How can we derive the asymptotic expansion for the second derivative of the gamma-function?

Question

We can expression the first derivative of the gamma function as: $$\Gamma'(s) \sim -\frac{1}{s^2}+\frac{6\gamma^2+\pi^2}{12}+O(s)$$ but what about the second derivative? I do not know how to approach the problem. Thank you.

Answer 1

You could obtain the expansion of any derivative of $\Gamma(s)$ using its own expansion and derivating term wise $$\Gamma(s)=\frac{1}{s}-\gamma +\frac{1}{12} \left(6 \gamma ^2+\pi ^2\right) s+\frac{1}{6} \left(-\gamma ^3-\frac{\gamma \pi ^2}{2}+\psi ^{(2)}(1)\right)s^2+\frac{1}{24} \left(8 \gamma \zeta (3)+\gamma ^4+\gamma ^2 \pi ^2+\frac{3 \pi ^4}{20}\right)s^3+\frac{ \left(-40 \left(6 \gamma ^2+\pi ^2\right) \zeta (3)-288 \zeta (5)-12 \gamma ^5-20 \gamma ^3 \pi ^2-9 \gamma \pi ^4\right)}{1440}s^4+O\left(s^5\right)$$ and truncate the result to $O(s)$.

Answer 2

• Start from the explicit formula for $\Gamma'(s)/\Gamma(s)$

• Expand $\Gamma'(s)/\Gamma(s)+\frac1s$ in power series, the coefficients are $(-1)^{k+1}\zeta(k+1)$,

• Integrate to get the power series of $\log \Gamma(s)+\log s$

• Exponentiate to get the power series of $s \Gamma(s)$

• Differentiate the obtained Laurent series of $\Gamma(s)$ to obtain the Laurent series of its derivatives.