I'm reading wiki page for Gamma function , where it says:

... Moreover, the gamma function has the following Laurent expansion in 1

$$\Gamma(z) = 1+\sum_{k=1}^\infty\frac{\Gamma^{(k)}(1)}{k!}(z-1)^{k},$$

valid for $|z-1| < 1$. In particular

$$\Gamma(z) = \frac{1}{z}-\gamma+\frac{1}{6}\left(3\gamma^2+\frac {\pi^2}2\right)z+O(z^2)$$

I have two questions here

  1. for the first equation, since $\Gamma(1)=1$, why there is a $\Gamma^{(k)}(1)$?
  2. for the second equation, what is $\gamma$?

1 Answer 1


(i) Recall the Taylor series at a point $x=a$

$$ f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n .$$

(ii) It is the Euler constant

$$ \gamma = \lim_{n \rightarrow \infty } \left( \sum_{k=1}^n \frac{1}{k} - \ln(n) \right).$$


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