2 dimensional variable subsitution How can you show the following? Where the variable substitution is $z=x-y$.
$$\int_0^t \textrm{d}x \int_0^t \textrm{d}y \;f(x-y) = 2 t \int_0^t \left(1-\frac{z}{t}\right) \; f(z) \; \textrm{d}z.$$
I feel a previous question of mine, which was never resolved fully, is related to this.
Update: I'm glad to see that it isn't so trivial, and so far both responses think to say it's not correct. The question arises from my efforts to read through McQuarrie's statistical mechanics book (Chapter 21-8 for anyone who owns a copy). Here is a screenshot of the passage, which claims it true.

 A: Assuming that $f$ is an even function.
Let's say you are using the substitution
$$x=x,z=x-y$$
and it can be shown that the Jacobian is $1$.
On the $xy$ plane, the domain is a square with vertex $(0,0), (t,0), (t,t), (0,t)$.
Transformed to the $xz$ plane, the domain becomes a parallelogram with vertex $(0,0),(0,-t),(t,0),(t,t)$.
So for $-t\le z\le 0$, $0 \le x \le t+z$. For $0 \le z \le t$, $z \le x \le t$.
So the integral becomes
$$\int_0^t \int_0^t f(x-y)dxdy=\int_{-t}^0 \int_0^{t+z}f(z)dxdz+\int_0^t\int_z^t f(z)dxdz$$
$$=\int_{-t}^0 (t+z)f(z)dz+\int_0^t(t-z)f(z)dz$$
$$=\int_0^t(t-z)f(z)dz+\int_0^t(t-z)f(z)dz$$
$$=2\int_0^t(t-z)f(z)dz$$
A: It's kind of hard to show it because it isn't true. Suppose $t=1$ and $f(x)=x$.
Then
$$\begin{align}\int_0^t\int_0^tf(x-y)dy\,dx&=\int_0^1\int_0^1(x-y)dy\,dx\\
&=\int_0^1\left.-\frac12(x-y)^2\right|_0^1dx\\
&=\int_0^1\left(\frac12x^2-\frac12(x-1)^2\right)dx\\
&=\left.\frac16x^3-\frac16(x-1)^3\right|_0^1\\
&=\frac16-\frac16=0\end{align}$$
But
$$\begin{align}2t\int_0^t\left(1-\frac zt\right)f(z)dz&=2\int_0^1(1-z)z\,dz\\
&=\left.z^2-\frac23z^3\right|_0^1\\
&=1-\frac23=\frac13\end{align}$$
Maybe you can show some of your steps so that we might point out where you went astray?
