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.

If we make a change of variable in higher dimensional integrals how do we decide the limits of integration. In two dimensions the problem is fairly simple using a geometric interpretation but in 3 and higher dimensions is there some way of doing this?

For eg. In the integral: $$\int_{0}^{a}dx\int_{x}^{a}dy\int_{y}^{a}dzf(x)g(z-y)$$ After making the change of variable $$x=\alpha;\,y+z=\beta;\, z-y=\gamma$$ how does the integration transform for the new variables $\alpha,\beta$ and $\gamma$? I'm more interested in the transformation of the limits of integration and any help/references will be appreciated?

share|improve this question

migrated from physics.stackexchange.com Jun 4 '12 at 16:55

This question came from our site for active researchers, academics and students of physics.

    
This is a math question ideally suited for math.stackexchange. –  Ron Maimon Jun 4 '12 at 7:37
add comment

1 Answer

You draw a picture in the y-z plane, and you rotate it by 45 degrees and blow up by a factor of $\sqrt{2}$. This gives you a triangle region, where the vertices are determined to be the image of the old vertices under the rotation: (x,x),(a,x),(a,a). The images in the $\beta,\gamma$ plane are $(2x,0),(a+x,a-x),(2a,0)$.

But I understand that you want a formal algorithm, so that you can do it mechanically. First, you make a one-variable transformation, where you trade in z for $\gamma= z-y$. If you really insist on doing the rotation (it's not the right thing), you can then trade in $y$ for $\beta$, using $\beta= 2y + \gamma$.

The answer for the first shift is that the z range is from y to a, so translating z by -y goes from 0 to a-y.

$$\int_0^a dx f(x) \int_x^a dy \int_0^{a-y} d\gamma g(\gamma) $$

To complete the mechanical change of variable, introduce an indicator function of one dimension $\phi(x)$, which is 1 for $x>0$ and 0 for $x<0$. Then you write the integral as follows:

$$ \int_0^a dx f(x) \int_x^a dy \int_0^\infty \phi(a-y-\gamma) g(\gamma) d\gamma$$

Rearrange the order of the y and $\gamma$ integrals and perform the y integral explicitly (using the fact that the indefinite integral of $\phi(x)$ is $x\phi(x)$) to get

$$\int_0^a dx f(x) \int_0^\infty (a-x-\gamma)\phi(a-x-\gamma)) g(\gamma) d\gamma $$

The $\phi$ function now gives you the new domain for $\gamma$

$$ \int_0^a dx f(x) \int_0^{a-x} (a-x-\gamma)g(\gamma)$$

and this is the best form. Using explicit indicator functions ($\phi$) this way is useful in many cases for getting explicit answers in closed form. If you really want to do the transformation of variables the way you said it, you introduce the indicator functions for the region, and then transform the region and find the new domain. This is just the same as drawing the triangle and rotating it.

share|improve this answer
    
Thanks for migrating my question to the math section and for the quick reply. Just another quick question since you are the expert. Is the derivative of the indicator function defined as the delta function? The reason I ask this is because in general I could have another function of $y$ accompanying the $\phi(a-y-\gamma)$ in the y-integral and in this case if the derivative can be defined then the y-integral could be performed by integration by parts. Am I right or am I missing something here? –  Juzar Jun 5 '12 at 6:36
    
@Juzar: Sorry for the late response, but yes, the derivative of the indicator function is a delta function at the ends, and you can use this for integration by parts. –  Ron Maimon Aug 2 '12 at 8:10
add comment

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.