Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

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

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

share|improve this question
    
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 Answer 1

up vote 2 down vote accepted

(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 $abc$.

share|improve this answer
    
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
1  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.