1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ What do you understand by $\;\Gamma(m,a)\;$ ? That's not the usual Gamma Function...\ $\endgroup$ – DonAntonio Nov 19 '18 at 1:06
  • $\begingroup$ Its the upper incomplete Gamma function $\endgroup$ – hakkunamattata Nov 19 '18 at 1:11
  • $\begingroup$ I think it'd be a rather good idea to explicitly say that, and not only "Gamma Functions", which can mislead. $\endgroup$ – DonAntonio Nov 19 '18 at 1:17
  • $\begingroup$ 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? $\endgroup$ – Eric Towers Nov 19 '18 at 1:19
  • $\begingroup$ yes m is positive, I have changed the title. Thanks for suggestion $\endgroup$ – hakkunamattata Nov 19 '18 at 1:23
0
$\begingroup$

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,

Gamma_Inc^2_1

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.