon)=1-\epsilon \gamma_E+\epsilon^2\zeta(2)/2+... $$ In some lecture notes the same is written as $$ \Gamma(1+a \varepsilon)=\exp \left(-a \varepsilon \gamma_E+\sum_{k=2}^{\infty} \frac{\zeta(k)}{k}(-a \varepsilon)^k\right) $$ Well, it is true we are getting the first order term of the $e^x$, but is there any proof for the same? By this, we can not make sure. $$ \Gamma(3+\epsilon)=1-\epsilon \gamma_E+\epsilon^2\zeta(2)/2+... $$ But from the Taylor expansion, it seems $$ \Gamma(3+\epsilon)=2-\epsilon \gamma_E+\epsilon^2\zeta(2)/2+... $$

Answer 1

Lot of mistakes. Here are fixed formulae:

$$\Gamma(1+\epsilon)=\Gamma(1)+\Gamma'(1)\epsilon+\Gamma''(1)\color{red}{\frac{\epsilon^2}2}+\dots$$

$$\Gamma'(1)=-\gamma,\quad\Gamma''(1)=\color{red}{\gamma^2+\frac{\pi^2}6}$$

$$\Gamma(3+\epsilon)=2+\underbrace{\color{red}{(3-2\gamma)}}_{=\Gamma'(3)}\epsilon+\underbrace{\color{red}{\left(\gamma^2-3\gamma+1+\frac{\pi^2}6\right)}}_{=\Gamma''(3)/2}\epsilon^2+\dots$$