Variation on Chen-iterated integrals Standard Chen-iterated integrals are for instance of the form
$$
\int\limits_a^b \!\! dx_1\!\!
\int\limits_a^{x_1} \!\! dx_2 \!\!
\int\limits_a^{x_2} \!\! dx_3 \cdots \!\!
\int\limits_a^{x_{n-1}} \!\!\! dx_n \;\;
f(x_1)\,f(x_2)\cdots f(x_n)
$$
which, if $\,f\,$ is not an operator (or an abelian operator), can be simplified into
$$
\frac{1}{n!}\left( \int_a^b \!\! dx \,\; f(x) \right)^n
$$
Now, in my calculations I stumbled upon a variation of this form, where only the odd or even factors of $\,f\,$ are kept. In other words, terms of the form:
$$
\int\limits_a^b \!\! dx_1\!\!
\int\limits_a^{x_1} \!\! dx_2 \!\!
\int\limits_a^{x_2} \!\! dx_3 \cdots \!\!
\int\limits_a^{x_{2n-1}} \!\!\! dx_{2n} \;\;
f(x_1)\,f(x_3)\cdots f(x_{2n-1})
$$
and
$$
\int\limits_a^b \!\! dx_1\!\!
\int\limits_a^{x_1} \!\! dx_2 \!\!
\int\limits_a^{x_2} \!\! dx_3 \cdots \!\!
\int\limits_a^{x_{2n-1}} \!\!\! dx_{2n} \;\;
f(x_2)\,f(x_4)\cdots f(x_{2n})
$$
I struggle to simplify these. Ideally, I'd like to factorise the integrals somehow (as these are terms in an infinite series; factorising them might lead to a full form expression). Any ideas?
 A: Let
$$
I_n(b)=\int\limits_a^b dx_1
\int\limits_a^{x_1} dx_2 
\int\limits_a^{x_2} dx_3 \cdots
\int\limits_a^{x_{n-1}} dx_n \;\;
f(x_1)\,f(x_2)\cdots f(x_n).
$$
We use the Mathematical Induction to show
$$ I_n(b)=\frac{1}{n!}I_1^n(b).\tag{1} $$
Note $f(x)dx=dI_1(x)$. For $n=2$, then
\begin{eqnarray}
I_2(b)&=&\int\limits_a^b dx_1
\int\limits_a^{x_1} dx_2 f(x_1)\,f(x_2)\\
&=&\int\limits_a^b\bigg[\int\limits_a^{x_1}f(x_2)dx_2\bigg]f(x_1)dx_1\\
&=&\int\limits_a^bI_1(x_1)I_1'(x)dx_1\\
&=&\frac12I_1^2(b).
\end{eqnarray}
Suppose for $k=n-1$, (1) holds, namely
$$
I_{n-1}(b)=\int\limits_a^b dx_1
\int\limits_a^{x_1} dx_2 
\int\limits_a^{x_2} dx_3 \cdots
\int\limits_a^{x_{n-2}} dx_{n-1} \;\;
f(x_1)\,f(x_2)\cdots f(x_{n-1})=\frac{1}{(n-1)!}I_1^{n-1}(b).
$$
Then 
\begin{eqnarray}
I_n(b)&=&\int\limits_a^b dx_1
\int\limits_a^{x_1} dx_2 
\int\limits_a^{x_2} dx_3 \cdots
\int\limits_a^{x_{n-1}} dx_n \;\;
f(x_1)\,f(x_2)\cdots f(x_n)\\
&=&\int\limits_a^bI_{n-1}(x_1)f(x_1)dx_1\\
&=&\frac{1}{(n-1)!}\int\limits_a^bI_1^{n-1}(x_1)f(x_1)dx_1\\
&=&\frac{1}{n!}I(b)_1^{n}.
\end{eqnarray}
Namely for $k=n$, (1) hold.
