# Behaviour of the Gamma function near zero

The Gamma function on the positive real half-line is defined via the reknown formula $$\Gamma(z)=\int_0^\infty x^{z-1}e^{-x}dx, \quad z>0.$$ A classical result is Stirling's formula, describing the behaviour of $$\Gamma(z)$$ as $$z$$ diverges to infinity, $$\Gamma(z)\sim \sqrt{\frac{2\pi}{z}} \left( \frac{z}{e}\right)^z, \quad z \to \infty.$$ Is there any such approximation formula for $$z \downarrow 0$$, describing the speed at which the Gamma function diverges near the origin?

• By the functional equation and Taylor's theorem, $$\Gamma (z) = \frac{1}{z}z\Gamma (z) = \frac{1}{z}\Gamma (z + 1) = \frac{1}{z}(\Gamma (0 + 1) + \mathcal{O}(z)) = \frac{1}{z} + \mathcal{O}(1)$$ as $z\to 0$. For a more precise statement, see math.stackexchange.com/q/1287555
– Gary
Nov 29, 2020 at 9:15
• $\gamma$ is Euler-Mascheroni constant. For $z\to 0$ we have $$\Gamma(z)=\frac{1}{z}-\gamma +\frac{1}{12} \left(6 \gamma ^2+\pi ^2\right) z+O\left(z^2\right)$$ Nov 29, 2020 at 13:21

Around $$z=0$$ $$\Gamma(z)=\frac{1}{z}-\gamma +\sum_{n=1} a_n z^n$$ $$a_1=\frac{1}{12} \left(6 \gamma ^2+\pi ^2\right)\qquad a_2=-\frac{1}{6} \left(2 \zeta (3)+\gamma ^3+\frac{\gamma \pi ^2}{2}\right)$$ $$a_3=\frac{1}{24} \left(8 \gamma \zeta (3)+\gamma ^4+\gamma ^2 \pi ^2+\frac{3 \pi ^4}{20}\right)$$
A quite good approximation is given by $$\Gamma(z)\sim \frac 1 z \frac{1+\frac{\left(\pi ^2-6 \gamma ^2\right) }{12 \gamma }z } {1+\frac{\left(\pi ^2+6 \gamma ^2\right) }{12 \gamma }z }$$
• Nice. The last formula is very pretty. How did you get the closed forms for $a_1,a_2,a_3$ ? Dec 1, 2020 at 16:15