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I have a specific integral $$\int_{\mathbb{R}^3} f(\varphi(x))dx,$$ for an integrable function $f$ and some smooth function $\varphi$. I want to apply the change of variable $x\mapsto y=\varphi(x)$. If I know that the Jacobian of my change of variable is a constant finite number independent of $x$ (say $a\neq 0$), but I can't prove that $\varphi$ is injective, does it remain true to say that

$$\int_{\mathbb{R}^3} f(x)dx= \frac{1}{a}\int_{Im(\varphi)}f(y)dy?$$ Or maybe up to a finite constant for example?

Thank you for your answers!

Marrie
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1 Answers1

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The correct formula is \begin{align} \int f(\phi(x)) |\det D\phi(x)| \,dx = \int f(y) \text{card}(\phi^{-1}(\{y\}))\,dy. \end{align} Intuitively, this follows from the version of the theorem where $\phi$ is injective because even if $\phi$ is not injective, the set of points where $\det D\phi(x) = 0$ has measure $0$, and at each point $x$ where $\det D\phi(x) \neq 0$ there is a neighborhood of $x$ where $\phi$ is injective and the usual theorem applies. The full proof can be found in "Measure Theory and Integration" by Taylor.

Mason
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