Derivation of Jacobian formula with Dirac delta function

In this subsection, we present our strategy to derive the Jacobian for a coordinate transformation or change of variables from the Cartesian coordinates x_i to the curvilinear coordinates q_i with transformation functions

$$\begin{equation}{q}_{i}={q}_{i}\left({x}_{1},\dots ,{x}_{n}\right)\end{equation} \tag{ 3 }$$

for i = 1 through n, where n is a positive integer. We assume that the two sets of coordinates describe a single point uniquely and, therefore, the two sets of coordinates have a one-to-one correspondence although the transformation is in general non-linear. Thus the transformation in equation (3) is invertible: the inverse transformation from the curvilinear coordinates to the Cartesian coordinates exists. If the curvilinear coordinates are a linear combination of the Cartesian coordinates, then the linear transformation is invertible if the transformation matrix is non-singular: the determinant of the matrix is not vanishing. Then, the inverse transformation can be written as

$$\begin{equation}{x}_{i}={x}_{i}\left({q}_{1},\dots ,{q}_{n}\right).\end{equation} \tag{ 4 }$$

The basic strategy to derive the Jacobian with Dirac delta functions is the same as that for the derivation of Cramer's rule for a partial set of a coordinate transformation or change of variables given in reference [4]. One could immediately apply the approach in reference [4] to find the Jacobian as long as the transformation (3) is linear. In general, the transformation functions (3) are non-linear. Here we develop a generalized version of the approach in reference [4] in order to consider arbitrary curvilinear coordinates.

We define an n-dimensional integral I_n ,

$$\begin{equation}{I}_{n}={\int }_{-\infty }^{\infty }{\mathrm{d}}^{n}x\enspace f\left({x}_{1},\dots ,{x}_{n}\right),\end{equation} \tag{ 5 }$$

where dⁿ x ≡ dx₁...dx_n is the n-dimensional differential volume element and the integrand f(x₁, ..., x_n ) is an arbitrary function of the Cartesian coordinates. We assume that every Cartesian coordinate is integrated over the region (−∞, ∞). We define the unity

$$\begin{equation}{\mathbb{1}}_{i}\equiv \int \mathrm{d}{q}_{i}\enspace \delta \left[{q}_{i}-{q}_{i}\left({x}_{1},\dots ,{x}_{n}\right)\right]=1\end{equation} \tag{ 6 }$$

for i = 1 through n. Multiplying the unity ${\mathbb{1}}_{i}$ to the integral I_n , one can integrate out the integration variable x_i by making use of the Dirac delta function keeping the q_i integral unevaluated. By applying this process to I_n recursively from i = 1 through n, we complete the change of the integration variables from the Cartesian coordinates to the curvilinear coordinates. While we have suppressed the bounds of the integration for the curvilinear coordinate q_i 's in equation (6), the curvilinear coordinates are assumed to be integrated over the entire region to cover the whole Euclidean space represented by the Cartesian coordinates by a single time.

We first compute ${\mathbb{1}}_{1}{\times}{I}_{n}$:

$$\begin{equation}{I}_{n}={\mathbb{1}}_{1}{\times}{I}_{n}=\int \mathrm{d}{q}_{1}{\int }_{-\infty }^{\infty }{\mathrm{d}}^{n}x\enspace f\left({x}_{1},\dots ,{x}_{n}\right)\delta \left[{q}_{1}-{q}_{1}\left({x}_{1},\dots ,{x}_{n}\right)\right].\end{equation} \tag{ 7 }$$

By definition, we integrate over x₁ by making use of the delta function δ[q₁ − q₁(x₁, ..., x_n )] keeping the q₁ integral unevaluated:

$$\begin{equation}{I}_{n}=\int \mathrm{d}{q}_{1}{\int }_{-\infty }^{\infty }\mathrm{d}{x}_{2}\dots \mathrm{d}{x}_{n}\enspace \frac{1}{{\mathcal{G}}_{1}}f\left({q}_{1},{x}_{2},\dots ,{x}_{n}\right),\end{equation} \tag{ 8 }$$

where x₁ is expressed in terms of q₁ and x_j 's for j = 2 through n satisfying the condition that the argument of the delta function vanishes, q₁ − q₁(x₁, ..., x_n ) = 0. Because the explicit forms of f(x_i ) and x_i vary depending on the integration step, we adopt the notation f(q₁, x₂, ..., x_n ) after the x₁ integration. We will present more detailed explanations for this notation in the later part of this subsection. The extra factor ${\mathcal{G}}_{1}$ is the remnant of the integration of the delta function and its explicit form will be given later in this paper.

Next, we multiply ${\mathbb{1}}_{2}$ to I_n in equation (8) to find that

$$\begin{align}\hfill {I}_{n}& ={\mathbb{1}}_{2}{\times}{I}_{n}=\int \mathrm{d}{q}_{1}\enspace \mathrm{d}{q}_{2}{\int }_{-\infty }^{\infty }\mathrm{d}{x}_{2}\dots \mathrm{d}{x}_{n}\enspace \frac{1}{{\mathcal{G}}_{1}}f\left({q}_{1},\dots ,{x}_{n}\right)\delta \left[{q}_{2}-{q}_{2}\left({q}_{1},{x}_{2},\dots ,{x}_{n}\right)\right],\hfill \end{align} \tag{ 9 }$$

where every x₁ in the argument of the delta function as well as the integrand function is replaced with the expression in terms of q₁ and x_j 's for j = 2 through n. Performing the integration over x₂ by making use of the delta function, we find

$$\begin{equation}{I}_{n}=\int \mathrm{d}{q}_{1}\enspace \mathrm{d}{q}_{2}{\int }_{-\infty }^{\infty }\mathrm{d}{x}_{3}\dots \mathrm{d}{x}_{n}\enspace \frac{1}{{\mathcal{G}}_{2}}f\left({q}_{1},{q}_{2},{x}_{3},\dots ,{x}_{n}\right).\end{equation} \tag{ 10 }$$

After the x₂ integration every x₂ in the integrand is expressed in terms of q₁, q₂ and x_j 's for j = 3 through n. Again, ${\mathcal{G}}_{2}$ is the remnant of the integration of the delta functions.

In this way, we integrate over x_k for k = 1 through n successively. Finally, after the integration over x_n we find that I_n reduces into the n-dimensional multiple integral over the curvilinear coordinates q_i for i = 1 through n only:

$$\begin{equation}{I}_{n}=\int \mathrm{d}{q}_{1}\dots \mathrm{d}{q}_{n}\enspace \frac{1}{{\mathcal{G}}_{n}}f\left({q}_{1},\dots ,{q}_{n}\right),\end{equation} \tag{ 11 }$$

where ${\mathcal{G}}_{n}$ is the remnant of the integration of the delta functions. At this stage, the integrand acquires an additional factor 1/${\mathcal{G}}_{n}$ in front of the original integrand f in equation (5). This extra factor is identified with the Jacobian.

In an intermediate step, for example, where the integration over x_j (1 ⩽ j ⩽ n) is carried out, x₁, ..., x_j in the integrand must be replaced with the expressions in terms of q₁, ..., q_j and x_j+1, ..., x_n as

$$\begin{equation}{x}_{k}={x}_{k}\left({q}_{1},\dots ,{q}_{j},{x}_{j+1},\dots ,{x}_{n}\right)\end{equation} \tag{ 12 }$$

for k = 1 through j. Each x_k in equation (12) is determined by the condition that the argument of the corresponding Dirac delta function vanishes:

$$\begin{align}\hfill & {q}_{k}-{q}_{k}\left[{x}_{1}\left({q}_{1},\dots ,{q}_{j},{x}_{j+1},\dots ,{x}_{n}\right),\dots ,{x}_{j}\left({q}_{1},\dots ,{q}_{j},{x}_{j+1},\dots ,{x}_{n}\right),{x}_{j+1},\dots ,{x}_{n}\right\}=0.\hfill \end{align} \tag{ 13 }$$

One must keep in mind that the explicit form of each x_k varies depending on the integration step as is displayed in equation (12). Thus, one must distinguish, for example, x_k (q₁, ..., q_j , x_j+1, ..., x_n ) from x_k (q₁, ..., q_j−1, x_j , ..., x_n ), where the former is the expression after the integration over x_j and the latter is that after the integration over x_j−1. This notation is also applied to the original integrand function f and the extra factor ${\mathcal{G}}_{i}$.

As an explicit example, we consider the coordinate transformation between the two-dimensional Cartesian coordinates and the polar coordinates with the transformation functions

$$\begin{equation}r=\sqrt{{x}^{2}+{y}^{2}}\quad \mathrm{a}\mathrm{n}\mathrm{d}\quad \theta =\mathrm{arctan}\enspace \frac{y}{x}\end{equation} \tag{ 14 }$$

and the inverse transformation functions

$$\begin{equation}x=r\mathrm{cos}\theta \quad \mathrm{a}\mathrm{n}\mathrm{d}\quad y=r\mathrm{sin}\theta .\end{equation} \tag{ 15 }$$

In an intermediate step, y can be expressed in terms of θ and x as y(θ, x) = x tan θ, which must be distinguished from y(θ, r) = r sin θ.

