# Evaluating product of Upper Incomplete Gamma functions

I have checked several posts but couldn't find the equivalent of $$\Gamma(m,a) \cdot \Gamma(m,b)$$, where '$$\cdot$$' means multiplication. I suspect that it can be solved by applying the equivalent of Gamma function $$(n-1)!e^{-x}\sum \limits_{k=0}^{m}\dfrac{x^k}{k!}$$ but then there will be two summations with same limits which I have no clue how to solve Any suggestions?

• What do you understand by $\;\Gamma(m,a)\;$ ? That's not the usual Gamma Function...\ – DonAntonio Nov 19 '18 at 1:06
• Its the upper incomplete Gamma function – hakkunamattata Nov 19 '18 at 1:11
• I think it'd be a rather good idea to explicitly say that, and not only "Gamma Functions", which can mislead. – DonAntonio Nov 19 '18 at 1:17
• Upper or lower incomplete? "*" means multiplication (or something else)? The formula you state only applies when the first argument is a positive integer; is $m$ a positive integer? – Eric Towers Nov 19 '18 at 1:19
• yes m is positive, I have changed the title. Thanks for suggestion – hakkunamattata Nov 19 '18 at 1:23

We have that \eqalign{ & \Gamma (m,a)\Gamma (m,b) = \Gamma (m)^{\,2} Q(m,a)Q(m,b) = \cr & = \Gamma (m)^{\,2} e^{\, - \left( {a + b} \right)} \sum\limits_{k = 0}^{m - 1} {{{a^{\,k} } \over {k!}}} \sum\limits_{j = 0}^{m - 1} {{{b^{\,j} } \over {j!}}} \cr}
The sum is over a square in $$k,j$$ and , also with the help of the following scheme,
we can re-write it as \eqalign{ & \sum\limits_{k = 0}^{m - 1} {{{a^{\,k} } \over {k!}}} \sum\limits_{j = 0}^{m - 1} {{{b^{\,j} } \over {j!}}} = \sum\limits_{k = 0}^{m - 1} {\sum\limits_{j = 0}^{m - 1} {{{a^{\,k} b^{\,j} } \over {k!j!}}} } = \cr & = \sum\limits_{s = 0}^{2m - 1} {\sum\limits_{k = 0}^s {{{a^{\,k} b^{\,s - k} } \over {k!\left( {s - k} \right)!}}} } - \sum\limits_{s = 0}^{m - 1} {\sum\limits_{k = 0}^s {{{a^{\,m + k} b^{\,s - k} } \over {\left( {m + k} \right)!\left( {s - k} \right)!}}} } - \sum\limits_{s = 0}^{m - 1} {\sum\limits_{k = 0}^s {{{a^{\,s - k} b^{\,m + k} } \over {\left( {m + k} \right)!\left( {s - k} \right)!}}} } = \cr & = \sum\limits_{s = 0}^{2m - 1} {{{\left( {a + b} \right)^{\,s} } \over {s!}}} - a^{\,m} \sum\limits_{s = 0}^{m - 1} {\sum\limits_{k = 0}^s {{{a^{\,k} b^{\,s - k} } \over {\left( {m + k} \right)!\left( {s - k} \right)!}}} } - b^{\,m} \sum\limits_{s = 0}^{m - 1} {\sum\limits_{k = 0}^s {{{a^{\,s - k} b^{\,k} } \over {\left( {m + k} \right)!\left( {s - k} \right)!}}} } \cr} note the summation extends to $$m-1$$, not to $$m$$.
The formula above can be managed in various other ways, but I cannot see a way of getting rid of the $$m+k$$ at denominator.