Evaluating product of Upper Incomplete Gamma functions












2














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?






summation gamma-function gamma-distribution







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  • What do you understand by $;Gamma(m,a);$ ? That's not the usual Gamma Function...
    – DonAntonio
    Nov 19 at 1:06










  • Its the upper incomplete Gamma function
    – hakkunamattata
    Nov 19 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 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 at 1:19












  • yes m is positive, I have changed the title. Thanks for suggestion
    – hakkunamattata
    Nov 19 at 1:23
















2














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?






summation gamma-function gamma-distribution







share|cite|improve this question
























  • What do you understand by $;Gamma(m,a);$ ? That's not the usual Gamma Function...
    – DonAntonio
    Nov 19 at 1:06










  • Its the upper incomplete Gamma function
    – hakkunamattata
    Nov 19 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 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 at 1:19












  • yes m is positive, I have changed the title. Thanks for suggestion
    – hakkunamattata
    Nov 19 at 1:23














2












2








2







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?






summation gamma-function gamma-distribution







share|cite|improve this 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?






summation gamma-function gamma-distribution




summation gamma-function gamma-distribution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 19 at 1:24









David G. Stork

9,81021232




9,81021232










asked Nov 19 at 1:01









hakkunamattata

475




475












  • What do you understand by $;Gamma(m,a);$ ? That's not the usual Gamma Function...
    – DonAntonio
    Nov 19 at 1:06










  • Its the upper incomplete Gamma function
    – hakkunamattata
    Nov 19 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 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 at 1:19












  • yes m is positive, I have changed the title. Thanks for suggestion
    – hakkunamattata
    Nov 19 at 1:23


















  • What do you understand by $;Gamma(m,a);$ ? That's not the usual Gamma Function...
    – DonAntonio
    Nov 19 at 1:06










  • Its the upper incomplete Gamma function
    – hakkunamattata
    Nov 19 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 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 at 1:19












  • yes m is positive, I have changed the title. Thanks for suggestion
    – hakkunamattata
    Nov 19 at 1:23
















What do you understand by $;Gamma(m,a);$ ? That's not the usual Gamma Function...
– DonAntonio
Nov 19 at 1:06




What do you understand by $;Gamma(m,a);$ ? That's not the usual Gamma Function...
– DonAntonio
Nov 19 at 1:06












Its the upper incomplete Gamma function
– hakkunamattata
Nov 19 at 1:11






Its the upper incomplete Gamma function
– hakkunamattata
Nov 19 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 at 1:17




I think it'd be a rather good idea to explicitly say that, and not only "Gamma Functions", which can mislead.
– DonAntonio
Nov 19 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 at 1:19






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 at 1:19














yes m is positive, I have changed the title. Thanks for suggestion
– hakkunamattata
Nov 19 at 1:23




yes m is positive, I have changed the title. Thanks for suggestion
– hakkunamattata
Nov 19 at 1:23










1 Answer
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We have that
$$
eqalign{
& Gamma (m,a)Gamma (m,b) = Gamma (m)^{,2} Q(m,a)Q(m,b) = cr
& = Gamma (m)^{,2} e^{, - left( {a + b} right)} sumlimits_{k = 0}^{m - 1} {{{a^{,k} } over {k!}}}
sumlimits_{j = 0}^{m - 1} {{{b^{,j} } over {j!}}} cr}
$$



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
$$
eqalign{
& sumlimits_{k = 0}^{m - 1} {{{a^{,k} } over {k!}}} sumlimits_{j = 0}^{m - 1} {{{b^{,j} } over {j!}}}
= sumlimits_{k = 0}^{m - 1} {sumlimits_{j = 0}^{m - 1} {{{a^{,k} b^{,j} } over {k!j!}}} } = cr
& = sumlimits_{s = 0}^{2m - 1} {sumlimits_{k = 0}^s {{{a^{,k} b^{,s - k} } over {k!left( {s - k} right)!}}} }
- sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,m + k} b^{,s - k} } over {left( {m + k} right)!left( {s - k} right)!}}} }
- sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,s - k} b^{,m + k} } over {left( {m + k} right)!left( {s - k} right)!}}} } = cr
& = sumlimits_{s = 0}^{2m - 1} {{{left( {a + b} right)^{,s} } over {s!}}}
- a^{,m} sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,k} b^{,s - k} } over {left( {m + k} right)!left( {s - k} right)!}}} }
- b^{,m} sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,s - k} b^{,k} } over {left( {m + k} right)!left( {s - k} right)!}}} } cr}
$$
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.






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    1 Answer
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    We have that
    $$
    eqalign{
    & Gamma (m,a)Gamma (m,b) = Gamma (m)^{,2} Q(m,a)Q(m,b) = cr
    & = Gamma (m)^{,2} e^{, - left( {a + b} right)} sumlimits_{k = 0}^{m - 1} {{{a^{,k} } over {k!}}}
    sumlimits_{j = 0}^{m - 1} {{{b^{,j} } over {j!}}} cr}
    $$



    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
    $$
    eqalign{
    & sumlimits_{k = 0}^{m - 1} {{{a^{,k} } over {k!}}} sumlimits_{j = 0}^{m - 1} {{{b^{,j} } over {j!}}}
    = sumlimits_{k = 0}^{m - 1} {sumlimits_{j = 0}^{m - 1} {{{a^{,k} b^{,j} } over {k!j!}}} } = cr
    & = sumlimits_{s = 0}^{2m - 1} {sumlimits_{k = 0}^s {{{a^{,k} b^{,s - k} } over {k!left( {s - k} right)!}}} }
    - sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,m + k} b^{,s - k} } over {left( {m + k} right)!left( {s - k} right)!}}} }
    - sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,s - k} b^{,m + k} } over {left( {m + k} right)!left( {s - k} right)!}}} } = cr
    & = sumlimits_{s = 0}^{2m - 1} {{{left( {a + b} right)^{,s} } over {s!}}}
    - a^{,m} sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,k} b^{,s - k} } over {left( {m + k} right)!left( {s - k} right)!}}} }
    - b^{,m} sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,s - k} b^{,k} } over {left( {m + k} right)!left( {s - k} right)!}}} } cr}
    $$
    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.