Since the dependence of a coordinate variable varies according to the integration step, one must take special care of dealing with their partial derivatives. In order to avoid such an ambiguity, we introduce a notation for the partial derivative with subscripts of variables that are held constant. In the above two-dimensional transformation, the partial derivative of θ with respect to y holding x fixed is denoted by

$$\begin{equation}{\left(\frac{\partial \theta }{\partial y}\right)}_{x}=\frac{x}{{x}^{2}+{y}^{2}},\end{equation} \tag{ 16 }$$

while the partial derivative of θ with respect to y holding r fixed is represented by

$$\begin{equation}{\left(\frac{\partial \theta }{\partial y}\right)}_{r}=\frac{1}{\sqrt{{r}^{2}-{y}^{2}}}.\end{equation} \tag{ 17 }$$

It is apparent from equations (16) and (17) that

$$\begin{equation}{\left(\frac{\partial \theta }{\partial y}\right)}_{x}\ne {\left(\frac{\partial \theta }{\partial y}\right)}_{r}.\end{equation} \tag{ 18 }$$

In a general case, for a variable q_j (q₁, ..., q_i , x_i+1, ..., x_n ), we denote the partial derivative of q_j with respect to x_a holding q₁, ..., q_i , x_i+1, ..., x_n fixed with the subscript (i) as

$$\begin{equation}{\left(\frac{\partial {q}_{j}}{\partial {x}_{a}}\right)}_{\left(i\right)}=\frac{\partial {q}_{j}\left({q}_{1},\dots ,{q}_{i},{x}_{i+1},\dots ,{x}_{n}\right)}{\partial {x}_{a}},\end{equation} \tag{ 19 }$$

where i + 1 ⩽ j ⩽ n and i + 1 ⩽ a ⩽ n. Similarly, for a variable x_k (q₁, ..., q_i , x_i+1, ..., x_n ), the partial derivative of x_k with respect to q_b holding q₁, ..., q_i , x_i+1, ..., x_n fixed is denoted by

$$\begin{equation}{\left(\frac{\partial {x}_{k}}{\partial {q}_{b}}\right)}_{\left(i\right)}=\frac{\partial {x}_{k}\left({q}_{1},\dots ,{q}_{i},{x}_{i+1},\dots ,{x}_{n}\right)}{\partial {q}_{b}},\end{equation} \tag{ 20 }$$

where 1 ⩽ k ⩽ i and 1 ⩽ b ⩽ i. We denote the partial derivative of x_ℓ with respect to x_c holding q₁, ..., q_i , x_i+1, ..., x_n fixed by

$$\begin{equation}{\left(\frac{\partial {x}_{\ell }}{\partial {x}_{c}}\right)}_{\left(i\right)}=\frac{\partial {x}_{\ell }\left({q}_{1},\dots ,{q}_{i},{x}_{i+1},\dots ,{x}_{n}\right)}{\partial {x}_{c}},\end{equation} \tag{ 21 }$$

where 1 ⩽ ℓ ⩽ i and i + 1 ⩽ c ⩽ n. Finally, we express the partial derivative of q_ℓ with respect to x_c holding x₁, ..., x_n without subscript:

$$\begin{equation}\frac{\partial {q}_{\ell }}{\partial {x}_{c}}=\frac{\partial {q}_{\ell }\left({x}_{1},\dots ,{x}_{n}\right)}{\partial {x}_{c}},\end{equation} \tag{ 22 }$$

where 1 ⩽ ℓ ⩽ n and 1 ⩽ c ⩽ n.

In the coordinate transformation from (x₁, ..., x_n ) to (q₁, ..., q_n ), we define the function ${\mathcal{G}}_{k}$, which is relevant for a partial set of integral variables corresponding to a transformation from (x₁, ..., x_k ) to (q₁, ..., q_k ), as

$$\begin{equation}{\mathcal{G}}_{k}=\left\vert \mathcal{D}\mathcal{e}\mathcal{t}\left(\begin{matrix}\hfill \frac{\partial {q}_{1}}{\partial {x}_{1}}\hfill & \hfill \frac{\partial {q}_{2}}{\partial {x}_{1}}\hfill & \hfill \dots \hfill & \hfill \frac{\partial {q}_{k}}{\partial {x}_{1}}\hfill \\ \hfill \frac{\partial {q}_{1}}{\partial {x}_{2}}\hfill & \hfill \frac{\partial {q}_{2}}{\partial {x}_{2}}\hfill & \hfill \dots \hfill & \hfill \frac{\partial {q}_{k}}{\partial {x}_{2}}\hfill \\ \hfill {\vdots}\hfill & \hfill {\vdots}\hfill & \hfill \ddots \hfill & \hfill {\vdots}\hfill \\ \hfill \frac{\partial {q}_{1}}{\partial {x}_{k}}\hfill & \hfill \frac{\partial {q}_{2}}{\partial {x}_{k}}\hfill & \hfill \dots \hfill & \hfill \frac{\partial {q}_{k}}{\partial {x}_{k}}\hfill \end{matrix}\right)\right\vert \end{equation} \tag{ 23 }$$

for k = 1 through n. Note that ${\mathcal{G}}_{n}$ is the inverse of the Jacobian ${\mathcal{J}}_{n}$ which is defined by

$$\begin{equation}{\mathcal{J}}_{n}=\left\vert \mathcal{D}\mathcal{e}\mathcal{t}\left(\begin{matrix}\hfill {\left(\frac{\partial {x}_{1}}{\partial {q}_{1}}\right)}_{\left(n\right)}\hfill & \hfill {\left(\frac{\partial {x}_{2}}{\partial {q}_{1}}\right)}_{\left(n\right)}\hfill & \hfill \dots \hfill & \hfill {\left(\frac{\partial {x}_{k}}{\partial {q}_{1}}\right)}_{\left(n\right)}\hfill \\ \hfill {\left(\frac{\partial {x}_{1}}{\partial {q}_{2}}\right)}_{\left(n\right)}\hfill & \hfill {\left(\frac{\partial {x}_{2}}{\partial {q}_{2}}\right)}_{\left(n\right)}\hfill & \hfill \dots \hfill & \hfill {\left(\frac{\partial {x}_{k}}{\partial {q}_{2}}\right)}_{\left(n\right)}\hfill \\ \hfill {\vdots}\hfill & \hfill {\vdots}\hfill & \hfill \ddots \hfill & \hfill {\vdots}\hfill \\ \hfill {\left(\frac{\partial {x}_{1}}{\partial {q}_{k}}\right)}_{\left(n\right)}\hfill & \hfill {\left(\frac{\partial {x}_{2}}{\partial {q}_{k}}\right)}_{\left(n\right)}\hfill & \hfill \dots \hfill & \hfill {\left(\frac{\partial {x}_{k}}{\partial {q}_{k}}\right)}_{\left(n\right)}\hfill \end{matrix}\right)\right\vert .\end{equation} \tag{ 24 }$$

In this subsection, we consider a two-dimensional integral I₂ for the integration of an arbitrary function f(x₁, x₂):

$$\begin{equation}{I}_{2}={\int }_{-\infty }^{\infty }\enspace \mathrm{d}{x}_{1}\mathrm{d}{x}_{2}\enspace f\left({x}_{1},{x}_{2}\right).\end{equation} \tag{ 25 }$$

The Cartesian coordinates x₁ and x₂ are transformed into curvilinear coordinates q₁ and q₂ with the transformation relations

$$\begin{equation}{q}_{1}={q}_{1}\left({x}_{1},{x}_{2}\right),\qquad {q}_{2}={q}_{2}\left({x}_{1},{x}_{2}\right),\end{equation} \tag{ 26 }$$

which are assumed to be invertible and non-singular. Thus, x₁ and x₂ can be expressed in terms of q₁ and q₂:

$$\begin{equation}{x}_{1}={x}_{1}\left({q}_{1},{q}_{2}\right),\qquad {x}_{2}={x}_{2}\left({q}_{1},{q}_{2}\right).\end{equation} \tag{ 27 }$$

We multiply the unities

$$\begin{equation}{\mathbb{1}}_{i}=\int \mathrm{d}{q}_{i}\enspace \delta \left[{q}_{i}-{q}_{i}\left({x}_{1},{x}_{2}\right)\right]=1\end{equation} \tag{ 28 }$$

to I₂ for i = 1, 2, sequentially. First, we multiply ${\mathbb{1}}_{1}$ in equation (28) to I₂. Then, I₂ can be expressed as

$$\begin{equation}{I}_{2}=\int \mathrm{d}{q}_{1}{\int }_{-\infty }^{\infty }\enspace \mathrm{d}{x}_{1}\mathrm{d}{x}_{2}\enspace \delta \left[{q}_{1}-{q}_{1}\left({x}_{1},{x}_{2}\right)\right]f\left({x}_{1},{x}_{2}\right).\end{equation} \tag{ 29 }$$

