Change of variables (Real Analysis)

Question

The change of variables formula is given as follows: For $\Omega, \Omega' \subset \mathbb{R}$ an open set, let $\phi: \Omega \to \Omega'$ such that $\phi$ is bijective, continuous, and $\phi^{-1}$ is continuous. Then for some integrable function $f$ (on $\Omega'$), $$ \int_{\Omega'} f(x') \, dx' = \int_{\Omega} (f \cdot \phi)(x) |\det(J(x))| \, dx. $$ I don't understand how to apply this to calculating the integral $$ \int_{\mathbb{R}^d} f(x-y) \, dy. $$ For instance, I don't know what I should take to be $\phi$. Is $\phi(x) = x+y$ because then $\phi(x-y) = x$? Or should $\phi(x) = x-y$ because we should take the change of variables $x-y = x'$? In the lectures, I think the change of variables $x-y = x'$ was taken.

Answer 1

If $\phi:\mathbb{R}^d\to \mathbb{R}^d$ is an arbitrary surjective map, then $\phi(\mathbb{R}^d)=\mathbb{R}^d$. Hence, $$ \int_{\mathbb{R}^d} f(x-y) \,\mathrm{d}y\;=\;\int_{\phi(\mathbb{R}^d)} f(x-y) \,\mathrm{d}y\ . $$ If $\phi:\mathbb{R}^d\to \mathbb{R}^d$ is also a diffeomorphism, then by the formula that you correctly write, you get $$ \int_{\phi(\mathbb{R}^d)} f(x-y) \,\mathrm{d}y\;=\; \int_{\mathbb{R}^d} f(x-\phi(z)) \cdot|\det D \phi(z)|\, \mathrm{d}z\, . $$ In particular, for the diffeomorphism of $\mathbb{R}^d$ given by $\phi(z)=x-z$, you obtain$$ \int_{\mathbb{R}^d} f(x-\phi(z))\cdot |\det D \phi(z)| \,\mathrm{d}z \, \;=\; \int_{\mathbb{R}^d} f(z) \,\mathrm{d}z \, $$ because of $|\det D \phi(z)|=|-1|=1$.