I'm attempting to read through Mehta's write-up of Selberg's proof of the formula for the Selberg integral formula given below:

$$ \begin{align} \operatorname{S}_n(\alpha,\ \beta,\ \gamma) & = \int_{[0,\ 1]^n}\prod_{i=1}^nx_i^{\alpha-1}(1-x_i)^{\beta-1}|\Delta(\vec x)|^{2\gamma}\text{d}\vec x \\ & = \prod_{j=0}^{n-1}\frac{\Gamma(1+(1+j)\gamma)\Gamma(\alpha+j\gamma)\Gamma(\beta+j\gamma)}{\Gamma(1+\gamma)\Gamma(\alpha + \beta + (n+j-1)\gamma)} \end{align} $$

The proof is from Mehta's book "Random Matrices" (Third Edition) on Page 311.

His proof, to me, uses a lot of hand-wavey arguments to derive bounds on the terms $j_m$ in the ordered partitions of $2\gamma{n\choose 2}$. He arrives at $\gamma(m-1)\le j_m\le \gamma(n+m-2)$.

To start, assume $\gamma$ is an integer. The first problem I have with the proof is that he expands $|\Delta(\vec x)|^{2\gamma}$ as a series: $$ \Delta(\vec x)^{2\gamma} = \sum_{\substack{j_1+\cdots+j_n=2\gamma{n\choose 2} \\ j_i\le j_{i+1}}}c(j_1,\cdots,\ j_n)x^{j_1}\cdots x^{j_n} $$ where the $c$'s are just integers. My problem with this is that he orders the $j_i$'s. If you do this for even a simple example ($n = 2$,\ $\gamma = 1$) you see that this can't possibly be true, unless I'm missing something brutally obvious: $$ \Delta_2(x_1,\ x_2)^2 = x_2^2 - 2x_1x_2 + x_1^2 \\ \sum_{\substack{j_1+j_2 = 2 \\ j_1\le j_2}}c(j_1,\ j_2)x_1^{j_1}x_2^{j_2} = c(0,\ 2)x_2^2 + c(1,\ 1)x_1x_2 $$ and clearly there are no integers $c(0,\ 2)$ and $c(1,\ 1)$ for which $c(0,\ 2)x_2^2 + c(1,\ 1)x_1x_2 = x_2^2 - 2x_1x_2 + x_1^2$.

He goes on to use this ordering of the $j_i$'s to prove other various things, but the whole point of this is to arrive at the bounds on $j_m$ given at the beginning of the question.

Now, nowhere else in the proof is it needed that $j_i\le j_{i+1}$, so I have two questions:

1. How can we justify ordering the $j_i$'s?

2. How can we arrive at the bounds $\gamma(m-1)\le j_m\le \gamma(n+m-2)$ more rigorously?

Edit: I realize now that there's a problem, and that's if we can't justify the ordering of the $j_i$'s then there aren't any bounds on $j_m$ (other than $0$ and $2\gamma{n\choose 2}$), so answering 1 immediately answers 2, and refuting 1 would, as far as I can tell, invalidate Selberg's proof of this formula. I'm confident the formula itself holds, since I've also read through Aomoto's proof and it seems much more solid, but I'm interested in seeing how, if possible, we could justify Selberg's original proof.

Disclaimer: I'm a first year at my university who was fortunate enough to attend pure math courses. I can read and understand generally higher level math, and more often than not understand the logic behind things, but sometimes I miss simple things. If this is a really simple question, please forgive me.


1 Answer 1


See Greg Anderson's two page proof.

  • $\begingroup$ This is impressive, thank you! I'll upvote you for the interesting result, but I'm still interested in a response regarding my questions above, so I'll resist accepting this as an answer yet. Thank you nonetheless. $\endgroup$ Dec 14, 2016 at 2:15
  • $\begingroup$ Did you answer this because the proof itself contains details which would answer my questions? If so, I'm sorry but I must have missed them, the proof went over my head at times. $\endgroup$ Dec 14, 2016 at 15:58

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