    share|cite|improve this answer




























      0














      We have that
      $$
      eqalign{
      & Gamma (m,a)Gamma (m,b) = Gamma (m)^{,2} Q(m,a)Q(m,b) = cr
      & = Gamma (m)^{,2} e^{, - left( {a + b} right)} sumlimits_{k = 0}^{m - 1} {{{a^{,k} } over {k!}}}
      sumlimits_{j = 0}^{m - 1} {{{b^{,j} } over {j!}}} cr}
      $$



      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
      $$
      eqalign{
      & sumlimits_{k = 0}^{m - 1} {{{a^{,k} } over {k!}}} sumlimits_{j = 0}^{m - 1} {{{b^{,j} } over {j!}}}
      = sumlimits_{k = 0}^{m - 1} {sumlimits_{j = 0}^{m - 1} {{{a^{,k} b^{,j} } over {k!j!}}} } = cr
      & = sumlimits_{s = 0}^{2m - 1} {sumlimits_{k = 0}^s {{{a^{,k} b^{,s - k} } over {k!left( {s - k} right)!}}} }
      - sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,m + k} b^{,s - k} } over {left( {m + k} right)!left( {s - k} right)!}}} }
      - sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,s - k} b^{,m + k} } over {left( {m + k} right)!left( {s - k} right)!}}} } = cr
      & = sumlimits_{s = 0}^{2m - 1} {{{left( {a + b} right)^{,s} } over {s!}}}
      - a^{,m} sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,k} b^{,s - k} } over {left( {m + k} right)!left( {s - k} right)!}}} }
      - b^{,m} sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,s - k} b^{,k} } over {left( {m + k} right)!left( {s - k} right)!}}} } cr}
      $$
      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.






      share|cite|improve this answer


























        0












        0








        0






        We have that
        $$
        eqalign{
        & Gamma (m,a)Gamma (m,b) = Gamma (m)^{,2} Q(m,a)Q(m,b) = cr
        & = Gamma (m)^{,2} e^{, - left( {a + b} right)} sumlimits_{k = 0}^{m - 1} {{{a^{,k} } over {k!}}}
        sumlimits_{j = 0}^{m - 1} {{{b^{,j} } over {j!}}} cr}
        $$



        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
        $$
        eqalign{
        & sumlimits_{k = 0}^{m - 1} {{{a^{,k} } over {k!}}} sumlimits_{j = 0}^{m - 1} {{{b^{,j} } over {j!}}}
        = sumlimits_{k = 0}^{m - 1} {sumlimits_{j = 0}^{m - 1} {{{a^{,k} b^{,j} } over {k!j!}}} } = cr
        & = sumlimits_{s = 0}^{2m - 1} {sumlimits_{k = 0}^s {{{a^{,k} b^{,s - k} } over {k!left( {s - k} right)!}}} }
        - sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,m + k} b^{,s - k} } over {left( {m + k} right)!left( {s - k} right)!}}} }
        - sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,s - k} b^{,m + k} } over {left( {m + k} right)!left( {s - k} right)!}}} } = cr
        & = sumlimits_{s = 0}^{2m - 1} {{{left( {a + b} right)^{,s} } over {s!}}}
        - a^{,m} sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,k} b^{,s - k} } over {left( {m + k} right)!left( {s - k} right)!}}} }
        - b^{,m} sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,s - k} b^{,k} } over {left( {m + k} right)!left( {s - k} right)!}}} } cr}
        $$
        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.






        share|cite|improve this answer














        We have that
        $$
        eqalign{
        & Gamma (m,a)Gamma (m,b) = Gamma (m)^{,2} Q(m,a)Q(m,b) = cr
        & = Gamma (m)^{,2} e^{, - left( {a + b} right)} sumlimits_{k = 0}^{m - 1} {{{a^{,k} } over {k!}}}
        sumlimits_{j = 0}^{m - 1} {{{b^{,j} } over {j!}}} cr}
        $$



        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
        $$
        eqalign{
        & sumlimits_{k = 0}^{m - 1} {{{a^{,k} } over {k!}}} sumlimits_{j = 0}^{m - 1} {{{b^{,j} } over {j!}}}
        = sumlimits_{k = 0}^{m - 1} {sumlimits_{j = 0}^{m - 1} {{{a^{,k} b^{,j} } over {k!j!}}} } = cr
        & = sumlimits_{s = 0}^{2m - 1} {sumlimits_{k = 0}^s {{{a^{,k} b^{,s - k} } over {k!left( {s - k} right)!}}} }
        - sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,m + k} b^{,s - k} } over {left( {m + k} right)!left( {s - k} right)!}}} }
        - sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,s - k} b^{,m + k} } over {left( {m + k} right)!left( {s - k} right)!}}} } = cr
        & = sumlimits_{s = 0}^{2m - 1} {{{left( {a + b} right)^{,s} } over {s!}}}
        - a^{,m} sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,k} b^{,s - k} } over {left( {m + k} right)!left( {s - k} right)!}}} }
        - b^{,m} sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,s - k} b^{,k} } over {left( {m + k} right)!left( {s - k} right)!}}} } cr}
        $$
        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 19 at 15:51

























        answered Nov 19 at 2:49









        G Cab

        17.9k31237




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