The integration over x₁ can be carried out by making use of the delta function for q₁ as

$$\begin{align}\hfill {\int }_{-\infty }^{\infty }\enspace \mathrm{d}{x}_{1}\enspace \delta \left[{q}_{1}-{q}_{1}\left({x}_{1},{x}_{2}\right)\right]& =\frac{1}{{\left\vert \frac{\partial {q}_{1}}{\partial {x}_{1}}\right\vert }_{{x}_{1}={x}_{1}\left({q}_{1},{x}_{2}\right)}}\hfill \\ \hfill & =\frac{1}{{\mathcal{G}}_{1}\left({q}_{1},{x}_{2}\right)},\hfill \end{align} \tag{ 30 }$$

which leads to

$$\begin{equation}{I}_{2}=\int \mathrm{d}{q}_{1}{\int }_{-\infty }^{\infty }\mathrm{d}{x}_{2}\enspace \frac{1}{{\mathcal{G}}_{1}\left({q}_{1},{x}_{2}\right)}f\left({q}_{1},{x}_{2}\right).\end{equation} \tag{ 31 }$$

After the integration over x₁, every x₁ on the right-hand side of equation (30) and δ[q₂ − q₂(x₁, x₂)]f(x₁, x₂) in equation (28) must be replaced with x₁(q₁, x₂) satisfying the condition that the argument of the delta function vanishes, q₁ − q₁(x₁, x₂) = 0. Then the delta function for q₂ and f(x₁, x₂) can be expressed as δ[q₂ − q₂(q₁, x₂)] and f(q₁, x₂), respectively.

After multiplying ${\mathbb{1}}_{2}$ in equation (28) to I₂ in equation (31), the integration over x₂ can be performed by making use of the remaining delta function for q₂ as

$$\begin{align}\hfill {\int }_{-\infty }^{\infty }\mathrm{d}{x}_{2}\enspace \delta \left[{q}_{2}-{q}_{2}\left({q}_{1},{x}_{2}\right)\right]& =\frac{1}{{\left\vert {\left(\frac{\partial {q}_{2}}{\partial {x}_{2}}\right)}_{\left(1\right)}\right\vert }_{{x}_{2}={x}_{2}\left({q}_{1},{q}_{2}\right)}}\hfill \\ \hfill & =\frac{{\mathcal{G}}_{1}\left({q}_{1},{q}_{2}\right)}{{\mathcal{G}}_{2}\left({q}_{1},{q}_{2}\right)}.\hfill \end{align} \tag{ 32 }$$

After the integration over x₂, every x₂ on the right-hand sides of equations (30) and (32) and in f(q₁, x₂) is replaced with x₂(q₁, q₂), which can be obtained from the condition that the argument of the delta function vanishes, q₂ − q₂(q₁, x₂) = 0. Then, we can express x₁ in terms of q₁ and q₂ by replacing x₂ with x₂(q₁, q₂) in x₁(q₁, x₂). We have obtained the last equality of equation (32) by making use of the chain rule for the partial derivatives. A rigorous proof of this formula is given in equation (A5) of appendix A.

After both x₁ and x₂ are integrated out, the integral I₂ reduces into

$$\begin{align}\hfill {I}_{2}& =\int \mathrm{d}{q}_{1}\enspace \mathrm{d}{q}_{2}\enspace \frac{1}{{\mathcal{G}}_{1}}{\times}\frac{{\mathcal{G}}_{1}}{{\mathcal{G}}_{2}}f\left[{x}_{1}\left({q}_{1},{q}_{2}\right),{x}_{2}\left({q}_{1},{q}_{2}\right)\right]\hfill \\ \hfill & =\int \mathrm{d}{q}_{1}\enspace \mathrm{d}{q}_{2}\enspace {\mathcal{J}}_{2}\enspace f\left[{x}_{1}\left({q}_{1},{q}_{2}\right),{x}_{2}\left({q}_{1},{q}_{2}\right)\right],\hfill \end{align} \tag{ 33 }$$

where the Jacobian for the change of variables is identified as $\mathcal{J}{=\mathcal{J}}_{2}=1{/\mathcal{G}}_{2}$. This completes the proof of the Jacobian for a two-dimensional coordinate transformation or change of variables.

In this subsection, we extend the results in the previous subsection to the three-dimensional case. This is a special case of the n-dimensional coordinate transformation or change of variables, which we will prove in the next subsection. However, it is worthwhile to prove the three-dimensional case in detail for a pedagogical purpose.

We consider a three-dimensional integral I₃ for an arbitrary function f(x₁, x₂, x₃):

$$\begin{equation}{I}_{3}={\int }_{-\infty }^{\infty }\mathrm{d}{x}_{1}\enspace \mathrm{d}{x}_{2}\enspace \mathrm{d}{x}_{3}\enspace f\left({x}_{1},{x}_{2},{x}_{3}\right).\end{equation} \tag{ 34 }$$

The Cartesian coordinates x_i for i = 1 through 3 are transformed into the curvilinear coordinates q_i 's as

$$\begin{equation}{q}_{1}={q}_{1}\left({x}_{1},{x}_{2},{x}_{3}\right),\qquad {q}_{2}={q}_{2}\left({x}_{1},{x}_{2},{x}_{3}\right),\qquad {q}_{3}={q}_{3}\left({x}_{1},{x}_{2},{x}_{3}\right).\end{equation} \tag{ 35 }$$

The inverse transformation can be expressed as

$$\begin{equation}{x}_{1}={x}_{1}\left({q}_{1},{q}_{2},{q}_{3}\right),\qquad {x}_{2}={x}_{2}\left({q}_{1},{q}_{2},{q}_{3}\right),\qquad {x}_{3}={x}_{3}\left({q}_{1},{q}_{2},{q}_{3}\right).\end{equation} \tag{ 36 }$$

We carry out the change of variables by multiplying the unites

$$\begin{equation}{\mathbb{1}}_{i}=\int \mathrm{d}{q}_{i}\enspace \delta \left[{q}_{i}-{q}_{i}\left({x}_{1},{x}_{2},{x}_{3}\right)\right]=1\end{equation} \tag{ 37 }$$

for i = 1 through 3 to I₃, sequentially. First, after multiplying ${\mathbb{1}}_{1}$ in equation (37) to I₃, we find that I₃ can be expressed as

$$\begin{equation}{I}_{3}=\int \mathrm{d}{q}_{1}{\int }_{-\infty }^{\infty }\mathrm{d}{x}_{1}\enspace \mathrm{d}{x}_{2}\enspace \mathrm{d}{x}_{3}\enspace \delta \left[{q}_{1}-{q}_{1}\left({x}_{1},{x}_{2},{x}_{3}\right)\right]f\left({x}_{1},{x}_{2},{x}_{3}\right).\end{equation} \tag{ 38 }$$

Similarly to the two-dimensional case, we perform the integration over x₁ by making use of the Dirac delta function for q₁ as

$$\begin{align}\hfill {\int }_{-\infty }^{\infty }\mathrm{d}{x}_{1}\enspace \delta \left[{q}_{1}-{q}_{1}\left({x}_{1},{x}_{2},{x}_{3}\right)\right]& =\frac{1}{{\left\vert \frac{\partial {q}_{1}}{\partial {x}_{1}}\right\vert }_{{x}_{1}={x}_{1}\left({q}_{1},{x}_{2},{x}_{3}\right)}}\hfill \\ \hfill & =\frac{1}{{\mathcal{G}}_{1}\left({q}_{1},{x}_{2},{x}_{3}\right)},\hfill \end{align} \tag{ 39 }$$

which leads to

$$\begin{equation}{I}_{3}=\int \mathrm{d}{q}_{1}{\int }_{-\infty }^{\infty }\mathrm{d}{x}_{2}\enspace \mathrm{d}{x}_{3}\enspace \frac{1}{{\mathcal{G}}_{1}\left({q}_{1},{x}_{2},{x}_{3}\right)}f\left({q}_{1},{x}_{2},{x}_{3}\right).\end{equation} \tag{ 40 }$$

After the integration over x₁, every x₁ in the integrand and remaining delta functions is replaced with x₁(q₁, x₂, x₃) satisfying the condition that the argument of the Dirac delta function vanishes, q₁ − q₁(x₁, x₂, x₃) = 0. Then the delta function for q₂ in equation (37) can be expressed as δ[q₂ − q₂(q₁, x₂, x₃)].

