Let $\gamma>0$. I would like a nice way to prove that $$\int_{\begin{array}{c} 0\leq s,t\leq1\\ s+t\leq1 \end{array}}t^{\gamma-1}s^{\gamma-1}(1-t-s)^{\gamma-1}dtds=\frac{\Gamma(\gamma)^3}{\Gamma(3\gamma)}.$$

Edit/Remark: zhoraster's answer along with induction allows one to prove the generalized identity when integrating over the simplex $$\int_{\begin{array}{c} 0\leq t_{i}\leq1\\ t_{1}+\cdots+t_{k}\leq1 \end{array}}t_{1}^{x_{1}-1}\cdots t_{k}^{x_{k}-1}\left(1-\sum_{i=1}^{k}t_{i}\right)^{x_{k+1}-1}dt_{1}\cdots dt_{k}=\frac{\Gamma(x_{1})\Gamma(x_{2})\cdots\Gamma(x_{k+1})}{\Gamma(x_{1}+\cdots+x_{k+1})} $$ for any $0<x_i<1$

This integral came about while I was reading through Balog and Friedlander's paper A Hybrid Theorem of Vinogradov and Piatetski-Shapiro. Let $0<\gamma<1$, and let$$P_{\gamma}=\left\{ p\ :\ p=\lfloor n^{1/\gamma}\rfloor\ \text{for some }n\right\} $$ be the set of Piatetski-Shapiro primes. While this set of primes is very sparse, it has many nice properties and it is known that a version of the Prime number theorem holds when $\gamma$ is not too small (see this paper of Heath-Brown) $$\sum_{\begin{array}{c} p\leq x\\ p\in P_{\gamma} \end{array}}1\sim\frac{x^{\gamma}}{\log x}. $$ In the paper of Balog and Friedlander, they prove that for $\gamma$ in some range close to $1$, we have $$\frac{1}{\gamma^{3}}\sum_{\begin{array}{c} p_{1}+p_{2}+p_{3}=N\\ p_{i}\in P_{\gamma} \end{array}}p_{1}^{1-\gamma}p_{2}^{1-\gamma}p_{3}^{1-\gamma}\log p_{1}\log p_{2}\log p_{3}\sim\frac{1}{2}\mathfrak{S}(N)N^{2}, $$ where $\mathfrak{S}(N)$ is a particular singular series. Note that this implies that every sufficiently large integer can be written as the sum of three Piatetski-Shapiro primes, which proves a variant of Goldbach's Ternary conjecture for all sufficient large integers. From this they deduce without proof that $$\sum_{\begin{array}{c} p_{1}+p_{2}+p_{3}=N\\ p_{i}\in P_{\gamma} \end{array}}1\sim\frac{1}{2}\frac{\mathfrak{S}(N)N^{3\gamma-1}}{\left(\log N\right)^{3}}\frac{\gamma^{3}\Gamma(\gamma)^{3}}{\Gamma(3\gamma)}.$$ When I applied partial summation, this deduction reduced to evaluating the integral written above.

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    $\begingroup$ Is the substitution $t =(1-s)u$ counted as a nice way? Anyway, I got $$\frac{\Gamma(2-\gamma)^3}{\Gamma(6-3\gamma)} $$ for your integral. Letting $\gamma \nearrow +\infty$ correctly reflects the fact that your integral diverges when $\gamma = 2$, which makes me doubt your claim... $\endgroup$ – Sangchul Lee Oct 22 '15 at 13:15
  • $\begingroup$ When you compare zhoraster's integral with your one, you will find that the exponents are all negated. That's why my answer is different. $\endgroup$ – Sangchul Lee Oct 22 '15 at 13:33
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    $\begingroup$ @SangchulLee: I just noticed that exactly as you posted the comment. (my mistake) I have edited things now. $\endgroup$ – Eric Naslund Oct 22 '15 at 13:34

Check the signs. $$ \int_{\substack{ 0\leq s,t\leq1\\ s+t\leq1}}t^{\gamma-1}s^{\gamma-1}(1-t-s)^{\gamma-1}dt\,ds= \int_0^1 t^{\gamma-1}\int_0^{1-t}s^{\gamma-1}(1-t-s)^{\gamma-1}ds\,dt \\ = \int_0^1 t^{\gamma-1}(1-t)^{2\gamma-1}\int_0^{1}u^{\gamma-1}(1-u)^{\gamma-1}du\,dt = B(\gamma,2\gamma)B(\gamma,\gamma) = \frac{\Gamma(2\gamma)\Gamma(\gamma)^3}{\Gamma(2\gamma)\Gamma(3\gamma)} = \frac{\Gamma(\gamma)^3}{\Gamma(3\gamma)}. $$

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  • $\begingroup$ Great, thanks! I thought I was missing simple substitution like this. $\endgroup$ – Eric Naslund Oct 22 '15 at 13:30
  • $\begingroup$ This substitution proves what looks like a $2$-dimensional beta function identity: $$\int_{\begin{array}{c} 0\leq s,t\leq1\\ s+t\leq1 \end{array}}t^{a-1}s^{b-1}(1-t-s)^{c-1}dtds=\frac{\Gamma(a)\Gamma(b)\Gamma(c)}{ \Gamma(a+b+c)} $$ $\endgroup$ – Eric Naslund Oct 22 '15 at 13:37
  • $\begingroup$ @EricNaslund, yes, it can be proved exactly as the usual beta function identity. $\endgroup$ – zhoraster Oct 22 '15 at 13:45
  • $\begingroup$ What is the usual way to prove the beta function identity? Are you suggesting writing $\Gamma(a)\Gamma(b)\Gamma(c)$ as an integral over $\mathbb{R}^3$? $\endgroup$ – Eric Naslund Oct 22 '15 at 13:49
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    $\begingroup$ @EricNaslund, yes, exactly. Integrate over $x+y+z = t$, getting $B(a,b,c) t^{a+b+c-1} e^{-t}$. Then integrate over $t$, which gives $B(a,b,c)\Gamma(a+b+c)$. $\endgroup$ – zhoraster Oct 22 '15 at 14:17

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