An integral involving determinant. 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:

*

*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).

*How other people proved this result?

 A: 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
$$
\int_{\Omega}\det\left(  A+\nabla\varphi\right)  =\det(A)\left\vert
\Omega\right\vert \text{.}
$$
A: 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 $I(t)=I(0)=\det(A) |\Omega|$.
