y_{m+1}), (x_{m+2}, y_{m+2}) \right)\\
 &\to& a_{m} ^{x} (x_{m+1},y_{m+2}) 1_{[x_{m+2}, 1] } (x_{m+1} )1_{[0,y_{m+2}] } (y_{m+1} )  1_{[0, x_{m+1} ] \times [0, y_{m+2}]} ^{\otimes m}\left( (x_{1}, y_{1}), (x_{2}, y_{2}),\ldots , (x_{m}, y_{m}) \right)\end{aligned}$$ with respect to the $m+2$ variables $$(x_{1}, y_{1}), (x_{2}, y_{2}),\ldots , (x_{m+}, y_{m+2}).$$ The above function is already symmetric in its first $m$ variables. Noting that for any function $f(a_{1}, a_{2},\ldots , a_{m}, x,y)$ which is symmetric with respect to $a_{1}, .., a_{m}$, its symmetrization with respect to its all $m+2$ variables is given by $$\begin{aligned}
\label{sim2}
\tilde{f}(a_{1},.., a_{m+2})= \frac{1}{(m+1)(m+2)} \sum_{k,l=1; k\not=l}^{m+2}f(a_{1},..,\hat{a_{k}}, ..,\hat{a_{l}}, ..a_{m+2}, a_{k}, a_{l})\end{aligned}$$ (the notation $\hat{a}$ means that the variable $a$ is missing above) we get, for every $x\in \mathbb{R}$, $$\begin{aligned}
\xi(x)
&=&
\sum_{m\geq 0}\frac{1}{(m+1)(m+2)}I_{m+2}\sum_{k,l=1; k\not= l} ^{m+2} a_{m} ^{x} (x_{k}, y_{l}) 1_{[0, x_{k}]\times [0, y_{l} ] } (x_{l}, y_{k})\nonumber \\
  &&\times 1_{[0, x_{k} ] \times [0, y_{l}]} ^{\otimes m}\left( (x_{1}, y_{1}), (x_{2}, y_{2}),\ldots , \overline{ (x_{k}, y_{k}) }, \overline{ (x_{l}, y_{l}) },  \ldots ,(x_{m+2}, y_{m+2}) \right) \nonumber \\
  &=& \sum_{m\geq 0} I_{m+2}\frac{1}{(m+1)(m+2)}\sum_{k,l=1; k\not= l} ^{m+2} a_{m} ^{x} (x_{k}, y_{l}) \nonumber\\
  &&1_{[0,x_{k}]}^{\otimes m+1} (x_{1}, .., \overline{x_{k}},.., x_{m+2} )1_{[0,y_{l}] }^{\otimes m+1} (y_{1},.., \overline{y_{l}},.., y_{m+2} )\label{xi-sim}\end{aligned}$$ where $\overline{ (x_{k}, y_{k}) }$ means, as above, that the variable 