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Recall that a $(r,s)$-shuffle $\sigma$ is a permutation of $r+s$ letters such that $\sigma^{-1}(1) < \cdots< \sigma^{-1}(r)$ and $\sigma^{-1}(r+1) < \cdots < \sigma^{-1}(r+s)$.

If $f_1(t), f_2(t),\ldots,f_r(t)$ are piecewise continuous functions, define inductively

$\int_a^bf_1\;dt\cdots f_r\;dt = \int_a^b \left(\int_a^t f_1\;dt \cdots f_{r-1}\;dt\right)f_r(t)\;dt.$

Can I have some help in showing the following formula:

$\left(\int_a^bf_1\;dt\cdots f_r\;dt\right)\left(\int_a^bf_{r+1}\;dt\cdots f_{r+s}\;dt\right)=\sum \int_a^bf_{\sigma(1)}\;dt\cdots f_{\sigma(r+s)}\;dt$ where the sum in the right hand side runs through all $(r,s)$ shuffles.

Thanks.

1 Answers 1

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We have

$\int_a^bf_1\,\mathrm dt\cdots f_r\,\mathrm dt=\int_{a\le t_1\le\cdots\le t_r\le b} f_1(t_1)\cdots f_r(t_r)\,\mathrm dt_1\ldots\mathrm dt_r\;,$

and thus

$ \begin{align} &\left(\int_a^bf_1\,\mathrm dt\cdots f_r\,\mathrm dt\right)\left(\int_a^bf_{r+1}\,\mathrm dt\cdots f_{r+s}\,\mathrm dt\right) \\ =& \left(\int_{a\le t_1\le\cdots\le t_r\le b} f_1(t_1)\cdots f_r(t_r)\,\mathrm dt_1\ldots\mathrm dt_r\right) \\ & \cdot\left(\int_{a\le t_{r+1}\le\cdots\le t_{r+s}\le b} f_{r+1}(t_{r+1})\cdots f_{r+s}(t_{r+s})\,\mathrm dt_{r+1}\ldots\mathrm dt_{r+s}\right) \\ =& \int_{{\scriptstyle a\le t_1\le\cdots\le t_r\le b}\atop{\scriptstyle a\le t_{r+1}\le\cdots\le t_{r+s}\le b}} f_1(t_1)\cdots f_{r+s}(t_{r+s})\,\mathrm dt_1\ldots\mathrm dt_{r+s} \\ =& \sum \int_{a\le t_1\le\cdots\le t_{r+s}\le b} f_{\sigma(1)}(t_1)\cdots f_{\sigma(r+s)}(t_{r+s})\,\mathrm dt_1\ldots\mathrm dt_{r+s} \\ =& \sum \int_a^bf_{\sigma(1)}\,\mathrm dt\cdots f_{\sigma(r+s)}\,\mathrm dt\;. \end{align} $

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    @Manuel: Not the possible ways of dividing, just one way, namely according to the order of the coordinates. The product of the integrals over the simplices is the integral over the product of the simplices, which is the sum over the integrals over all the simplicial parts of the product of the simplices, one part for each ordering of the coordinates.2012-08-05