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Jan 13, 2022 at 15:51 history bounty ended Marcel
Jan 10, 2022 at 18:35 vote accept Marcel
Jan 7, 2022 at 21:15 history edited Carlo Beenakker CC BY-SA 4.0
added 57 characters in body
Jan 7, 2022 at 21:13 comment added Carlo Beenakker concerning the $V=-a\log(1-X)$ potential: to be able to use the RH approach in the large-$N$ limit the coefficient $a=\alpha N$ must increase linearly with $N$; the calculation then proceeds along the same lines as for the quadratic potential, it's just a more complicated RH equation that one has to solve, in particular the support of the solution will be an $\alpha$-dependent sub-interval $(0,c)$ of $(0,1)$, with a $1/\sqrt{c-x}$ singularity at the upper limit.
Jan 7, 2022 at 17:08 comment added Carlo Beenakker I fixed the error in the coefficient, the asymptotic result now agrees with the Selberg expression in the large-$N$ limit.
Jan 7, 2022 at 17:07 history edited Carlo Beenakker CC BY-SA 4.0
remaining error fixed
Jan 7, 2022 at 16:08 history edited Carlo Beenakker CC BY-SA 4.0
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Jan 7, 2022 at 15:39 history edited Carlo Beenakker CC BY-SA 4.0
added 42 characters in body
Jan 7, 2022 at 13:34 comment added Marcel Ok, I still wonder if the case $V=\log(1-X)$ can be done, but thank you for your answer.
Jan 7, 2022 at 13:26 comment added Carlo Beenakker I have added an explicit comparison between the RH asymptotic result and the large-N limit of the Selberg integral for the case of a quadratic $V$ (where I know how to solve the RH equation). There is probably still an error in my calculation, and I'm unable to carry out the large-N limit of the product of Gamma functions, so I have compared them numerically.
Jan 7, 2022 at 13:24 history edited Carlo Beenakker CC BY-SA 4.0
added 1288 characters in body
Jan 7, 2022 at 11:49 comment added Marcel If you consider $0<X<1$ from the start, you can take $V$ to be the well defined function $a\log(1-X)$. Although there is no direct connection between the two techniques, the question is whether R-H is capable of reproducing the Selberg result, at least to leading orders.
Jan 7, 2022 at 11:27 history edited Carlo Beenakker CC BY-SA 4.0
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Jan 7, 2022 at 11:20 history answered Carlo Beenakker CC BY-SA 4.0