After multiplying ${\mathbb{1}}_{2}$ in equation (37) to I₃ in equation (40), the integration over x₂ can be carried out by making use of the Dirac delta function for q₂ as

$$\begin{align}\hfill {\int }_{-\infty }^{\infty }\mathrm{d}{x}_{2}\enspace \delta \left[{q}_{2}-{q}_{2}\left({q}_{1},{x}_{2},{x}_{3}\right)\right]& =\frac{1}{{\left\vert {\left(\frac{\partial {q}_{2}}{\partial {x}_{2}}\right)}_{\left(1\right)}\right\vert }_{{x}_{2}={x}_{2}\left({q}_{1},{q}_{2},{x}_{3}\right)}}\hfill \\ \hfill & =\frac{{\mathcal{G}}_{1}\left({q}_{1},{q}_{2},{x}_{3}\right)}{{\mathcal{G}}_{2}\left({q}_{1},{q}_{2},{x}_{3}\right)},\hfill \end{align} \tag{ 41 }$$

where the last equality comes from equation (B5). Then, equation (41) is expressed as

$$\begin{equation}{I}_{3}=\int \mathrm{d}{q}_{1}\enspace \mathrm{d}{q}_{2}{\int }_{-\infty }^{\infty }\mathrm{d}{x}_{3}\enspace \frac{1}{{\mathcal{G}}_{1}\left({q}_{1},{q}_{2},{x}_{3}\right)}{\times}\frac{{\mathcal{G}}_{1}\left({q}_{1},{q}_{2},{x}_{3}\right)}{{\mathcal{G}}_{2}\left({q}_{1},{q}_{2},{x}_{3}\right)}f\left({q}_{1},{q}_{2},{x}_{3}\right).\end{equation} \tag{ 42 }$$

x₂(q₁, q₂, x₃) is determined from the condition that the argument of the Dirac delta function vanishes, q₂ − q₂(q₁, x₂, x₃) = 0. Substituting x₂(q₁, q₂, x₃) into x₁(q₁, x₂, x₃), we obtain x₁ = x₁(q₁, q₂, x₃) and the argument of the Dirac delta function for q₃ in equation (37) is expressed as δ[q₃ − q₃(q₁, q₂, x₃)].

Finally, after multiplying ${\mathbb{1}}_{3}$ in equation (37) to I₃ in equation (42), we integrate over x₃ by taking into account the delta function for q₃ as

$$\begin{align}\hfill {\int }_{-\infty }^{\infty }\mathrm{d}{x}_{3}\enspace \delta \left[{q}_{3}-{q}_{3}\left({q}_{1},{q}_{2},{x}_{3}\right)\right]& =\frac{1}{{\left\vert {\left(\frac{\partial {q}_{3}}{\partial {x}_{3}}\right)}_{\left(2\right)}\right\vert }_{{x}_{3}={x}_{3}\left({q}_{1},{q}_{2},{q}_{3}\right)}}\hfill \\ \hfill & =\frac{{\mathcal{G}}_{2}\left({q}_{1},{q}_{2},{q}_{3}\right)}{{\mathcal{G}}_{3}\left({q}_{1},{q}_{2},{q}_{3}\right)},\hfill \end{align} \tag{ 43 }$$

where the last equality comes from equation (B11). After the integration over x₃, every x₃ in the integrand and the right-hand sides of equations (39), (41) and (43) is replaced with x₃(q₁, q₂, q₃) which is determined from the condition that the argument of the delta function vanishes, q₃ − q₃(q₁, q₂, x₃) = 0. Substituting x₃(q₁, q₂, q₃) into x₁(q₁, q₂, x₃) and x₂(q₁, q₂, x₃), we can obtain the expressions for x₁ = x₁(q₁, q₂, q₃) and x₂ = x₂(q₁, q₂, q₃), respectively. Then, we can express the integrand f(x₁, x₂, x₃) in terms of q₁, q₂ and q₃ and the integral I₃ is expressed as

$$\begin{align}\hfill {I}_{3}& =\int \mathrm{d}{q}_{1}\enspace \mathrm{d}{q}_{2}\enspace \mathrm{d}{q}_{3}\enspace \frac{1}{{\mathcal{G}}_{1}}{\times}\frac{{\mathcal{G}}_{1}}{{\mathcal{G}}_{2}}{\times}\frac{{\mathcal{G}}_{2}}{{\mathcal{G}}_{3}}f\left[{x}_{1}\left({q}_{1},{q}_{2},{q}_{3}\right),{x}_{2}\left({q}_{1},{q}_{2},{q}_{3}\right),{x}_{3}\left({q}_{1},{q}_{2},{q}_{3}\right)\right]\hfill \\ \hfill & =\int \mathrm{d}{q}_{1}\enspace \mathrm{d}{q}_{2}\enspace \mathrm{d}{q}_{3}\enspace {\mathcal{J}}_{3}\enspace f\left[{x}_{1}\left({q}_{1},{q}_{2},{q}_{3}\right),{x}_{2}\left({q}_{1},{q}_{2},{q}_{3}\right),{x}_{3}\left({q}_{1},{q}_{2},{q}_{3}\right)\right],\hfill \end{align} \tag{ 44 }$$

where the Jacobian for the change of variables is identified as $\mathcal{J}{=\mathcal{J}}_{3}=1{/\mathcal{G}}_{3}$. This completes the proof of the Jacobian for a three-dimensional coordinate transformation or change of variables.

In this subsection, we consider the n-dimensional integral I_n for a function f(x₁, ..., x_n ) defined in equation (5) by multiplying the unities ${\mathbb{1}}_{i}$ in equation (6) to I_n in equation (5) sequentially. Then, we integrate I_n , which is multiplied by the unity, over x_i for i = 1 through n successively by making use of the Dirac delta function

$$\begin{equation}{\int }_{-\infty }^{\infty }\mathrm{d}{x}_{i}\enspace \delta \left[{q}_{i}-{q}_{i}\left({x}_{1},\dots ,{x}_{n}\right)\right].\end{equation} \tag{ 45 }$$

After the integration of all x_i variables, the corresponding Jacobian formula is obtained by employing mathematical induction.

First, the integration over x₁ can be carried out from equation (7) and the result for the integration over x₁ can easily be generalized from the two-dimensional version in equation (30) as

$$\begin{align}\hfill {\int }_{-\infty }^{\infty }\mathrm{d}{x}_{1}\enspace \delta \left[{q}_{1}-{q}_{1}\left({x}_{1},\dots ,{x}_{n}\right)\right]& =\frac{1}{{\left\vert \frac{\partial {q}_{1}}{\partial {x}_{1}}\right\vert }_{{x}_{1}={x}_{1}\left({q}_{1},{x}_{2},\dots ,{x}_{n}\right)}}\hfill \\ \hfill & =\frac{1}{{\mathcal{G}}_{1}\left({q}_{1},{x}_{2},\dots ,{x}_{n}\right)},\hfill \end{align} \tag{ 46 }$$

which leads to

$$\begin{equation}{I}_{n}=\int \mathrm{d}{q}_{1}\enspace {\int }_{-\infty }^{\infty }\mathrm{d}{x}_{2}\dots \mathrm{d}{x}_{n}\frac{1}{{\mathcal{G}}_{1}\left({q}_{1},{x}_{2},\dots ,{x}_{n}\right)}f\left({q}_{1},{x}_{2},\dots ,{x}_{n}\right).\end{equation} \tag{ 47 }$$

After the x₁ integration, every x₁ in the integrand of equation (7) and ${\mathcal{G}}_{1}$ in equation (46) is replaced with

$$\begin{equation}{x}_{1}={x}_{1}\left({q}_{1},{x}_{2},\dots ,{x}_{n}\right).\end{equation} \tag{ 48 }$$

The constraint equation coming from the convolution with the Dirac delta function in equation (46) is

$$\begin{equation}{q}_{1}-{q}_{1}\left[{x}_{1}\left({q}_{1},{x}_{2},\dots ,{x}_{n}\right),{x}_{2},\dots ,{x}_{n}\right]=0.\end{equation} \tag{ 49 }$$

