I've encountered a problem in my PhD. I would greatly appreciate any suggestions, tips, or comments you might have. The problem is

Let $\Gamma(s,x)$ be the incomplete gamma function. Given integers $n > m$, there exist functions $f,g$ such that

$$ \Gamma(n,m) = f(n)g(m) ? $$

In other words, can we separate Gamma in two independent functions, in this specific case?

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    $\begingroup$ Is that an incomplete gamma function? You should probably explain your notation. $\endgroup$ Apr 15 at 12:28
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    $\begingroup$ Welcome to MathOverflow. This was already posted about $2$ hours earlier on Math SE at Separating Gamma in two independent functions. Note that cross-posting, especially if it's done less than about a week later, is generally frowned on. Also, if you do post on another site, please at least include a link to the other post to help avoid duplication of efforts. $\endgroup$ Apr 15 at 13:39
  • $\begingroup$ Yes, that's the incomplete gamma function, fixed now. Thank you all for the time helping me. I noticed there's a typo in the notation. I edited it on the other site, very sorry about the duplication thing. math.stackexchange.com/questions/4899440/… $\endgroup$ Apr 15 at 15:49
  • $\begingroup$ @curiosity96 : Do you have a response to the answer below? $\endgroup$ Apr 16 at 21:18

1 Answer 1


$\newcommand\Ga\Gamma$If this factorization were true, we would get $$1=\frac{f(3)g(1)\,f(4)g(2)}{f(4)g(1)\,f(3)g(2)} =\frac{\Ga(3,1)\Ga(4,2)}{\Ga(4,1)\Ga(3,2)}=\frac{19}{16},$$ a contradiction.

So, your desired factorization cannot hold.


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