Define treta$^*$ function as $$ \tau(\alpha_1,\alpha_2,\alpha_3) = \iint_{0< x_1< x_2<1} x_1^{\alpha_1-1}(x_2-x_1)^{\alpha_2-1}(1-x_2)^{\alpha_3-1}\, d(x_1,x_2).\tag{1} $$ Similarly to the beta function, it can be shown that $$ \tau(\alpha_1,\alpha_2,\alpha_3) = \frac{\Gamma(\alpha_1)\Gamma(\alpha_2)\Gamma(\alpha_3)}{\Gamma(\alpha_1+\alpha_2+\alpha_3)}.\tag{2} $$

Now I am interested in anti-treta function $$ \tau_a(\alpha_1,\alpha_2,\alpha_3) = \iint_{0< \color{red}{x_2< x_1}<1} x_1^{\alpha_1-1}(x_1-x_2)^{\alpha_2-1}(1-x_2)^{\alpha_3-1} d(x_1,x_2).\tag{3} $$ This looks innocently similar to $(1)$, however, applying the technique used in deriving of $(2)$, I was able (if I didn't mess up) only to come to $$ \tau_a(\alpha_1,\alpha_2,\alpha_3) = \frac{1}{\Gamma(\alpha_1+\alpha_2+\alpha_3)} \int_0^\infty \Gamma(\alpha_1,x)\Gamma(\alpha_3,x)x^{\alpha_2-1}dx, $$ where $\Gamma(\alpha,x) = \int_x^\infty u^{\alpha-1} e^{-u}du$ denotes the incomplete gamma function.

Now I wonder why $(3)$ is so different from $(1)$ and whether there is a nice formula like $(2)$ for it, at least in some particular cases, like $\alpha_1=\alpha_2=\alpha_3$ or similar.

$^*$ I use the word "treta" since, on one hand, "gamma" is reserved (as well as "delta" for the fourth-order version of this object); on the other hand, "treta" relates to "beta" in a manner similar to that relating "trinomial" to "binomial". There also may be a standard terminology I don't know, I welcome corrections!

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    $\begingroup$ While I see the motivation for using 'treta', you might want to find something else if only because "treta function" is easily misinterpreted as a typo of "theta function" when reading the list of problems. $\endgroup$ – Semiclassical May 19 '16 at 15:17
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    $\begingroup$ Maybe the title should be changed to "Anti-treta (not to be confused with $\theta$)...", or just "Anti-treta (sic!)..." $\endgroup$ – Ivan Neretin May 19 '16 at 15:19
  • $\begingroup$ I think it is safe, since there is no "theta-functions" tag :) $\endgroup$ – zhoraster May 19 '16 at 17:20
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    $\begingroup$ @zhoraster You mean, besides this one? math.stackexchange.com/questions/tagged/theta-functions $\endgroup$ – Semiclassical May 19 '16 at 20:08
  • $\begingroup$ Yep, I meant that, exactly. $\endgroup$ – zhoraster May 19 '16 at 20:11

Why (3) is so different from (1) ?

Because if we rewrite the double integrals as iterated ones, for (1) we get $$\int_0^1 \left(1-x_2\right)^{\alpha_3-1}\left({\color{blue}{\int_0^{x_2}x_1^{\alpha_1-1}\left(x_2-x_1\right)^{\alpha_2-1}dx_1}}\right)dx_2,$$ and for (3) we find $$\int_0^1 x_1^{\alpha_1-1}\left({\color{red}{\int_0^{x_1} \left(1-x_2\right)^{\alpha_3-1}\left(x_1-x_2\right)^{\alpha_2-1}dx_2}}\right)dx_1.$$ The change of variables $x_1=x_2 t$ reduces the blue integral to beta function (more precisely, to $x_2^{\alpha_1+\alpha_2-1}B\left(\alpha_1,\alpha_2\right)$, which gives another beta function for the integral with respect to $x_2$). There is no obvious change of variables that would lead to similar simplifications in the red integral.

Whether there exists a nice formula like (2) ?

The red integral is in fact given by a hypergeometric function $$\color{red}{\frac{x_1^{\alpha_2}}{\alpha_2}{}_2F_1\left(1,1-\alpha_3;1+\alpha_2;x_1\right)}$$ Subsequent integration with respect to $x_1$ yields $$\tau_a\left(\alpha_1,\alpha_2,\alpha_3\right)=\frac{{}_3F_2\biggl[\begin{array}{c}1,\alpha_1+\alpha_2,1-\alpha_3\\ 1+\alpha_2,1+\alpha_1+\alpha_2\end{array};1\biggr]}{\alpha_2\left(\alpha_1+\alpha_2\right)}.$$ This does not seem to have simpler expression for general parameter values - however, putting some restrictions there is a huge number of cases where the answer does simplify, see e.g. Prudnikov-Brychkov-Marychev, Vol. III, Section 7.4.4. For example, $$\tau_a\left(\alpha_1,\alpha_2,1-\alpha_2\right)=\frac{\psi\left(\alpha_1+\alpha_2\right)-\psi\left(\alpha_2\right)}{\alpha_1},$$ where $\psi(z)$ denotes the digamma function.

  • $\begingroup$ This is pretty helpful, thanks. $\endgroup$ – zhoraster May 23 '16 at 11:49
  • $\begingroup$ Great answer! +1 $\endgroup$ – Yuriy S Aug 2 '16 at 18:40

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