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The lower incomplete gamma function for positive $s$ is defined by the integral $$ \gamma(s,x)=\int_0^{x} t^{s-1} e^{-t} dt. $$ Taylor expansion of the exponential function and term by term integration give the following expansion $$ \gamma(s,x)=\sum_{n=0}^\infty \frac{(-1)^n x^{n+s}}{n! (n+s)} $$ Here $\gamma(s,x)$ can be analytically continued for complex $s$ except some singular points. Does the above expansion still hold in this case? Especially I'd like to know if the relation holds for $s < 0$. Each term of the series is well defined for $s < 0$. Is this enough to ensure the validity of the relation or does it need more arguments?

Amzoti
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asofas
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  • You need more arguments. If you could show that the sum is an analytic function of $s$ for $s \neq 0, -1, -2, \ldots$ then the equality would hold for $s \neq 0, -1, -2, \ldots$ by analytic continuation. For such an argument one could probably make use of the dominated convergence theorem. – Antonio Vargas Mar 13 '14 at 16:59

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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.

gammatester
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  • Thank you for the answer. Ok, the expansion is right for complex $s$. I understand the $\gamma^*$ approach. By the way, do you have any comment about my last question? – asofas Mar 13 '14 at 14:43
  • I guess with validity you mean (well-)defined. This is necessary but you still need the convergence of the sum. – gammatester Mar 13 '14 at 15:09