Integral involving a Dirac delta

Question

The following result is provided in the Matrix Cookbook (Equation 547): If $\mathbf{A}$ is an $m \times k$ matrix with linearly independent columns and $m > k$, then \begin{align} \int f(\mathbf{x}) \, \delta( \mathbf{y} - \mathbf{A} \mathbf{x}) d \mathbf{x} = \begin{cases} \dfrac{1}{\det ( \mathbf{A}^{T} \mathbf{A} )^{\frac{1}{2}}} f( \mathbf{A}^{+} \mathbf{y} ) & \text{if } (\mathbf{I} - \mathbf{A} \mathbf{A}^{+}) \mathbf{y} = \mathbf{0}, \\ 0 & \text{otherwise.} \end{cases} \end{align} Here $f: \mathbb{R}^{m} \rightarrow \mathbb{R}$ is a well-behaved function, $\delta(\mathbf{y})$ is a multivariate Dirac delta, and $\mathbf{A}^{+} = (\mathbf{A}^{T} \mathbf{A})^{-1} \mathbf{A}^T$ is the Moore-Penrose pseudoinverse of $\mathbf{A}$. I found the cited proof of this result (Equation 6), but I don't understand it. In particular, I do not understand how the Dirac delta is factorised and moved outside the integral during the proof. How do I prove this result?

Answer 1

We proceed as following $$ \int f(\mathbf{x}) \, \delta( \mathbf{y} - \mathbf{A} \mathbf{x}) d \mathbf{x} {= \int f(\mathbf{x}) \, \delta( \mathbf{y} - \mathbf{A} \mathbf{x}) d \mathbf{x}. } $$ If $y$ doss not belong to the column space of $A$, then $\mathbf{y} - \mathbf{A} \mathbf{x}$ will never become zero. Otherwise, there is some $\mathbf{u}$ such that $\mathbf{y} = \mathbf{A} \mathbf{u}$. By using Moore-Penrose inverse, we obtain $$ \mathbf{A}^+\mathbf{y} = \mathbf{A}^+\mathbf{A} \mathbf{u}=\mathbf{I}_k \mathbf{u}= \mathbf{u}. $$ Therefore, $$ \int f(\mathbf{x}) \, \delta( \mathbf{y} - \mathbf{A} \mathbf{x}) d \mathbf{x}. { = \int f(\mathbf{x}+\mathbf{u}) \, \delta( \mathbf{y} - \mathbf{A} (\mathbf{x}+\mathbf{u})) d \mathbf{x} \\= \int f(\mathbf{x}+\mathbf{u}) \, \delta( \mathbf{y} - \mathbf{A} \mathbf{x}-\mathbf{A}\mathbf{u}) d \mathbf{x} \\= \int f(\mathbf{x}+\mathbf{u}) \, \delta( \mathbf{A} \mathbf{x}) d \mathbf{x}. } $$ By definition of Dirac's delta function, we can write $\delta(\mathbf{x})=\delta(|\mathbf{x}|)$, where $|\mathbf{x}|$ is the norm of $\mathbf{x}$. Therefore, $$ \int f(\mathbf{x}+\mathbf{u}) \, \delta( \mathbf{A} \mathbf{x}) d \mathbf{x} { = \int f(\mathbf{x}+\mathbf{u}) \, \delta( |\mathbf{A} \mathbf{x}|) d \mathbf{x} \\= \int f(\mathbf{x}+\mathbf{u}) \, \delta\left( \sqrt{\mathbf{x}^T\mathbf{A}^T\mathbf{A} \mathbf{x}}\right) d \mathbf{x}. } $$ Now, it is time for a coordinate transformation. By defining the new coordinate system $\mathbf{\tilde x}=(\mathbf{A}^T\mathbf{A})^{\frac{1}{2}}\mathbf{x}$, the Jacobian matrix will be $$ \mathbf{J}(\mathbf{\tilde x})=\frac{d\mathbf{x}}{d\mathbf{\tilde x}}=(\mathbf{A}^T\mathbf{A})^{-\frac{1}{2}}. $$ Hence, $$ \int f(\mathbf{x}+\mathbf{u}) \, \delta\left( \sqrt{\mathbf{x}^T\mathbf{A}^T\mathbf{A} \mathbf{x}}\right) d \mathbf{x} { = \int d \mathbf{\tilde x} \det(J( \mathbf{\tilde x}))f((\mathbf{A}^T\mathbf{A})^{-\frac{1}{2}}\mathbf{\tilde x}+\mathbf{u}) \\\times \delta\left( \sqrt{\mathbf{\tilde x}^T(\mathbf{A}^T\mathbf{A})^{-\frac{1}{2}}\mathbf{A}^T\mathbf{A} (\mathbf{A}^T\mathbf{A})^{-\frac{1}{2}}\mathbf{\tilde x}}\right) \\= \int f((\mathbf{A}^T\mathbf{A})^{-\frac{1}{2}}\mathbf{\tilde x}+\mathbf{u}) \, \delta\left( \sqrt{\mathbf{\tilde x}^T\mathbf{\tilde x}}\right) \det(\mathbf{A}^T\mathbf{A})^{-\frac{1}{2}} d \mathbf{\tilde x} \\= \int f((\mathbf{A}^T\mathbf{A})^{-\frac{1}{2}}\mathbf{\tilde x}+\mathbf{u}) \, \delta\left( \mathbf{\tilde x}\right) \det(\mathbf{A}^T\mathbf{A})^{-\frac{1}{2}} d \mathbf{\tilde x} \\= f(\mathbf{u})\det(\mathbf{A}^T\mathbf{A})^{-\frac{1}{2}} \\= f(\mathbf{A}^+\mathbf{y})\det(\mathbf{A}^T\mathbf{A})^{-\frac{1}{2}} \blacksquare } $$