An integral involving determinant.

Question

I learned the following identity (without proof) from the book "Lectures on Elliptic Partial Differential Equations" by L. Ambrosio et.al. $$\int_\Omega \det(A+\nabla\varphi)=\det(A)|\Omega|,$$ where $A$ is $n\times n$ matrix, $\varphi\in C^2_0(\Omega)$ and $\Omega\subset\mathbb{R}^n$ is bounded smooth domain. After some consideration, I have an elegant proof of this result (using divergent theorem and Hadamard identity). My question is:

1. What is the name of this identity? I believe that it should be discovered by some mathematician(s), thus should be named by the mathematician(s).
2. How other people proved this result?

Answer 1

We write $$ A=\left( \begin{array} [c]{c} A^{1}\\ \vdots\\ A^{n} \end{array} \right) \text{,} $$ and $M=\left( M^{1},\ldots,M^{n}\right) $, where $M^{i}$ is the $\left( n,i\right) $-cofactor of $\det(A+\nabla\varphi)$. Applying the Hadamard identity to $f:x\mapsto Ax+\varphi(x)$, we get $\operatorname{div}M=0$. By the divergence theorem, \begin{align*} & \int_{\Omega}\det\left( \begin{array} [c]{c} A^{1}+\nabla\varphi^{1}\\ \vdots\\ A^{n-1}+\nabla\varphi^{n-1}\\ \nabla\varphi^{n} \end{array} \right) =\int_{\Omega}\sum_{i=1}^{n}\partial_{i}\varphi^{n}M^{i}=\int _{\Omega}\nabla\varphi^{n}\cdot M\\ & =\int_{\Omega}\left( \operatorname{div}\left( \varphi^{n}M\right) -\varphi^{n}\operatorname{div}M\right) \\ & =\int_{\Omega}\operatorname{div}\left( \varphi^{n}M\right) \,\mathrm{d} x=\int_{\partial\Omega}\varphi^{n}M\cdot\nu\,\mathrm{d}\sigma=0\text{.} \end{align*} Therefore \begin{align*} & \int_{\Omega}\det\left( A+\nabla\varphi\right) =\int_{\Omega}\det\left( \begin{array} [c]{c} A^{1}+\nabla\varphi^{1}\\ \vdots\\ A^{n-1}+\nabla\varphi^{n-1}\\ A^{n}+\nabla\varphi^{n} \end{array} \right) \\ & =\int_{\Omega}\det\left( \begin{array} [c]{c} A^{1}+\nabla\varphi^{1}\\ \vdots\\ A^{n-1}+\nabla\varphi^{n-1}\\ A^{n} \end{array} \right) +\int_{\Omega}\det\left( \begin{array} [c]{c} A^{1}+\nabla\varphi^{1}\\ \vdots\\ A^{n-1}+\nabla\varphi^{n-1}\\ \nabla\varphi^{n} \end{array} \right) \\ & =\int_{\Omega}\det\left( \begin{array} [c]{c} A^{1}+\nabla\varphi^{1}\\ \vdots\\ A^{n-1}+\nabla\varphi^{n-1}\\ A^{n} \end{array} \right) \text{.} \end{align*} Similarly, we can apply Hadamard identity and the divergence theorem to the map $$ x\mapsto Ax+\left( \begin{array} [c]{c} \varphi^{1}(x)\\ \vdots\\ \varphi^{n-1}(x)\\ 0 \end{array} \right) $$ to eliminate $\nabla\varphi^{n-1}$ from the right hand side. Repeating the above argument, we eliminate $\nabla\varphi^{n-2}$, $\ldots$, $\nabla\varphi ^{1}$, and get $$ \int_{\Omega}\det\left( A+\nabla\varphi\right) =\det(A)\left\vert \Omega\right\vert \text{.} $$

Answer 2

The following proof is inspired by the transgression formula of Chern classes.

Note that $I(t)=\int_{\Omega}\det(A+t\,\nabla \varphi)d x=\int_{\Omega}d y^1\wedge\cdots\wedge d y^n$, where $y=Ax+t\,\varphi$, so we have $$I'(t)=\int_{\Omega}d\left(\varphi^1\wedge d y^2\wedge\cdots\wedge d y^n+\cdots+d y^1\wedge\cdots\wedge dy^{n-1}\wedge \varphi^n\right)=0,$$ Thus $I(t)=I(0)=\det(A) |\Omega|$.