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.