Then, we carry out the integration over x_i for i = 1 through n − 1 by multiplying ${\mathbb{1}}_{i}$ to I_n in equation (47) sequentially. We assume that, after the integration over x_i for i = 1 through n − 1, the result of the integration of Dirac delta functions is

$$\begin{align}\hfill {\int }_{-\infty }^{\infty }\mathrm{d}{x}_{1}\dots \mathrm{d}{x}_{i}\enspace \prod\limits _{j=1}^{i}\delta \left[{q}_{j}-{q}_{j}\left({x}_{1},\dots ,{x}_{n}\right)\right]& =\frac{1}{{\mathcal{G}}_{1}}{\times}\frac{{\mathcal{G}}_{1}}{{\mathcal{G}}_{2}}{\times}\cdots {\times}\frac{{\mathcal{G}}_{i-1}}{{\mathcal{G}}_{i}}\hfill \\ \hfill & =\frac{1}{{\mathcal{G}}_{i}\left({q}_{1},\dots ,{q}_{i},{x}_{i+1},\dots ,{x}_{n}\right)},\hfill \end{align} \tag{ 50 }$$

which leads to

$$\begin{align}\hfill {I}_{n}& =\int \mathrm{d}{q}_{1}\dots \mathrm{d}{q}_{i}{\int }_{-\infty }^{\infty }\mathrm{d}{x}_{i+1}\dots \mathrm{d}{x}_{n}\hfill \\ \hfill & \quad {\times}\frac{1}{{\mathcal{G}}_{i}\left({q}_{1},\dots ,{q}_{i},{x}_{i+1},\dots ,{x}_{n}\right)}f\left({q}_{1},\dots ,{q}_{i},{x}_{i+1},\dots ,{x}_{n}\right).\hfill \end{align} \tag{ 51 }$$

For any j ⩽ i every x_j in the integrand of equation (5) and ${\mathcal{G}}_{j}$'s in equation (50) is replaced with

$$\begin{equation}{x}_{j}={x}_{j}\left({q}_{1},\dots ,{q}_{i},{x}_{i+1},\dots ,{x}_{n}\right)\end{equation} \tag{ 52 }$$

after the integrations over x₁ through x_i . There are i constraint equations coming from the convolution with Dirac delta functions in equation (50):

$$\begin{align}\hfill & {q}_{j}-{q}_{j}\left[{x}_{1}\left({q}_{1},\dots ,{q}_{i},{x}_{i+1},\dots ,{x}_{n}\right),\dots ,{x}_{i}\left({q}_{1},\dots ,{q}_{i},{x}_{i+1},\dots ,{x}_{n}\right),\right.\hfill \\ \hfill & \left.\quad {\times}{x}_{i+1},\dots ,{x}_{n}\right]=0,\hfill \end{align} \tag{ 53 }$$

where j runs from 1 through i.

After multiplying ${\mathbb{1}}_{i+1}$ to I_n in equation (51), we integrate out one more Cartesian coordinate x_i+1 to find that

$$\begin{align}\hfill & {\int }_{-\infty }^{\infty }\mathrm{d}{x}_{i+1}\delta {\left[{q}_{i+1}-{q}_{i+1}\left({q}_{1},\dots ,{q}_{i},{x}_{i+1},\dots ,{x}_{n}\right)\right]}_{{x}_{j}={x}_{j}\left({q}_{1},\dots ,{q}_{i},{x}_{i+1},\dots ,{x}_{n}\right)}\hfill \\ \hfill & =\frac{1}{{\left\vert {\left(\frac{\partial {q}_{i+1}}{\partial {x}_{i+1}}\right)}_{\left(i\right)}\right\vert }_{{x}_{k}={x}_{k}\left({q}_{1},\dots ,{q}_{i+1},{x}_{i+2},\dots ,{x}_{n}\right)}}\hfill \\ \hfill & =\frac{{\mathcal{G}}_{i}\left({q}_{1},\dots ,{q}_{i+1},{x}_{i+2},\dots ,{x}_{n}\right)}{{\mathcal{G}}_{i+1}\left({q}_{1},\dots ,{q}_{i+1},{x}_{i+2},\dots ,{x}_{n}\right)},\hfill \end{align} \tag{ 54 }$$

where 1 ⩽ j ⩽ i and 1 ⩽ k ⩽ i + 1. The proof of the last equality can be found in equation (C8) in appendix C. In combination with equations (50) and (54), we find that

$$\begin{align}\hfill & {\int }_{-\infty }^{\infty }\mathrm{d}{x}_{1}\dots \mathrm{d}{x}_{i}\enspace \mathrm{d}{x}_{i+1}\prod\limits _{j=1}^{i+1}\delta \left[{q}_{j}-{q}_{j}\left({x}_{1},\dots ,{x}_{n}\right)\right]\hfill \\ \hfill & \qquad =\frac{1}{{\mathcal{G}}_{1}}{\times}\frac{{\mathcal{G}}_{1}}{{\mathcal{G}}_{2}}{\times}\cdots {\times}\frac{{\mathcal{G}}_{i-1}}{{\mathcal{G}}_{i}}{\times}\frac{{\mathcal{G}}_{i}}{{\mathcal{G}}_{i+1}}=\frac{1}{{\mathcal{G}}_{i+1}\left({q}_{1},\dots ,{q}_{i+1},{x}_{i+2},\dots ,{x}_{n}\right)}.\hfill \end{align} \tag{ 55 }$$

For any j ⩽ i + 1 every x_j in the integrand of equation (7) and ${\mathcal{G}}_{j}$'s in equation (55) is replaced with

$$\begin{equation}{x}_{j}={x}_{j}\left({q}_{1},\dots ,{q}_{i+1},{x}_{i+2},\dots ,{x}_{n}\right)\end{equation} \tag{ 56 }$$

after the integrations over x₁ through x_i+1. There are i + 1 constraint equations coming from the convolution with Dirac delta functions in equation (55):

$$\begin{align}\hfill & {q}_{j}-{q}_{j}\left[{x}_{1}\left({q}_{1},\dots ,{q}_{i+1},{x}_{i+2},\dots ,{x}_{n}\right),\dots ,{x}_{i}\left({q}_{1},\dots ,{q}_{i+1},{x}_{i+2},\dots ,{x}_{n}\right),{x}_{i+1},\dots ,{x}_{n}\right]=0,\hfill \end{align} \tag{ 57 }$$

where j runs from 1 through i + 1. According to mathematical induction, this proves that the assumption in equation (50) with the constraints (52) and (53) is true for all i = 1 through n.

Finally, the integral I_n can be expressed in terms of q₁, ..., q_n as

$$\begin{align}\hfill {I}_{n}& =\int {\mathrm{d}}^{n}q\frac{1}{{\mathcal{G}}_{n}}f\left({x}_{1},\dots ,{x}_{n}\right){\vert }_{{x}_{k}={x}_{k}\left({q}_{1},\dots ,{q}_{n}\right)}\hfill \\ \hfill & =\int {\mathrm{d}}^{n}{q\mathcal{J}}_{n}f\left({x}_{1},\dots ,{x}_{n}\right){\vert }_{{x}_{k}={x}_{k}\left({q}_{1},\dots ,{q}_{n}\right)},\hfill \end{align} \tag{ 58 }$$

where 1 ⩽ k ⩽ n. This completes the proof of the Jacobian $\mathcal{J}{=\mathcal{J}}_{n}=1{/\mathcal{G}}_{n}$ for an n-dimensional coordinate transformation or change of variables.

In this subsection, we consider the change of variables from the three-dimensional Cartesian coordinates (x, y, z) to the spherical coordinates (r, θ, ϕ). The Cartesian coordinates can be expressed in terms of the radius r, the polar angle θ and the azimuthal angle ϕ as

$$\begin{equation}x=r\enspace \mathrm{sin}\enspace \theta \enspace \mathrm{cos}\enspace \phi ,\qquad y=r\enspace \mathrm{sin}\enspace \theta \enspace \mathrm{sin}\enspace \phi ,\qquad z=r\enspace \mathrm{cos}\enspace \theta .\end{equation} \tag{ 59 }$$

We use cos θ instead of θ as the integration variable and reorganize the order of multiple integrations in order to simplify the computation. That is, (x₁, x₂, x₃) in section 2.3 corresponds to (z, y, x) while (q₁, q₂, q₃) corresponds to (cos θ, ϕ, r), respectively. However, it turns out that the integral is invariant under this reordering. We also note that one can use, for instance, sin θ instead of cos θ as an integration variable and it will not alter the result of the integration. Conventionally, the arctangent function is defined in the region $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$. The period of the tangent function is π, while the azimuthal angle ranges from 0 to 2π. Thus the angle ϕ for x > 0 is set to be $\mathrm{arctan}\enspace \frac{y}{x}\in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ while that for x < 0 is set to be $\pi +\mathrm{arctan}\enspace \frac{y}{x}\in \left[\frac{\pi }{2},\frac{3\pi }{2}\right]$ in order to make the transformation function invertible in the entire range. Then the inverse transformation of equation (59) is expressed as

$$\begin{equation}\mathrm{cos}\enspace \theta =\frac{z}{\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}},\end{equation} \tag{ 60a }$$

$$\begin{equation}\phi =\begin{cases}\mathrm{arctan}\enspace \frac{y}{x}\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right),\quad \hfill & \mathrm{f}\mathrm{o}\mathrm{r}\enspace x{ >}0,\hfill \\ \pi +\mathrm{arctan}\enspace \frac{y}{x}\in \left(\frac{\pi }{2},\frac{3\pi }{2}\right),\quad \hfill & \mathrm{f}\mathrm{o}\mathrm{r}\enspace x{< }0,\hfill \\ \frac{\pi }{2},\quad \hfill & \mathrm{f}\mathrm{o}\mathrm{r}\enspace x=0\enspace \mathrm{a}\mathrm{n}\mathrm{d}\enspace y{ >}0,\hfill \\ -\frac{\pi }{2},\quad \hfill & \mathrm{f}\mathrm{o}\mathrm{r}\enspace x=0\enspace \mathrm{a}\mathrm{n}\mathrm{d}\enspace y{< }0,\hfill \\ 0,\quad \hfill & \mathrm{f}\mathrm{o}\mathrm{r}\enspace x=y=0,\hfill \end{cases}\end{equation} \tag{ 60b }$$

$$\begin{equation}r=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}.\end{equation} \tag{ 60c }$$

