# From Selberg integral to Dyson integral

My question is about the derivation from Selberg integral to Dyson integral in this paper:

Selberg integral : $$S_n(\alpha,\beta,\gamma) := \int_0 ^1 \cdots \int_0 ^1 \prod_{j=1}^n t_j^{\alpha-1}(1-t_j)^{\beta-1} \prod_{1\leqslant j<k\leqslant n} \lvert {t_j - t_k} \rvert ^{2\gamma} dt_1\cdots d t_n = \prod_{j=0}^{n-1} \frac{\Gamma (\alpha+j\gamma) \Gamma(\beta+j\gamma)\Gamma(1+(j+1)\gamma)} {\Gamma(\alpha+\beta+(n+j-1)\gamma)\Gamma(1+\gamma)}.$$

Dyson integral:

$$C_n(\gamma) := \frac{1}{(2\pi)^n} \int_{-\pi}^{\pi} \cdots \int_{-\pi}^{\pi} \prod_{1 \leqslant j<k \leqslant n} |e^{i \theta_j} - e^{i \theta_k}|^{2\gamma} d \theta_1 \cdots d\theta_n$$

Due to R. Askey, the Selberg integral can be used to express the Dyson integral directly.
Askey's observation is based on the easily established general identity:

$$\int_0^1 \cdots \int_0^1 (t_1\cdots t_n)^{z-1} f(t_1,\dots,t_n) \, d t_1 \cdots d t_n \\ =\left( \frac{1}{2\sin \pi z} \right)^n \int_{-\pi}^{\pi} \cdots \int_{-\pi}^{\pi} e^{iz(\theta_1+\cdots+\theta_n)} f(- e^{i\theta_1} ,\dots, - e^{i\theta_n}) \, d\theta_1 \cdots d \theta_n,$$ which is valid for $f$ a Laurent polynomial and $\Re(z)$ large enough so that the left-hand side exists. Applying the identity to the Selberg integral with $\beta$ a positive integer and $\gamma$ a nonnegative integer shows that $$S_n(\alpha,\beta,\gamma)=(-1)^{n+\binom{n}{2}\gamma} \Bigl(\frac{\pi}{\sin \pi b}\Bigr)^n M_n(a,b,\gamma),$$ where $\alpha:=-b-(n-1)\gamma,\ \ \beta:=a+b+1$ and $$M_n(a,b,\gamma) := \frac{1}{(2\pi)^n} \int_{-\pi}^{\pi} \cdots \int_{-\pi}^{\pi} \prod_{j=1}^n e^{\frac{1}{2}i\theta_j (a-b)} |{1+e^{i\theta_j}}|^{a+b} \\ \times \prod_{1 \leqslant j<k \leqslant n} |{e^{i \theta_j}-e^{i \theta_k}}|^{2\gamma} \, d\theta_1 \cdots d \theta_n.$$ From the Selberg integral, the reflection formula and finally Carlson's theorem, it follows that $$M_n(a,b,\gamma) = \prod_{j=0}^{n-1} \frac{\Gamma (1+a+b+j\gamma) \Gamma(1+(j+1)\gamma)} {\Gamma (1+a+j\gamma)\Gamma (1+b+j\gamma) \Gamma (1+\gamma)},$$ for $a,b,\gamma \in\mathbb C$ such that $\Re (a+b+1)>0,\Re(\gamma)>-\min\{\frac1n, \frac{\Re(a+b+1)}{n-1}\}$.
For $a=b=0$ this is the desired Dyson integral.

1) About $\alpha$, they identify $\alpha:=-b-(n-1)\gamma$. From my calculation, the term $-(n-1)\gamma$ in $\alpha$ comes from $\prod_{1 \leqslant j<k \leqslant n} |t_j - t_k|^{2\gamma}$ when changing variables from $t_r$ to $e^{i\theta_r}$, i.e. $$t_j-t_k= e^{i\theta_j}-e^{i\theta_k}= |e^{i\theta_j}-e^{i\theta_k}| i e^{i\frac{\theta_j+\theta_k}{2}}.$$ If the Vandermonde product term in the Selberg integral is in the absolute value, there seems no such term $-(n-1)\gamma$ in $\alpha$. Anything wrong with my calculation?
• if $\gamma$ is non-negative integer, you should not care on absolute value, yes? Jun 29, 2017 at 12:51