Changing variables in multiple integral (commutative of convolution operation) In the space $\mathbb{R}^n$, $n\geq 1$, the Lebesgue measure is denoted by $dx=dx_1\dots dx_n$ and $\int_{\mathbb{R}^n}f(x)dx$ stnads for $\int_{\mathbb{R}^n}f(x_1\dots x_n)dx_1\dots dx_n$. 
I want to prove that convolution operation is commutative, i.e. $f\ast g=g\ast f$. More precisely, 
$$
\int_{\mathbb{R}^n}f(x-y)g(y)dy=\int_{\mathbb{R}^n}f(y)g(x-y)dy
$$
This is a consequence of changing variables. But if i say $x-y=u$ then $(-1)^n dy=du$ and the result is
$$
\int_{\mathbb{R}^n}f(x-y)g(y)dy=(-1)^n\int_{\mathbb{R}^n}f(y)g(x-y)dy
$$
Where is my mistake?

If $T: \mathbb {R}^n\to  \mathbb {R}^n$ is a nice smooth bijection with $\det J_T(x)\ne 0$ for all $x,$ then
$$\int_{ \mathbb {R}^n} f(x)\,dx = \int_{ \mathbb {R}^n} f(T(y))|\det J_T(y)|\,dy.$$
