Evaluating product of Upper Incomplete Gamma functions

Question

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?

Answer 1

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.