The incomplete gamma function (upper) is defined as $\Gamma(a,x)=\int_x^{\infty}t^{a-1}e^{-t}dt$. Is there a series expansion for small non-integer $a << 1$ and small $x$?
[Math] Series Expansion of the Incomplete Gamma Function
gamma functionsequences-and-series
Related Solutions
From http://dlmf.nist.gov/8.7.E3 we have the series expansion $$\Gamma(s,x) = \Gamma(s) - \sum_{n=0}^\infty \frac{(-1)^n x^{n+s}}{n! (n+s)}, \qquad s \ne 0, -1, -2, \ldots $$ Combine this with the relation for the gamma functions (http://dlmf.nist.gov/8.2.E3) $$\gamma(s,x) + \Gamma(s,x) = \Gamma(s).$$ Therefore the series expansions remains valid for all non-integer $s<0$ $$ \gamma(s,x) = \sum_{n=0}^\infty \frac{(-1)^n x^{n+s}}{n! (n+s)}, \qquad s \ne 0, -1, -2, \ldots $$ Another route is via Tricomi's entire incomplete gamma function $\gamma^{*}$, see http://dlmf.nist.gov/8.7.E1.
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.
Best Answer
Thanks to this question, we know that
$$\Gamma(a,x)=\Gamma(a)-\sum_{n=0}^\infty\frac{(-1)^nx^{a+n}}{n!(a+n)}$$
Which follows from the Taylor expansion of $e^x$.