Originally Posted by

**shawsend** Enough with the taus and stuff and should $\displaystyle \rho(x)$ not be an even function for that? Suppose so then how about just x, and y and u and v and just a real example with T=2 which in that case:

$\displaystyle \frac{1}{4}\int_{-1}^{1} \int_{-1}^{1} f(y-x)dx dy$

which is a square. Now if I let $\displaystyle u=y-x$ and $\displaystyle v=y+x$ then that transforms the region in x and y to a diamond shape in the u-v plane going from (2,0) to (0,2) then (-2,0) to (0,-2) and back to (2,0). Then calculate the Jacobian $\displaystyle \frac{\partial (x,y)}{\partial (u,v)}=-1/2$ but the positive direction over the u-v contour goes in the negative direction over the x-y contour so drop the negative sign. Then I get:

$\displaystyle \frac{2}{2}\int_0^{2} (1-\frac{u}{2}) f(u)du$

assuming $\displaystyle f(x)$ is an even function and is in the form you stated. Looks to me anyway.