We consider a three-dimensional integral J₃ with an arbitrary integrand f(x, y, z)

$$\begin{equation}{J}_{3}={\int }_{-\infty }^{\infty }\mathrm{d}x\enspace \mathrm{d}y\enspace \mathrm{d}z\enspace f\left(x,y,z\right).\end{equation} \tag{ 61 }$$

We multiply the unities

$$\begin{equation}{\mathbb{1}}_{1}={\int }_{-1}^{1}\mathrm{d}\enspace \mathrm{cos}\enspace \theta \enspace \delta \left(\mathrm{cos}\enspace \theta -\frac{z}{\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}}\right)=1,\end{equation} \tag{ 62a }$$

$$\begin{equation}{\mathbb{1}}_{2}={\int }_{-\pi /2}^{3\pi /2}\mathrm{d}\phi \left[\delta \left(\phi -\mathrm{arctan}\enspace \frac{y}{x}\right){\Theta}\left(x\right)+\delta \left(\phi -\mathrm{arctan}\enspace \frac{y}{x}-\pi \right){\Theta}\left(-x\right)\right]=1,\end{equation} \tag{ 62b }$$

$$\begin{equation}{\mathbb{1}}_{3}={\int }_{0}^{\infty }\mathrm{d}r\enspace \delta \left(r-\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}\right)=1\end{equation} \tag{ 62c }$$

to J₃ without changing the value of the integral sequentially.

$$\begin{equation}{\Theta}\left(x\right)=\begin{cases}1,\quad \hfill & \mathrm{f}\mathrm{o}\mathrm{r}\enspace x{ >}0,\hfill \\ \frac{1}{2},\quad \hfill & \mathrm{f}\mathrm{o}\mathrm{r}\enspace x=0,\hfill \\ 0,\quad \hfill & \mathrm{f}\mathrm{o}\mathrm{r}\enspace x{< }0.\hfill \end{cases}\end{equation} \tag{ 63 }$$

Here, the Heaviside step function Θ(x) is defined by

First, we integrate out the x₁ = z coordinate by multiplying ${\mathbb{1}}_{1}$ in equation (62a) to J₃ in equation (61). The integration over z can be performed as

$$\begin{equation}{\int }_{-\infty }^{\infty }\mathrm{d}z\enspace \delta \left(\mathrm{cos}\enspace \theta -\frac{z}{\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}}\right)=\frac{\sqrt{{x}^{2}+{y}^{2}}}{{\mathrm{sin}}^{3}\enspace \theta }.\end{equation} \tag{ 64 }$$

After the integration over z, every z in the integrand of J₃ is replaced with the expression in terms of cos θ, y and x: $z=\sqrt{{x}^{2}+{y}^{2}}/\mathrm{tan}\enspace \theta$, where we omit the simple conversion between trigonometric functions here and after. The integrand of the remaining double integral over x and y is a function of θ, x and y:

$$\begin{equation}{J}_{3}={\int }_{-1}^{1}\mathrm{d}\enspace \mathrm{cos}\enspace \theta {\int }_{-\infty }^{\infty }\mathrm{d}x\enspace \mathrm{d}y\enspace f\left(x,y,\sqrt{{x}^{2}+{y}^{2}}/\mathrm{tan}\enspace \theta \right)\frac{\sqrt{{x}^{2}+{y}^{2}}}{{\mathrm{sin}}^{3}\enspace \theta }.\end{equation} \tag{ 65 }$$

This can also be obtained by substituting cos θ, y and x into equation (39).

Next, we integrate out the x₂ = y coordinate by multiplying ${\mathbb{1}}_{2}$ in equation (62b) into J₃ in equation (65). The integration over y can be performed as

$$\begin{align}\hfill & {\int }_{-\infty }^{\infty }\mathrm{d}y\left[\delta \left(\phi -\mathrm{arctan}\enspace \frac{y}{x}\right){\Theta}\left(x\right)+\delta \left(\phi -\mathrm{arctan}\enspace \frac{y}{x}-\pi \right){\Theta}\left(-x\right)\right]\hfill \\ \hfill & \qquad \qquad =\frac{1}{\vert \frac{\mathrm{d}}{\mathrm{d}y}\enspace \mathrm{arctan}\enspace \frac{y}{x}\vert }\left[{\Theta}\left(x\right)+{\Theta}\left(-x\right)\right]{\left\vert \right.}_{y=x\mathrm{tan}\phi }=\frac{\vert x\vert }{{\mathrm{cos}}^{2}\enspace \phi },\hfill \end{align} \tag{ 66 }$$

where we have used the identity

$$\begin{equation}{\Theta}\left(x\right)+{\Theta}\left(-x\right)=1.\end{equation} \tag{ 67 }$$

Equation (66) can also be obtained by substituting cos θ, ϕ and x into equation (41). After the integration over y, every y in the integrand of J₃ is replaced with the expression in terms of cos θ, ϕ and x. The integrand of the remaining integral over x is a function of cos θ, ϕ and x:

$$\begin{align}\hfill {J}_{3}& ={\int }_{-1}^{1}\mathrm{d}\enspace \mathrm{cos}\enspace \theta {\int }_{0}^{2\pi }\mathrm{d}\phi {\int }_{-\infty }^{\infty }\mathrm{d}xf\left(x,y,\sqrt{{x}^{2}+{y}^{2}}/\mathrm{tan}\enspace \theta \right)\frac{\sqrt{{x}^{2}+{y}^{2}}}{{\mathrm{sin}}^{3}\enspace \theta }\frac{\vert x\vert }{{\mathrm{cos}}^{2}\enspace \phi }{\left\vert \right.}_{y=x\mathrm{tan}\phi }\hfill \\ \hfill & ={\int }_{-1}^{1}\mathrm{d}\enspace \mathrm{cos}\enspace \theta {\int }_{0}^{2\pi }\mathrm{d}\phi {\int }_{-\infty }^{\infty }\mathrm{d}xf\left[x,x\enspace \mathrm{tan}\enspace \phi ,\vert x\vert /\left(\vert \mathrm{cos}\enspace \phi \vert \mathrm{tan}\enspace \theta \right)\right]\frac{{x}^{2}}{{\mathrm{sin}}^{3}\enspace \theta \vert \mathrm{cos}\enspace \phi {\vert }^{3}},\hfill \end{align} \tag{ 68 }$$

where y = x tan ϕ. After the integrations over both z and y, y = x tan ϕ and z = x/(cos ϕ  tan θ).

Finally, after multiplying ${\mathbb{1}}_{3}$ in equation (62c) into J₃ in equation (68), the integration over x₃ = x can be performed as

$$\begin{align}\hfill & {\int }_{-\infty }^{\infty }\mathrm{d}x\enspace \delta \left(r-\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}\right){\left\vert \right.}_{y=x\enspace \mathrm{tan}\enspace \phi ,\;z=\frac{x}{\mathrm{cos}\enspace \phi \enspace \mathrm{tan}\enspace \theta }}\hfill \\ \hfill & \qquad \qquad ={\int }_{-\infty }^{\infty }\mathrm{d}x\enspace \delta \left[r-\left\vert \frac{x}{\mathrm{cos}\enspace \phi \enspace \mathrm{sin}\enspace \theta }\right\vert \right]=\vert \mathrm{cos}\enspace \phi \vert \mathrm{sin}\enspace \theta .\hfill \end{align} \tag{ 69 }$$

This can also be obtained by substituting cos θ, ϕ and r into equation (43). Here, the radius r is non-negative because the Dirac delta function requires that $r=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$ and x² + y² + z² is non-negative. After the integration over all of the Cartesian coordinates, x, y and z are expressed as in equation (59).

