My question is from the drivation from Slberg integral to Dyson integral in this paper:

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

Dyson integral:

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

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

$$ \int_0^1 \cdots \int_0^1 (t_1\cdots t_n)^{\zeta-1} f(t_1,\dots,t_n) \, d t_1 \cdots d t_n =( \frac{1}{2\sin \pi \zeta} )^n \int_{-\pi}^{\pi} \cdots \int_{-\pi}^{\pi} e^{i\zeta(\theta_1+\cdots+\theta_n)} f(- e^{i\theta_1} ,\dots, - e^{i\theta_n}) d\theta_1 \cdots d \theta_n, $$ which is valid for $f$ a Laurent polynomial and Re$(\zeta)$ 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 $$ M_n(a,b,\gamma) := \frac{1}{(2\pi)^n} \int_{-\pi}^{\pi} \cdots \int_{-\pi}^{\pi} \prod_{i=1}^n e^{\frac{1}{2}i\theta_i (a-b)} |{1+e^{i\theta_i}}|^{a+b} \\ \times \prod_{1 \le i < j \le n} |{e^{i \theta_i}-e^{i \theta_j}}|^{2\gamma} \, d\theta_1 \cdots d \theta_n. $$ 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 $a,b,\gamma \in C$ such that $\rm{Re} (a+b+1)>0,\rm{Re}(\gamma)>-\min\{1/n,\rm{Re}(a+b+1)/(n-1) $. For $a=b=0$ this is Dyson integral.

**I have some questions about this calculation:**

1) About $\alpha$, they identify $\alpha:=-b-(n-1)\gamma$. From my calculation, the term $-(n-1)\gamma$ in $\alpha$ is from $\prod_{1 \le i < j \le n} |t_i - t_j|^{2\gamma} $ when changing variables from $t_i$ to $e^{i\theta_i}$, i.e $$ t_i-t_j= e^{i\theta_i}-e^{i\theta_j}= |e^{i\theta_i}-e^{i\theta_j}| i e^{i\frac{\theta_i+\theta_j}{2}}$$ If the Vandermonde product term in Selberg integral is in the absolute value, there seems no such a factor $-(n-1)\gamma$ in $\alpha$ . Anything wrong with my calculation?

2) Is the the absolute value outside Vandermonde product necessary in Selberg integral? What the integral is if there is no the absolute value outside Vandermonde product?