What is the Taylor series expansion of the lower incomplete gamma function $\gamma(s,-x)$?
The standard answer from wikipedia is interms of $x^k$:
$$\gamma(s,-x)=\sum_{k=0}^{\infty}\frac{(-x)^se^{x}(-x)^k}{s(s+1)\cdots(s+k)}=(-x)^se^{x}\Gamma(s)\sum_{k=0}^{\infty}\frac{(-1)^k x^k}{\Gamma(s+k+1)}\tag{1}$$
Mathematica 10.0 gives a series in terms of $x^{-k}$:
$$\gamma(s,-x)=\Gamma(s)+e^x x^s (-1)^s\left(\frac{1}{x}+\frac{s-1}{x^2}+\frac{(s-1)(s-2)}{x^3}+\cdots\right)\tag{2}$$
The series in (2) suits our needs because it is in terms of $1/x$ so we can truncate it at $1/x^3$ for example.
The series in (1) does not suits our need because we have to keep large enough terms, say $K$ terms, to make $\frac{x^K}{|\Gamma(s+K+1)|}<1$ as $x\to+\infty$.
The expression (2) is actually from the asymptotic series for $\Gamma(s,-x)$ because
$$\gamma(s,-x)=\Gamma(s)-\Gamma(s,-x)\tag{3}$$.
It is not obvious to me how one can derive from (1) to (2). That is why I ask this question.
Thanks-
mike
Best Answer
Both (1) and (2) can be derived from the integral definition of the lower gamma function. By definition: $$ \gamma(s,-x) = \int_0^{-x} t^{s-1} e^{-t} dt $$ To see either, it suffices to perform repeated integration by parts on this definition. Letting: $$ u = e^{-t} \qquad dv=t^{s-1}dt \\ du = -e^{-t}dt \qquad v=\frac{t^s}{s} $$ One obtains: $$ \gamma(s,-x) = \left [\frac{e^{-t}t^s}{s} \right|_0^{-x}+\frac{1}{s} \int_0^{-x} t^s e^{-t}dt \\= \frac{e^{x}(-x)^s}{s}+\frac{1}{s} \gamma(s+1,-x) \\= e^{x}(-x)^s \left ( \frac{1}{s} + \frac{x}{s(s+1)} + \frac{x^2}{s(s+1)(s+2)} + \cdots \right) $$ This relation may be repeated to obtain (1). To obtain a similar version of (2). Simply invert the IBP and let: $$ u = t^{s-1} \qquad dv=e^{-t}dt \\ du = (s-1) t^{s-2}dt \qquad v=-e^{-t} $$ Then: $$ \gamma(s,-x) = \left [-e^{-t}t^{s-1} \right|_0^{-x}+(s-1) \int_0^{-x} t^{s-2} e^{-t}dt \\= -e^{x}(-x)^{s-1}+(s-1) \gamma(s-1,-x) \\= -e^{x}(-x)^s \left(\frac{1}{x} + \frac{(s-1)}{x^2} + \frac{(s-1)(s-2)}{x^3} + \cdots \right) $$ Your form (2) is simply then from applying a similar analysis on the integral for $\Gamma(s,-x)$ through (3).
Alternatively, one can avoid all of the integral definitions by simply using the recurrences thus derived, as in here. To truly go from (1) to (2) though, one would have to effectively work back to a more general definition, before progressing.