Substituting equation (59) into (64), (66), (69), and f(x, y, z), we find that J₃ can be expressed in terms of the spherical coordinates as

$$\begin{align}\hfill {J}_{3}& ={\int }_{0}^{\infty }\mathrm{d}r{\int }_{-1}^{1}\mathrm{d}\enspace \mathrm{cos}\enspace \theta {\int }_{0}^{2\pi }\mathrm{d}\phi \hfill \\ \hfill & \quad {\times}\frac{r\enspace \mathrm{sin}\enspace \theta }{{\mathrm{sin}}^{3}\enspace \theta }\frac{\vert r\enspace \mathrm{sin}\enspace \theta \enspace \mathrm{cos}\enspace \phi \vert }{{\mathrm{cos}}^{2}\enspace \phi }\vert \mathrm{cos}\enspace \phi \vert \mathrm{sin}\enspace \theta f\left(r\enspace \mathrm{sin}\enspace \theta \enspace \mathrm{cos}\enspace \phi ,r\enspace \mathrm{sin}\enspace \theta \enspace \mathrm{sin}\enspace \phi ,r\enspace \mathrm{cos}\enspace \theta \right)\hfill \\ \hfill & ={\int }_{0}^{\infty }\mathrm{d}r{\int }_{-1}^{1}\mathrm{d}\enspace \mathrm{cos}\enspace \theta {\int }_{0}^{2\pi }\mathrm{d}\phi \enspace {r}^{2}f\left(r\enspace \mathrm{sin}\enspace \theta \enspace \mathrm{cos}\enspace \phi ,r\enspace \mathrm{sin}\enspace \theta \enspace \mathrm{sin}\enspace \phi ,r\enspace \mathrm{cos}\enspace \theta \right)\hfill \\ \hfill & ={\int }_{0}^{\infty }\mathrm{d}r{\int }_{0}^{\pi }\mathrm{d}\theta {\int }_{0}^{2\pi }\mathrm{d}\phi \enspace {r}^{2}\enspace \mathrm{sin}\enspace \theta f\left(r\enspace \mathrm{sin}\enspace \theta \enspace \mathrm{cos}\enspace \phi ,r\enspace \mathrm{sin}\enspace \theta \enspace \mathrm{sin}\enspace \phi ,r\enspace \mathrm{cos}\enspace \theta \right).\hfill \end{align} \tag{ 70 }$$

The overall factor r²  sin θ is identified with the Jacobian

$$\begin{equation}\mathcal{J}={r}^{2}\enspace \mathrm{sin}\enspace \theta .\end{equation} \tag{ 71 }$$

This exactly reproduces the result which can be obtained by applying the formula (44) [1].

In this subsection, we extend the three-dimensional case to the n-dimensional coordinate transformation from the n-dimensional Cartesian coordinates (x₁, ..., x_n ) to the n-dimensional polar coordinates (r, θ₁, θ₂, ..., θ_n−2, ϕ). Here, r is the radius and there are n − 2 polar angles θ_i 's and a single azimuthal angle ϕ. The corresponding transformation functions are expressed as

$$\begin{align}\hfill {x}_{1}& =r\enspace \mathrm{sin}\enspace {\theta }_{1}\enspace \mathrm{sin}\enspace {\theta }_{2}\dots \mathrm{sin}\enspace {\theta }_{n-3}\enspace \mathrm{sin}\enspace {\theta }_{n-2}\enspace \mathrm{cos}\enspace \phi ,\hfill & \hfill \\ \hfill {x}_{2}& =r\enspace \mathrm{sin}\enspace {\theta }_{1}\enspace \mathrm{sin}\enspace {\theta }_{2}\dots \mathrm{sin}\enspace {\theta }_{n-3}\enspace \mathrm{sin}\enspace {\theta }_{n-2}\enspace \mathrm{sin}\enspace \phi ,\hfill & \hfill \\ \hfill {x}_{3}& =r\enspace \mathrm{sin}\enspace {\theta }_{1}\enspace \mathrm{sin}\enspace {\theta }_{2}\dots \mathrm{sin}\enspace {\theta }_{n-3}\enspace \mathrm{cos}\enspace {\theta }_{n-2},\hfill & \hfill \\ \hfill & \;\;{\vdots}\hfill & \hfill \\ \hfill {x}_{n-1}& =r\enspace \mathrm{sin}\enspace {\theta }_{1}\enspace \mathrm{cos}\enspace {\theta }_{2},\hfill & \hfill \\ \hfill {x}_{n}& =r\enspace \mathrm{cos}\enspace {\theta }_{1}.\hfill & \hfill \end{align} \tag{ 72 }$$

We reorganize the order of integrations of the Cartesian coordinates as (x_n , x_n−1, ..., x₁) for simplicity and the corresponding curvilinear coordinates are reorganized as (cos θ₁, cos θ₂, ..., cos θ_n−2, ϕ, r). We consider an n-dimensional integral J_n with an arbitrary integrand f(x₁, x₂, ..., x_n ):

$$\begin{equation}{J}_{n}={\int }_{-\infty }^{\infty }{\mathrm{d}}^{n}x\enspace f\left({x}_{1},{x}_{2},\dots ,{x}_{n}\right),\end{equation} \tag{ 73 }$$

where dⁿ x = dx₁dx₂...dx_n . We multiply the unities

$$\begin{equation}{\mathbb{1}}_{i}={\int }_{-1}^{1}\mathrm{d}\enspace \mathrm{cos}\enspace {\theta }_{i}\enspace \delta \left(\mathrm{cos}\enspace {\theta }_{i}-\frac{{x}_{n-i+1}}{\sqrt{{x}_{1}^{2}+\cdots +{x}_{n-i+1}^{2}}}\right)=1,\quad i=1,\enspace \dots ,n-2,\end{equation} \tag{ 74a }$$

$$\begin{align}\hfill {\mathbb{1}}_{n-1}& ={\int }_{-\frac{\pi }{2}}^{\frac{3}{2}\pi }\mathrm{d}\phi \left[\delta \left(\phi -\mathrm{arctan}\enspace \frac{{x}_{2}}{{x}_{1}}\right){\Theta}\left({x}_{1}\right)+\delta \left(\phi -\mathrm{arctan}\enspace \frac{{x}_{2}}{{x}_{1}}-\pi \right){\Theta}\left(-{x}_{1}\right)\right]=1,\hfill \end{align} \tag{ 74b }$$

$$\begin{equation}{\mathbb{1}}_{n}={\int }_{0}^{\infty }\mathrm{d}r\enspace \delta \left(r-\sqrt{{x}_{1}^{2}+\cdots +{x}_{n}^{2}}\right)=1\end{equation} \tag{ 74c }$$

for i = 1 through n to J_n , sequentially, keeping the integral invariant. Θ(x) in equation (74b) is the Heaviside step function defined in equation (60b). There are numerous ways to perform the multiple integrations over the n Cartesian coordinates. Our strategy to integrate out x_i 's is as follows: according to the integrand of the right-hand side of equation (74a), the integration over x_n−i+1 in the integral J_n provides the constraint to the polar angle θ_i . Thus we choose to integrate over from x_n to x₃ to express them in terms of the polar angles from θ₁ through θ_n−2. Then we integrate out x₂ to express x_n through x₂ in terms of the n − 2 polar angles and the azimuthal angle ϕ by making use of equation (74b). As the last step, we integrate out x₁ to determine all of the Cartesian coordinates in terms of the spherical polar coordinates by making use of (74c).

First, after multiplying ${\mathbb{1}}_{i}$ in equation (74a) into J_n , the integration over x_n−i+1 for i = 1 through n − 2 can be performed as

$$\begin{equation}{\int }_{-\infty }^{\infty }\mathrm{d}{x}_{n-i+1}\enspace \delta \left(\mathrm{cos}\enspace {\theta }_{i}-\frac{{x}_{n-i+1}}{\sqrt{{x}_{1}^{2}+\cdots +{x}_{n-i+1}^{2}}}\right)=\frac{\sqrt{{x}_{1}^{2}+\cdots +{x}_{n-i}^{2}}}{{\mathrm{sin}}^{3}\enspace {\theta }_{i}},\end{equation} \tag{ 75 }$$

where one can obtain the same results from equations (46) and (54) taking care of the order of integration. After the integration over x_n−i+1, we make the replacement

$$\begin{equation}{x}_{n-i+1}=\frac{\sqrt{{x}_{1}^{2}+\cdots +{x}_{n-i}^{2}}}{\mathrm{tan}\enspace {\theta }_{i}}.\end{equation} \tag{ 76 }$$

Then, J_n is expressed as

$$\begin{align}\hfill {J}_{n}& ={\int }_{-1}^{1}\mathrm{d}\enspace \mathrm{cos}\enspace {\theta }_{1}\dots {\int }_{-1}^{1}\mathrm{d}\enspace \mathrm{cos}\enspace {\theta }_{n-2}\hfill \\ \hfill & \quad {\times}{\int }_{-\infty }^{\infty }\mathrm{d}{x}_{1}\enspace \mathrm{d}{x}_{2}f\left({x}_{1},{x}_{2},\mathrm{cos}\enspace {\theta }_{1},\dots ,\mathrm{cos}\enspace {\theta }_{n-2}\right)\prod\limits _{i=1}^{n-2}\frac{\sqrt{{x}_{1}^{2}+\cdots +{x}_{n-i}^{2}}}{{\mathrm{sin}}^{3}\enspace {\theta }_{i}},\hfill \end{align} \tag{ 77 }$$

where every x_i in the last factor for i = 3 through n is replaced by that in equation (76).

