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Just as the product rule can be generalised to the product of more than two functions, i.e. $$\frac{d}{dx} \left [ \prod_{i=1}^k f_i(x) \right ] = \sum_{i=1}^k \left(\frac{d}{dx} f_i(x) \prod_{j\ne i} f_j(x) \right) = \left( \prod_{i=1}^k f_i(x) \right) \left( \sum_{i=1}^k \frac{f'_i(x)}{f_i(x)} \right),$$ is there a way to generalise integration by parts to evaluate $$\int f_1(x) f_2(x) \cdots f_n(x) dx \qquad ?$$

For contextual purposes only, I'm trying to evaluate by hand

$$\int x \cosh(x+1)e^x\sin(x) dx \quad . $$

I know that I could let $u=x \cosh(x+1)$ and $v \ '=e^x\sin(x)$ but that would require integration by parts to be performed at least three times.

Is there a more-efficient way to evaluate this integral using the proposed 'generalisation'?

Thanks!

beep-boop
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    This doesn't use the generalization, but you could write both cosh and sin as a linear combination of exponentials and then break up the integral into three or four, each of which only require one integration by parts (or in fact, it's just the integral of $x e^{ax}$ with different values of $a$). – msteve Jul 25 '14 at 19:58
  • @msteve Thank you. That's a nice observation that I otherwise wouldn't have noticed. But I am looking for a more-general method not specific to this example. e.g. integrating the product of six functions, say, only one of which is expressible in the form $e^x$, I might not be so lucky! – beep-boop Jul 25 '14 at 20:00

2 Answers2

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The Wikipedia article on integration by parts gives you the generalization you're looking for:

$$\Bigl[ \prod_{i=1}^n u_i(x) \Bigr]_a^b = \sum_{j=1}^n \int_a^b \prod_{i\neq j}^n u_i(x) \, du_j(x),$$

where $u_i(x)$ are your $n$ functions of $x$ that are terms of the product that comprise your integrand.

John
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    And why does the product on LHS have limits? Should it just be $$\int_a^b \prod_{i=1}^n u_i(x)dx=...$$? – beep-boop Jul 25 '14 at 20:19
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    Actually, I take that back. It was correct as it was. One of the terms in the integration by parts formula is the product of all of the terms, without the integral. For IBP with two functions the form is $uv = \int udv + \int vdu.$ – John Jul 25 '14 at 20:27
  • Got it now. Thanks a lot! – beep-boop Jul 25 '14 at 20:32
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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?

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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