Jacobian of the transformation

[Niles]

Homework Statement

Hi

I wish to perform an integral of the form

<br /> \int_0^a {\int_0^b {f\left( {x - y} \right)dxdy} } <br />

What I do first is to define s := x-y, and ds = dx. Then we get

<br /> \int_0^a {\int_{-y}^{b-y} {f\left( {s} \right)dsdy} } <br />

Then I can define t := x+y, so dt = dy. Then I get

<br /> \int_{x}^{x+a} {\int_{-y}^{b-y} {f\left( {s} \right)dsdt} } <br />

I also have to multiply by 2, since it is the Jacobian of the transformation. But look at the limits: It doesn't seem to make things easier. Where am I going wrong?

 

[HallsofIvy]

Well, what, exactly do you want to do with that? You obviously can't "do" the integral without knowing what f is. However, your limits of integration in the final integral makes no sense. From the first form, since the limits of integration are constants, the integral does not depend on x or y but your last form for the integral clearly does.

The region of integration in the x,y plane is a rectangle with sides parallel to the x and y axes. In the u,v plane it is still a rectangle but with sides at 45 degrees to the u and v axes. Depending on what a and b are, you would probably have to divide the area into two separate integrations. In any case, I don't see that change of variable helping.

 

[Niles]

The reason why I am asking is because in my book they write the following

<br /> I = \int_0^a {dt_1 \int_0^a {dt_2 \,F(x,t_1 - t_2 )} } F(y,t_1 - t_2 ) = a\int_0^a {dt\,\,F(x,t)F(y, - t)} <br />

I cannot quite see how the second equality-sign comes about. I thought it came from substitution, as above.