Result: Let $T\in GL(d, \mathbb R).$ If $f$ is a Lebesgue measurable function on $\mathbb R^d, $ so is $f\circ T.$ If $f\geq 0$ or $f\in L^1(\mathbb R^d)$, then $$\int_{\mathbb R^d} f(x) dx = |\text{det} \ T| \int_{\mathbb R^d} f\circ T (x) dx.$$
e.g., Let $T:\mathbb R^2 \to \mathbb R^2, T(x,y)=f(x+y, x-y).$ Then by above result we have $$\int_{\mathbb R^2} f(x,y) dx dy = 2 \int_{\mathbb R^2} f(x+y, x-y) dx dy.$$
Put $h(x,y)= f(x+y, x-y).$
Question:I'm trying to prove the above formula without using any result: $I=\int_{\mathbb R^2} h(x,y) dx dy= \int_{\mathbb R^2} f(x+y, x-y) dx dy. $ If take substitution $u=x+y, v=x-y,$ then $I=\int_{\mathbb R^2} f(u,v) dx dy$ I think, I should transform my $dx, dy$ and in terms of $du, dv$. I do not know how to do this.
There's two ways to read your question. You might be asking "How can I prove the change-of-variables formula for double integrals?" Or you might be asking "Is there a more convenient way to calculate these things than writing down a matrix and calculating the determinant?" I'll assume what you want is the second thing.
The language of differential forms provides a convenient means of expressing these kinds of calculations. Instead of $dxdy$, we write $dx \wedge dy$. The wedge is a product, but it is no longer commutative. Instead it obeys the rule $dx \wedge dy = -dy \wedge dx$. Instead of representing a scalar quantity, $dx\wedge dy$ is a thing called a 2-form, which you can think of as a quantity that can be integrated over an oriented 2 dimensional domain.
We can use the rules of $d$ and $\wedge$ to calculate $du \wedge dv$:
$$ du \wedge dv = d(x+y)\wedge d(x-y) = (dx + dy)\wedge(dx - dy) \\ =(dx\wedge dx) - (dx\wedge dy) + (dy\wedge dx) - (dy \wedge dy) \\ = 0 - (dx\wedge dy) - (dx\wedge dy) - 0 \\ = -2 dx \wedge dy $$