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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?

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  • $\begingroup$ I guess the "odd one" is $$ {\left(b - a\right)^{n} \over \left(2n\right)!}\,\left[\int_{a}^{b}\mathrm{f}\left(x\right)\,\mathrm{d}x\right]^{n} $$ I'm still checking it$\ldots$ $\endgroup$ Jun 19, 2017 at 4:38

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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.

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  • $\begingroup$ Thanks, this I know. But I'm looking for a similar relation with only odd or only even iterations of f.. $\endgroup$ May 31, 2017 at 21:33

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