Puiseux expansion of logarithm of lower incomplete gamma function

I'm currently exploring different methods of evaluating the lower incomplete gamma function: $$\gamma(s,x)=\int_0^x{t^{s-1}}e^{-t}\;\mathrm{d}t.$$ When I look at series representations on WolframAlpha, it returns a generalized Puiseux series expansion around $$x=0$$ (at least if $$s$$ is not too large). For example, $$\gamma(100,x)=\log\bigg(\frac{x^{100}}{100}\bigg) - \frac{100}{101}x + \frac{25}{520251}x^2 + \frac{550}{1804057051}x^3 + \frac{12367525}{5798643470693424}x^4 + O(x^5)$$ This series is a complete mystery to me. How is such a thing derived? Is there a formula for generating this series for general $$s$$?

$$I=\int_0^x{t^{s-1}}e^{-t}\, dt=\Gamma (s)-\Gamma (s,x)$$ So, around $$x=0$$, we have $$I=x^s \left(\frac{1}{s}-\frac{x}{s+1}+\frac{x^2}{2 (s+2)}-\frac{x^3}{6 (s+3)}+\frac{x^4}{24 (s+4)}+O\left(x^5\right)\right)$$ that is to say $$I=\frac {x^s}s\left(1-\frac{s x}{s+1}+\frac{s x^2}{2 (s+2)}-\frac{s x^3}{6 (s+3)}+\frac{s x^4}{24 (s+4)}+O\left(x^5\right) \right)$$
$$\log(I)=\log \left(\frac{x^s}{s}\right)+$$ $$\log \left(1-\frac{s x}{s+1}+\frac{s x^2}{2 (s+2)}-\frac{s x^3}{6 (s+3)}+\frac{s x^4}{24 (s+4)}+O\left(x^5\right)\right)$$ Now, expand the logarithm to make $$\log(I)=\log \left(\frac{x^s}{s}\right)+\sum_{n=1}^p a_n x^n$$ $$a_1=-\frac s{1+s}$$ $$a_2=\frac{s}{2 (s+1)^2 (s+2)}$$ $$a_3=\frac{(s-1) s}{3 (s+1)^3 (s+2) (s+3)}$$ $$a_4=\frac{s \left(s^3-s^2-6 s+2\right)}{4 (s+1)^4 (s+2)^2 (s+3) (s+4)}$$ and so on.
• Thanks! This is perfect. For my future reference: the general term $a_n$ can be computed using Faa di Bruno's formula – David M. Dec 12 '20 at 19:24