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$$\mathrm {PV}\int_{\mathbb{R}^n} \frac{e^{i(b_1z_1+\cdots+b_nz_n)}}{\prod_{j<k}(z_k-z_j)}dz=?$$ Other than integrate this term by term? It is in fact the Fourier transform of the inverse Vandermonde determinant.

I also found this question: Fourier transform of the inverse of Vandermonde determinant which is asking for the $n=3$ case.

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    $\begingroup$ @Clayton I forgot to mention that it is a real integral. $\endgroup$ – Jiyuan Zhang Mar 6 '19 at 3:02
  • $\begingroup$ What do you mean with a real integral ? $F(z)=\frac{e^{i(b_1z_1+\cdots+b_nz_n)}}{\prod_{j<k}(z_k-z_j)}$ is holomorphic in several variables on simply connected domains $U$ where the denominator doesn't vanish, there for each $z_j$ there is an holomorphic function such that $F = \partial_{z_j} f_j$ and $\int_a^b F(z)dz_j= f_j(b)-f_j(a)$ doesn't depend on the path. $\endgroup$ – reuns Mar 6 '19 at 3:18
  • $\begingroup$ @reuns all $z_i$ are real variables.This integral is the principal value integral. $\endgroup$ – Jiyuan Zhang Mar 6 '19 at 4:26
  • $\begingroup$ @reuns $\int_{x_1\ne x_2}\frac{e^{ib_1x_1+ib_2x_2}}{x_2-x_1}\mathrm dx_1\mathrm d x_2$. The support is when the Vandermonde determinant is non-zero. $\endgroup$ – Jiyuan Zhang Mar 6 '19 at 4:34
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So finally you meant the Fourier transform of the distribution $$F(x) = PV(\frac{1}{\prod_{j < k} (x_k-x_j)})$$ that is $$\hat{F}(\omega) = \int_{\mathbb{R}^n} F(x) e^{-i \langle \omega,x \rangle}d^n x$$ Then $$1=F(x)\prod_{j < k} (x_k-x_j)$$

implies $$(2\pi)^n i^{n(n-1)/2} \delta(\omega)=\prod_{j < k}(\partial_j-\partial_k) \widehat{F}(\omega) $$

so that, with $\ast$ the convolution and $T_{j,k}$ the Dirac delta distribution on the subspace orthogonal to $x_j-x_k$ $$\widehat{F}(\omega) = (2\pi)^n i^{n(n-1)/2} \quad T_{1,2} 1_{x_1-x_2 > 0} \ast \ldots \ast T_{j,k} 1_{x_j-x_k > 0} \ast \ldots \ast T_{n-1,n} 1_{x_{n-1}-x_n > 0} + P(\omega)$$

for some polynomial $P$ such that $(\prod_{j < k}(\partial_j-\partial_k))P = 0$, and we may deduce the coefficients of $P$ from symmetry considerations.

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