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Mar 14, 2020 at 5:07 comment added Yemon Choi +1 (10 years after the fact) for the convolution semigroup observation - this semigroup property seems to be quite useful for some work I've recently been doing on positive definite functions on the "ax+b group" of the reals.
Mar 12, 2020 at 14:02 comment added Novice C @PietroMajer while I also cannot follow how one arrives at a binomial coefficient, it hardly corroborates that the Beta function implies Gamma is superior. The point still stands that a modified Beta function could easily be written in terms of the Pi function—in parallel to the example you have given. If you want to argue the Beta function we have is more natural than the modified Beta function that corresponds to Pi you can, but just reduces to a weaker argument in this case of saying Gamma is more natural than Pi.
Jan 30, 2016 at 18:55 comment added Pietro Majer @Stephen Montgomery-Smith. In fact the formula you wrote is wrong, corroborating my claim that $\Gamma(t)\Gamma(s)/\Gamma(t+s)$ is easier to deal with ;)
Jul 16, 2015 at 1:53 comment added Tom Copeland $\int_0^z \frac{x^{t-1}}{(t-1)!} \frac{(z-x)^{s-1}}{(s-1)!} \, dx = D_{z}^{-s} \frac{z^{t-1}}{(t-1)!} = \frac{z^{s+t-1}}{(s+t-1)!}$ is easy to remember if one can remember to associate $s=1$ with integration and $s=0$ with the identity or Dirac delta function.
Jul 16, 2015 at 0:09 history edited Michael Hardy CC BY-SA 3.0
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Jul 15, 2015 at 21:00 comment added Stephen Montgomery-Smith You would define $\tilde B(t,s) = \int_0^1 x^t (1-x)^s \, dx$. Then $\tilde B(m,n) = 1\big/ \binom{m+n}m$.
May 26, 2010 at 7:29 history edited Pietro Majer CC BY-SA 2.5
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May 26, 2010 at 7:23 history answered Pietro Majer CC BY-SA 2.5