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"Classic" integration by parts has two functions ($u$ and $dv$; $\int udv = uv - \int vdu$). What if there is a product of $n$ functions? In other words, what's the solution to the following?

\begin{equation}\label{eq:tosolve} F(x)=\int_{a}^{b}\prod_{i=1}^{n}u_{i}(x)\:dx \end{equation}

Wikipedia's integration by parts article mentions this problem and offers an equation, but I think a few extra steps are necessary for a full solution. Wikipedia leaves us with the following product rule for $n$ functions: \begin{equation} \bigg(\prod_{i=1}^{n}u_{i}(x)\bigg)'=\sum_{j=1}^{n}u_{j}'(x)\prod_{i\neq j}^{n}u_{i}(x). \end{equation}

Integrating, this leads to

\begin{equation}\label{eq:onwiki} \bigg[\prod_{i=1}^{n}u_{i}(x)\bigg]_{a}^{b}=\sum_{j=1}^{n}\int_{a}^{b}u_{j}'(x)\prod_{i\neq j}^{n}u_{i}(x)\:dx \end{equation}

The object we want to solve for (the RHS of the first equation at the top) does not appear in the expression immediately above. How does one recover it?

ABC123
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    You can apply integration by parts taking $u = \prod_{i=1}^{n-1} u_i$ and $dv = u_n$ – Ben Grossmann Jun 05 '23 at 19:36
  • Could you complete the solution? I don't think I follow how to this leads back to $\int_{a}^{b}\prod_{i=1}^{n}u_{i}(x) dx$. – ABC123 Jun 05 '23 at 20:11
  • See my answer below. By the way, it doesn't make sense to write $F(x)=\int_{a}^{b}\prod_{i=1}^{n}u_{i}(x):dx$. Either you want the function (indefinite integral) $F(x)=\int\prod_{i=1}^{n}u_{i}(x):dx$ or the quantity (definite integral) $\int_{a}^{b}\prod_{i=1}^{n}u_{i}(x):dx$ – Ben Grossmann Jun 05 '23 at 20:45

2 Answers2

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Let $U_n$ denote an antiderivative of $u_n$. We have \begin{align} \int \prod_{i=1}^n u_i(x)\,dx &= \int \left(\prod_{i=1}^{n-1} u_i(x)\right)u_n(x)\,dx \\ & = \int \left(\prod_{i=1}^{n-1} u_i(x)\right)U_n'(x)\,dx \\ & = \left(\prod_{i=1}^{n-1} u_i(x)\right)U_n(x) - \int \frac d{dx}\left(\prod_{i=1}^{n-1} u_i(x)\right)U_n(x)\,dx \\ & = \left(\prod_{i=1}^{n-1} u_i(x)\right)U_n(x) - \int \left(\sum_{j=1}^{n-1}u_{j}'(x)\prod_{i\neq j}^{n-1}u_{i}(x)\right)U_n(x)\,dx. \end{align}

Ben Grossmann
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  • Thank you @Ben; I posted another approach which I think is the same. Do you agree it is the same? – ABC123 Jun 14 '23 at 03:42
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Here is another approach that I think returns the same result as the @Ben-Grossman's, but with a different route. Suppose we examine the sum over $n+1$ functions where the final $u_{n+1}(x)$ function equals $x$, such that $u'_{n+1}(x)=1$. We now have:

\begin{align} \bigg[\prod_{i=1}^{n+1}u_{i}(x)\bigg]_{a}^{b} &= \sum_{j=1}^{n+1}\int_{a}^{b}u_{j}'(x)\prod_{i\neq j}^{n+1}u_{i}(x)dx \\ & = \sum_{j=1}^{n}\int_{a}^{b}u_{j}'(x)\prod_{i\neq j}^{n}u_{i}(x)dx +\int_{a}^{b}\prod_{i=1}^{n}u_{i}(x)dx \end{align}

The $j=n+1$ term appears to be ``missing'' a $u'(x)$ term because this value is 1. A simple rearrangement yields:

\begin{align} \int_{a}^{b}\prod_{i=1}^{n}u_{i}(x)dx=\bigg[\prod_{i=1}^{n+1}u_{i}(x)\bigg]_{a}^{b}-\sum_{j=1}^{n}\int_{a}^{b}u_{j}'(x)\prod_{i\neq j}^{n}u_{i}(x)dx \end{align} Note the LHS is what we wanted to solve for.

ABC123
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