# Lower incomplete gamma function format in series representation and R [closed]

1. As known that the lower incomplete gamma function can be written as $$\gamma(a,x) = x^{a}e^{-x}\sum_k^\infty{{x^{k}}\over a^{k+1}}.$$ What is the format for $$\sum_j^\infty{\gamma(v/p-j,rx^{p})}$$ in series representation? (where $$v$$, $$p$$ and $$r$$ are parameters)

2. $$\gamma(a,x)$$ in R is zipfR::Igamma(a, x, lower=TRUE, log=FALSE). If $$\sum_j^\infty{\gamma(v/p-j,rx^{p})}$$, what is the $$a$$ and $$b$$ should be in Igamma()?

## closed as off-topic by jbowman, kjetil b halvorsen, Peter Flom♦Nov 8 '18 at 12:25

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• You probably mean $\gamma(a,x) = x^a e^{-x} \sum_{k=0}^\infty \frac{x^k}{\Gamma(a+k+1)/\Gamma(a) }$ note that the summation is specified as $\sum_{k=0}^\infty$ instead of $\sum_{k}^\infty$, and probably your term $a^{k+1}$ is not right. – Martijn Weterings Nov 8 '18 at 8:56
• You are looking for the sum $\sum_{j=0}^\infty \gamma(v/p-j,rx^p)$. Based on your earlier question I guess that this relates to the generalized incomplete gamma function introduced by Chaudhry and Zubair. For this you should use: $$\gamma(a,x;b) = \sum_{k=0}^\infty \frac{(-b)^k}{k!} \gamma(a-k,x)$$ it will be different in your case with other parameters, but the absence of a term like $\frac{(-b)^k}{k!}$ is strange. – Martijn Weterings Nov 8 '18 at 9:13

$$\begin{array}{rl} \Gamma \left( -v, \frac{1}{x};\frac{z^2}{4} \right) = & \Gamma\left( - v , \frac{1}{x} \right) \Gamma(1+c) \left( \frac{2}{z} \right)^v I_v(z) \\ & - \frac{z^2}{4} \frac{x^{1+v}}{1+v} e^{-1/x} F \mathstrut_{2:0;0}^{0:2;1} \left[ \begin{array}{ r r r } \dfrac{\phantom{2,2+v}}{\phantom{2,2+v}} : & 1,1+v ; & 1 ; \\ 2, 2+v : & \dfrac{\phantom{1,1+v}}{\phantom{1,1+v}} ; & \dfrac{\phantom{1}}{\phantom{1}} ; & \end{array} -x \dfrac{z^2}{4}, \dfrac{z^2}{4} \right] \end{array}$$ where now $$v \neq -1, -2, -3, ...$$
$$\begin{array}{rl} \Gamma \left( n, x; z \right) = & \frac{(-z)^n}{n!} \lbrace \sum_{k=1}^n \frac{(-n)_k}{z^k} \Gamma(k,x) + _0F_1[-;n+1;z]\Gamma(0,x) \\& - \frac{z}{n+1} \frac{e^{-x}}{x} F \mathstrut_{2:0;0}^{0:2;1} \left[ \begin{array}{ r r r } \dfrac{\phantom{2,n+2}}{\phantom{2,2+v}} : & 1,1 ; & 1 ; \\ 2, n+2 \;: & \dfrac{\phantom{1,1}}{\phantom{1,1}} ; & \dfrac{\phantom{1}}{\phantom{1}} ; & \end{array} -x^{-1} z, z \right] \rbrace \end{array}$$