Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integral over standard n-simplex

  1. Sep 10, 2010 #1
    Given a http://en.wikipedia.org/wiki/Simplex#The_standard_simplex" in [itex]\mathbb{R}^n[/itex]:
    [tex]\Omega = \left\{ {{\textbf{y}}:\,\,\,\,{y_i} \ge 0,\,\,\,\,1 = \sum\limits_{i = 1}^n {{y_i}} } \right\}[/tex]
    where n is a positive integer, and a vector [itex]\textbf{a}[/itex] with n elements [itex]a_i>1[/itex],

    I need to evaluate the integral

    (1) [tex]G({\textbf{a}},m,s,t) = \int\limits_{\textbf{y} \in \Omega } {{{\left( {\sum\limits_{i = 1}^m {{y_i}} } \right)}^s}{{\left( {\sum\limits_{i = m}^n {{y_i}} } \right)}^t}\left( {\prod\limits_{i = 1}^n {{y_i}^{{a_i} - 1}} } \right)dy} [/tex]

    where t and s are real non-negative numbers, and m is an integer between 1 and n. The integral is taken over the standard n-1 simplex [itex]\Omega[/itex] with respect to [itex]dy = dy_1 dy_2...dy_n[/itex].



    Similar integrals that have well-known solutions are:

    (2) [tex]\int\limits_{y \in \Omega } {\prod\limits_{i = 1}^n {{y_i}^{{a_i} - 1}} dy} = \frac{{\prod\limits_{i = 1}^n {\Gamma ({a_i})} }}{{\Gamma \left( {\sum\limits_{i = 1}^n {{a_i}} } \right)}}[/tex]

    and

    (3) [tex]\int\limits_{y \in \Omega } {{{\left( {\sum\limits_{i = 1}^m {{y_i}} } \right)}^s}{{\left( {\sum\limits_{i = m + 1}^n {{y_i}} } \right)}^t}\left( {\prod\limits_{i = 1}^n {{y_i}^{{a_i} - 1}} } \right)dy} = \frac{{\prod\limits_{i = 1}^n {\Gamma ({a_i})} }}{{\Gamma \left( {s + t + \sum\limits_{i = 1}^n {{a_i}} } \right)}}\frac{{\Gamma \left( {s + \sum\limits_{i = 1}^m {{a_i}} } \right)}}{{\Gamma \left( {\sum\limits_{i = 1}^m {{a_i}} } \right)}}\frac{{\Gamma \left( {t + \sum\limits_{i = m + 1}^n {{a_i}} } \right)}}{{\Gamma \left( {\sum\limits_{i = m + 1}^n {{a_i}} } \right)}}[/tex]

    where [itex]\Gamma[/itex] is the Gamma function.

    Note that the only difference between (1) and (3) is that the second sum in (3) goes from m+1 to n instead of from m to n. This difference is however important, surely making the solution to (1) more complicated than that of (3).
     
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. Sep 12, 2010 #2
    If it is of any help, I am in particular interested in [itex]\partial^2/\partial s \partial t[/itex] of [itex]\ln \left( {G({\bf{a}},m,s,t)} \right)[/itex] when both t and s equal zero, that is, I want to calculate

    [tex]{{{\left. {\frac{\partial }{{\partial s}}} \right|}_{s = 0}}{{\left. {\frac{\partial }{{\partial t}}} \right|}_{t = 0}}\ln \left( {G({\bf{a}},m,s,t)} \right)}[/tex]
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook