Motivation: Consider a distribution $\frac{1}{x+i\epsilon}$ with $\epsilon \to 0^+$. For a test function $f\in \mathcal S(\mathbb R)$, the integral $$\lim_{\epsilon\to 0^+} \int_{-\infty}^\infty \frac{1}{x+i\epsilon} f(x) dx$$ is well-defined.
More generally, the distribution $\frac{1}{(x+i\epsilon)^\alpha}$ is well-defined for $\alpha>0$. If $\alpha$ is an integer, integration by parts gives (up to prefactor) $$\int_{-\infty}^\infty \frac{1}{(x+i\epsilon)^\alpha} f(x) dx \sim (-1)^{\alpha-1} \int_{-\infty}^\infty \frac{1}{x+i\epsilon} f^{(\alpha-1)}(x) dx,$$ hence reduces to the previous case. If $\alpha$ is not an integer, we have $$\int_{-\infty}^\infty \frac{1}{(x+i\epsilon)^\alpha} f(x) dx \sim (-1)^{\lfloor\alpha\rfloor} \int_{-\infty}^\infty \frac{1}{(x+i\epsilon)^{\alpha-\lfloor\alpha\rfloor}} f^{(\lfloor\alpha\rfloor)}(x) dx,$$ where $\lfloor \cdots \rfloor$ is the floor function. Since $0<\alpha-\lfloor\alpha\rfloor<1$, the integral is well-defined.
I want to consider a product of such distributions.
Question: Consider $\mathbb R^n$ with coordinates $(x_1,\ldots, x_n)$. Given $\alpha_{jk}>0$ for each $1\leq j<k \leq n$, is the distribution $$\lim_{\epsilon_1 \to 0^+} \lim_{\epsilon_{2} \to 0^+} \cdots \lim_{\epsilon_{n} \to 0^+} \prod_{1\leq j<k\leq n} \frac{1}{(\tilde x_j - \tilde x_k)^{\alpha_{jk}}}$$ well-defined? Here, $\tilde x_j = x_j + i\epsilon_j$ for each $j$ and assume $\epsilon_1>\epsilon_2>\cdots \epsilon_n$. In other words, for $f\in \mathcal S(\mathbb R^n)$, is the integral $$\lim_{\epsilon_1 \to 0^+} \lim_{\epsilon_{2} \to 0^+} \cdots \lim_{\epsilon_{n} \to 0^+} \int_{\mathbb R^n} \frac{1}{(\tilde x_j - \tilde x_k)^{\alpha_{jk}}} f(x_1,\ldots, x_n) dx_1\cdots dx_n \tag{$*$}$$ well-defined?
Some thoughts: If $f$ can be analytically continued on a strip $\{(x_1,\ldots, x_n): 0<\Im x_j <\beta \textrm{ for each $j$} \}$ for some $\beta>0$, then I think ($*$) is well-defined. In this case, we use the Cauchy's theorem to shift the contour as:
(The contours are vertically shifted so that the contour for $x_1$ has the largest imaginary part, and those for $x_n$ has the smallest. The shifts are possible since $\epsilon_1>\cdots >\epsilon_n$. Each of the contours are separated by finite distances.) Now, the singularity for coincident point $x_j=x_k$ is tamed by "imaginary separation" (hence the limit $\epsilon_j \to 0^+$ can be safely taken). Is my reasoning correct? What about Schwartz function that is not-analytically continuable? Is there a more general class of function such that ($*$) is well-defined? Is there any general theory related to this problem?
I remark that this problem is motivated from physics.