While playing with some PDE I came across a singular integral that looks something like
$$T(f\_1,f\_2,\ldots,f\_n)(x)=p.v.\int\_{-\infty}^\infty\frac{(f\_1(x)-f\_1(y))(f\_2(x)-f\_2(y))\cdots(f\_n(x)-f\_n(y))}{(x-y)^n}dy$$
where the functions are really nice, say Schwartz. Clearly if $n=1$ then $T$ is just a multiple of the Hilbert transform. Therefore I'm hoping one could write this as some kind of combination of $n$ Hilbert transforms, but I don't see how. The closest thing I've found is the Calderon commutators which differ in that there are $n+1$ factors of $(x-y)$ in the denominator. In that case, Coifman, McIntosh, Meyer were able to write it in terms of $H$ (except they did it in French).


Note: I posted this question on stackexchange first but no one had an answer.



A singular integral of several functions
Without loss of generality, $x=0$. 
Since $T(f\_1,\dots,f\_n)$ is $n$-linear in $(f\_1,\dots,f\_n)$, you can express each $f\_i$ as a mixture of harmonics $\exp(it\cdot)$, $t\in\mathbb R$ (with, in general, complex coefficients), so that $T(f\_1,\dots,f\_n)$ is expressed as a mixture of the values $T(\exp(it\_1\cdot),\dots,\exp(it\_n\cdot))$ with real $t\_1,\dots,t\_n$. 
In turn, to express $T(\exp(it\_1\cdot),\dots,\exp(it\_n\cdot))$, you can expand the product $\prod\_{j=1}^n(e^{it\_jx}-e^{it\_jy})=\prod\_{j=1}^n(1-e^{it\_jy})$ in the numerator of the integrand in $T(\exp(it\_1\cdot),\dots,\exp(it\_n\cdot))$ and then proceed as in Appendix A in <https://arxiv.org/abs/1603.07365>, based on Lemma A.1 there (cf. Proposition A.4 there). 


**Addendum:** 
What **[[Coifman et al.](http://www.jstor.org/stable/2007065?seq=1#page_scan_tab_contents)]** seem to be doing concerning their formula (12) and what can be done in your situation seems really simple, much simpler than what was suggested above. For further simplicity, assume that $n=2$. Up to a constant factor, the Hilbert transform $H$ of a smooth enough function $g$ (say with a compact support) is given by the formula 
$$(Hg)(x)=\int\frac{g(x+t)-g(x)}t\,dt$$
for real $x$; the integrals here are over $\mathbb R$ and understood in the principal value sense. So, differentiating in $x$ and then integrating by parts in $t$, one has 
$$(Hg)'(x)=\int\frac{g'(x+t)-g'(x)}t\,dt
=\int\frac{g(x+t)-g(x)-g'(x)t}{t^2}\,dt.$$
Letting now $g(y):=(f\_1(y)-f\_1(0))(f\_2(y)-f\_2(0))$, one has $g(0)=g'(0)=0$, whence 
$$T(f\_1,f\_2)(0)=(Hg)'(0).$$
The general case of any natural $n$ is similar. 


