Integrating by substitution of more than one variable - How?

Question

I want to integrate a function: Int[ 1/(x-y)^2 dx dy ] I want to change x-y to a new variable z. Can I do this? And if so how do I change the variables I am integrating over? I guess dz = dx - dy, but this doesn't go back into the equation. Is this way possible or am going in the wrong direction? And if I am going in the wrong direction, how should I go about tackling this problem? My full integral is actually over at least 6 variables paired up like: Int[ 1/sqrt((x1-y1)^2+(x2-y2)^2+(x3-y3)^2) dx1 dx2 dx3 dy1 dy2 dy3] Any pointers greatfully received!

The integration is across all of space for all of the variables.

Actually, I think integrating the y co-ordinates across all space will leave you with infinites, which is probably not so good. If this is the case, how do I get round these, since I need to do this integral for a path integral problem.

Answer 1

Int[ 1/(x-y)^2 dx dy ]
int[int(1/(x-y)^2)dx]dy
do the inner integration first,
treating the y as a constant
int(1/(x-y)^2)dx= 1/(y-x)+C

then do the outer integration
treating the x as a constant

Int[ 1/(x-y)^2 dx dy ]
=int[1/(y-x)+C]dy
=Cy+ln(y-x)+D

where C and D are constants

i hope that this helps

Answer 2

The word you are looking for is the Jacobian, J. It is the natural extension of substitution to more than one variable.

Given an integral of a function of x1,...xn. You make up new variables y1,...,yn which are functions of the x's. The Jacobian is the ratio of volumes from the x space to the y space. It is also a determinant made up of the partial derivatives of the y's with respect to the x's. Then

the integral of f(y1(x1,...),y2(x1,..),...) dy1...dyn = the integral of f(x1,...,xn) * J dx1...dxn

Any book in multivariable calculus will cover the topic.

Answer 3

It all depends on the domain of integration. looks like you want to go to spherical coordinates. Give me more details, I'll do the same.

Well if you integrate f(u-v) over all of space in (u,v), where u and v are valued in some vector space, the integral is always going to be infinite. You don't need to compute it.