After multiplying ${\mathbb{1}}_{n-1}$ in equation (74b) into J_n in equation (77), the integration over x₂ can be carried out in a similar manner as is done in equation (66). The result is

$$\begin{align}\hfill & {\int }_{-\infty }^{\infty }\mathrm{d}{x}_{2}\enspace \left[\delta \left(\phi -\mathrm{arctan}\enspace \frac{{x}_{2}}{{x}_{1}}\right){\Theta}\left({x}_{1}\right)+\delta \left(\phi -\mathrm{arctan}\enspace \frac{{x}_{2}}{{x}_{1}}-\pi \right){\Theta}\left(-{x}_{1}\right)\right]\hfill \\ \hfill & \qquad \qquad \quad =\frac{\vert {x}_{1}\vert }{{\mathrm{cos}}^{2}\enspace \phi },\hfill \end{align} \tag{ 78 }$$

where x₂ = x₁  tan ϕ after the integration. This can also be obtained from equation (54) while keeping the results in equations (75) and (76). Then, x_n−i+1 for i = 1 through n − 1 can be expressed as

$$\begin{equation}{x}_{n-i+1}=\frac{\vert {x}_{1}\vert }{\vert \mathrm{cos}\phi \vert \mathrm{sin}\enspace {\theta }_{n-2}\dots \mathrm{sin}\enspace {\theta }_{i+1}\enspace \mathrm{tan}\enspace {\theta }_{i}}.\end{equation} \tag{ 79 }$$

Then, we find that

$$\begin{align}\hfill {J}_{n}& ={\int }_{-1}^{1}\enspace \mathrm{d}\enspace \mathrm{cos}\enspace {\theta }_{1}\dots {\int }_{-1}^{1}\mathrm{d}\enspace \mathrm{cos}\enspace {\theta }_{n-2}{\int }_{0}^{2\pi }\mathrm{d}\phi \enspace \hfill \\ \hfill & \quad {\times}{\int }_{-\infty }^{\infty }\mathrm{d}{x}_{1}f\left({x}_{1},\phi ,\mathrm{cos}\enspace {\theta }_{1},\dots ,\mathrm{cos}\enspace {\theta }_{n-2}\right)\enspace \frac{\vert {x}_{1}\vert }{{\mathrm{cos}}^{2}\enspace \phi }\prod\limits _{i=1}^{n-2}\frac{\sqrt{{x}_{1}^{2}+\cdots +{x}_{n-i}^{2}}}{{\mathrm{sin}}^{3}\enspace {\theta }_{i}},\hfill \end{align} \tag{ 80 }$$

where every x_i in the last factor for i = 2 through n is replaced by that in equation (79).

Finally, after multiplying ${\mathbb{1}}_{n}$ in equation (74c) to J_n in equation (80), the integration over x₁ can be performed like equation (69) and we obtain

$$\begin{align}\hfill {\int }_{-\infty }^{\infty }\mathrm{d}{x}_{1}\enspace \delta \left(r-\sqrt{{x}_{1}^{2}+\cdots +{x}_{n}^{2}}\right)& ={\int }_{-\infty }^{\infty }\mathrm{d}{x}_{1}\enspace \delta \left[r-\left\vert \frac{{x}_{1}}{\mathrm{cos}\enspace \phi \enspace \mathrm{sin}\enspace {\theta }_{n-2}\dots \mathrm{sin}\enspace {\theta }_{1}}\right\vert \right]\hfill \\ \hfill & =\left\vert \mathrm{cos}\enspace \phi \right\vert \mathrm{sin}\enspace {\theta }_{n-2}\dots \mathrm{sin}\enspace {\theta }_{1},\hfill \end{align} \tag{ 81 }$$

where we have omitted the replacements of x₂ = x₁  tan ϕ and x₃ through x_n that can be obtained from equation (79) on the left-hand side. This result can also be obtained from equation (54) while keeping the results in equations (75), (76), (78) and (79). After integrating out all of the Cartesian coordinates, we reproduce the expression for every x_i that is given in equation (72).

Combining all of the results listed above, we find that the n-dimensional coordinate transformation or change of variables is carried out as

$$\begin{align}\hfill {J}_{n}& ={\int }_{0}^{\infty }\mathrm{d}r{\int }_{-1}^{1}\mathrm{d}\enspace \mathrm{cos}\enspace {\theta }_{1}\dots {\int }_{-1}^{1}\mathrm{d}\enspace \mathrm{cos}\enspace {\theta }_{n-2}{\int }_{0}^{2\pi }\mathrm{d}\phi \;\;\frac{r\enspace \mathrm{sin}\enspace {\theta }_{1}}{{\mathrm{sin}}^{3}\enspace {\theta }_{1}}\dots \frac{r\enspace \mathrm{sin}\enspace {\theta }_{1}\dots \mathrm{sin}\enspace {\theta }_{n-2}}{{\mathrm{sin}}^{3}\enspace {\theta }_{n-2}}\hfill \\ \hfill & \quad {\times}\frac{r\enspace \mathrm{sin}\enspace {\theta }_{1}\dots \mathrm{sin}\enspace {\theta }_{n-2}\vert \mathrm{cos}\enspace \phi \vert }{{\mathrm{cos}}^{2}\enspace \phi }\vert \mathrm{cos}\enspace \phi \vert \mathrm{sin}\enspace {\theta }_{1}\dots \mathrm{sin}\enspace {\theta }_{n-2}\hfill \\ \hfill & \quad {\times}f\left(r\enspace \mathrm{sin}\enspace {\theta }_{1}\dots \mathrm{sin}\enspace {\theta }_{n-2}\enspace \mathrm{cos}\enspace \phi ,r\enspace \mathrm{sin}\enspace {\theta }_{1}\dots \mathrm{sin}\enspace {\theta }_{n-2}\enspace \mathrm{sin}\enspace \phi ,\right.\hfill \\ \hfill & \left.\quad {\times}r\enspace \mathrm{sin}\enspace {\theta }_{1}\dots \mathrm{sin}\enspace {\theta }_{n-3}\enspace \mathrm{cos}\enspace {\theta }_{n-2},\dots ,r\enspace \mathrm{sin}\enspace {\theta }_{1}\enspace \mathrm{cos}\enspace {\theta }_{2},r\enspace \mathrm{cos}\enspace {\theta }_{1}\right)\hfill \\ \hfill & ={\int }_{0}^{\infty }\mathrm{d}r{\int }_{0}^{\pi }\mathrm{d}{\theta }_{1}\dots {\int }_{0}^{\pi }\mathrm{d}{\theta }_{n-2}{\int }_{0}^{2\pi }\mathrm{d}\phi {r}^{n-1}\enspace {\mathrm{sin}}^{n-2}\enspace {\theta }_{1}\dots {\mathrm{sin}}^{2}\enspace {\theta }_{n-3}\enspace \mathrm{sin}\enspace {\theta }_{n-2}\hfill \\ \hfill & \quad {\times}f\left(r\enspace \mathrm{sin}\enspace {\theta }_{1}\dots \mathrm{sin}\enspace {\theta }_{n-2}\enspace \mathrm{cos}\enspace \phi ,r\enspace \mathrm{sin}\enspace {\theta }_{1}\dots \mathrm{sin}\enspace {\theta }_{n-2}\enspace \mathrm{sin}\enspace \phi ,\right.\hfill \\ \hfill & \left.\quad {\times}r\enspace \mathrm{sin}\enspace {\theta }_{1}\dots \mathrm{sin}\enspace {\theta }_{n-3}\enspace \mathrm{cos}\enspace {\theta }_{n-2},\dots ,r\enspace \mathrm{sin}\enspace {\theta }_{1}\enspace \mathrm{cos}\enspace {\theta }_{2},r\enspace \mathrm{cos}\enspace {\theta }_{1}\right).\hfill \end{align} \tag{ 82 }$$

The extra factor rⁿ⁻¹  sinⁿ⁻² θ₁...sin² θ_n−3  sin θ_n−2 in front of the original integrand f is identified with the Jacobian

$$\begin{equation}\mathcal{J}={r}^{n-1}\enspace {\mathrm{sin}}^{n-2}\enspace {\theta }_{1}\dots {\mathrm{sin}}^{2}\enspace {\theta }_{n-3}\enspace \mathrm{sin}\enspace {\theta }_{n-2}.\end{equation} \tag{ 83 }$$

This exactly reproduces the result in references [5, 6], which can be obtained by applying the general formula (58).