# distributive property of integration

This is a very basic question about how integrals distribute over multiple variables. Suppose one has functions $f(x_1)$, $g(x_2)$, and $h(x_3)$ with antiderivatives $F(x_1)$, $G(x_2)$, and $H(x_3)$. Which of the following two expressions, if either, concerning $\iiint \mathrm{d}x_1 \mathrm{d}x_2 \mathrm{d}x_3 \left( f(x_1) + g(x_2) + h(x_3)\right)$ is correct?

\begin{align} \text{(1) } &\iiint \mathrm{d}x_1 \mathrm{d}x_2 \mathrm{d}x_3 \left( f(x_1) + g(x_2) + h(x_3)\right) \\ = &\iiint \mathrm{d}x_1 \mathrm{d}x_2 \mathrm{d}x_3 f(x_1) + \iiint \mathrm{d}x_1 \mathrm{d}x_2 \mathrm{d}x_3g(x_2) + \iiint \mathrm{d}x_1 \mathrm{d}x_2 \mathrm{d}x_3h(x_3) \\ = &F(x_1) \cdot x_2 \cdot x_3 + G(x_2) \cdot x_1 \cdot x_3 + H(x_3) \cdot x_1 \cdot x_2 + C \end{align}

\begin{align} \text{(2) } &\iiint \mathrm{d}x_1 \mathrm{d}x_2 \mathrm{d}x_3 \left( f(x_1) + g(x_2) + h(x_3)\right) \\ = &\int \mathrm{d}x_1 f(x_1) + \int \mathrm{d}x_2 g(x_2) + \int \mathrm{d}x_3 h(x_3) \\ = &F(x_1) + G(x_2) + H(x_3) + C' \end{align}

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Make it simple: which one is correct if $f(x_1)=1$, $g(x_2)=0$, and $h(x_3)=0$ and you are integrating over $[0,2]\times[0,2]\times[0,2]$? Is it $x_1x_2x_3$, or is it $x_1$? –  Arturo Magidin Mar 4 '12 at 3:55
@Arturo It seems we had similar thoughts. –  Alex Becker Mar 4 '12 at 3:57
@AlexBecker: No doubt a case of "professional deformation"; our training has warped our thoughts to run along certain unnatural lines. (-: –  Arturo Magidin Mar 4 '12 at 3:58

(1) is correct, as integration is a linear operator. One easy way to see that (2) is false is to test it with $f=g=h=\frac{1}{3}$, so that the definite integral should give you the area of the region of integration. But this gives you $a+b+c$ when you integrate over $[0,a]\times[0,b]\times[0,c]$, while the area of this cube is clearly $abc$.
Thank you, Alex. Could you also show why $\iiint \mathrm{d}x_1\mathrm{d}x_2\mathrm{d}x_3 f(x_1)g(x_2)h(x_3) = \int \mathrm{d}x_1 f(x_1) \int \mathrm{d}x_2 g(x_2) \int \mathrm{d}x_3 h(x_3) = F(x_1)G(x_2)H(x_3) + C$ rather than $\iiint \mathrm{d}x_1\mathrm{d}x_2\mathrm{d}x_3 f(x_1)g(x_2)h(x_3) = \iiint \mathrm{d}x_1\mathrm{d}x_2\mathrm{d}x_3 f(x_1) \iiint \mathrm{d}x_1\mathrm{d}x_2\mathrm{d}x_3 g(x_2) \iiint \mathrm{d}x_1\mathrm{d}x_2\mathrm{d}x_3 h(x_3) = F(x_1) \cdot x_2 \cdot x_3 \cdot G(x_2) \cdot x_1 \cdot x_3 \cdot H(x_3) \cdot x_1 \cdot x_2 + C' \text{ ?}$ –  user001 Mar 4 '12 at 4:28
@user001 The first case occurs because $g,h$ are independent of $x_1$, $f,h$ of $x_2$, and $f,g$ of $x_3$. The second can easily be eliminated by considering $f=g=h=1$. –  Alex Becker Mar 4 '12 at 4:30