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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  • $\begingroup$ 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. $\endgroup$ – Martijn Weterings Nov 8 '18 at 8:56
  • $\begingroup$ 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. $\endgroup$ – Martijn Weterings Nov 8 '18 at 9:13

If your question is about the generalized incomplete gamma function that you mention in an earlier question, then note that the path you try to follow here has been done before (I have now added this also to my previous answer to your previous question):

Miller, Allen R., and Ira S. Moskowitz. "On certain generalized incomplete gamma functions." Journal of computational and applied mathematics 91.2 (1998): 179-190

They eventually express the function in terms of a Kampé de Fériet function (a two-variable generalization of the generalized hypergeometric series)

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

and for non negative integers

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


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