# I. Triangle Surface Element
To evaluate the integral of $\exp(i(\mathbf{q} \cdot \mathbf{r}))$ over the triangle defined by three vertices $\mathbf{r}_1, \mathbf{r}_2, \mathbf{r}_3$ in 3D space, we will perform the following steps:

## 1. **Define the Integration**
The integral is over the triangle, which is a two-dimensional region embedded in 3D space. We use the parametrization of the triangle to express $\mathbf{r}$ as a combination of the vertices:
$$
\mathbf{r}(u, v) = (1-u-v)\mathbf{r}_1 + u\mathbf{r}_2 + v\mathbf{r}_3,
$$
where $u, v \geq 0$ and $u+v \leq 1$.

## 2. **Compute the Integral**
The integral is given by
$$
\int_{\text{triangle}} e^{i (\mathbf{q} \cdot \mathbf{r})} \, dA,
$$
The integral over the triangle can be transformed into an integral over $u$ and $v$:
$$
\int_{\text{triangle}} e^{i (\mathbf{q} \cdot \mathbf{r})} \, dA = \int_0^1 \int_0^{1-u} e^{i (\mathbf{q} \cdot \mathbf{r}(u,v))} |\mathbf{n}| \, dv \, du.
$$
Here, $|\mathbf{n}| = 2A$, where:
$$
A = \frac{1}{2} \| (\mathbf{r}_2 - \mathbf{r}_1) \times (\mathbf{r}_3 - \mathbf{r}_1) \|
$$
is the area of the triangle.

## 3. Express the result as $\mathbf{q} \cdot \mathbf{r}_1, \mathbf{q} \cdot \mathbf{r}_2, \mathbf{q} \cdot \mathbf{r}_3$
If we denote the known values of $\mathbf{q} \cdot \mathbf{r}_1, \mathbf{q} \cdot \mathbf{r}_2, \mathbf{q} \cdot \mathbf{r}_3$ as $\text{qr1}, \text{qr2}, \text{qr3}$, the expression for the integral simplifies to:

$$
|\mathbf{n}|\int_0^1 \int_0^{1-u} e^{i \big((1-u-v)\text{qr1} + u\text{qr2} + v\text{qr3}\big)} \, dv \, du.
$$

Now we focus on the double integral.

### 3.1 **Substitute the exponent**
$$
(1-u-v)\text{qr1} + u\text{qr2} + v\text{qr3} = \text{qr1} + u(\text{qr2} - \text{qr1}) + v(\text{qr3} - \text{qr1}).
$$

### 3.2 **Factorize the exponential**
$$
e^{i\text{qr1}} \int_0^1 \int_0^{1-u} e^{i \big(u(\text{qr2} - \text{qr1}) + v(\text{qr3} - \text{qr1})\big)} \, dv \, du.
$$

### 3.3 **Perform the $v$-integration**
Since $(v$ is independent of $u$ within the exponential term:
$$
\int_0^{1-u} e^{i v (\text{qr3} - \text{qr1})} dv = \frac{e^{i(1-u)(\text{qr3} - \text{qr1})} - 1}{i(\text{qr3} - \text{qr1})}.
$$

### 3.4. **Substitute back and integrate over $u$**
The remaining integral over $u$ involves the term:
$$
\int_0^1 e^{i u (\text{qr2} - \text{qr1})} \frac{e^{i(1-u)(\text{qr3} - \text{qr1})} - 1}{i(\text{qr3} - \text{qr1})} du.
$$

## 4. Evaluate the integral

$$
\int_0^1 \frac{e^{i u (\text{qr2} - \text{qr1})} \big(e^{i (1-u)(\text{qr3} - \text{qr1})} - 1\big)}{i (\text{qr3} - \text{qr1})} \, du.
$$

$$
= \frac{e^{i u(\text{qr3} - \text{qr1})}}{i(\text{qr3} - \text{qr1})}\int_0^1 e^{i u(\text{qr2} - \text{qr3})}\, du + \int_0^1 e^{i u(\text{qr2} - \text{qr1})}\, du
$$

<!-- ### 4.1. **Simplify the exponentials**:
Expand $e^{i(1-u)(\text{qr3} - \text{qr1})}$:
$$
e^{i(1-u)(\text{qr3} - \text{qr1})} = e^{i (\text{qr3} - \text{qr1})} e^{-i u (\text{qr3} - \text{qr1})}.
$$

Substitute this into the integral:
$$
\int_0^1 \frac{e^{i u (\text{qr2} - \text{qr1})} \big(e^{i (\text{qr3} - \text{qr1})} e^{-i u (\text{qr3} - \text{qr1})} - 1\big)}{i (\text{qr3} - \text{qr1})} \, du.
$$

Factor out constants:
$$
= \frac{e^{i (\text{qr3} - \text{qr1})}}{i (\text{qr3} - \text{qr1})} \int_0^1 e^{i u (\text{qr2} - \text{qr3})} \, du- \frac{1}{i (\text{qr3} - \text{qr1})} \int_0^1 e^{i u (\text{qr2} - \text{qr1})} \, du.
$$ -->

### 4.2. **Evaluate the basic exponential integrals**:
For any $a \neq 0$,
$$
\int_0^1 e^{i a u} \, du = \frac{e^{i a} - 1}{i a}.
$$

Apply this formula to both terms:
- For $\int_0^1 e^{i u (\text{qr2} - \text{qr3})} \, du$:
$$
\frac{e^{i (\text{qr2} - \text{qr3})} - 1}{i (\text{qr2} - \text{qr3})}.
$$
- For $\int_0^1 e^{i u (\text{qr2} - \text{qr1})} \, du$:
$$
\frac{e^{i (\text{qr2} - \text{qr1})} - 1}{i (\text{qr2} - \text{qr1})}.
$$

### 4.3. **Combine terms**:
Substitute these results back into the original expression:
$$
\frac{e^{i (\text{qr3} - \text{qr1})}}{i (\text{qr3} - \text{qr1})} \cdot \frac{e^{i (\text{qr2} - \text{qr3})} - 1}{i (\text{qr2} - \text{qr3})}- \frac{1}{i (\text{qr3} - \text{qr1})} \cdot \frac{e^{i (\text{qr2} - \text{qr1})} - 1}{i (\text{qr2} - \text{qr1})}.
$$
$$
=\frac{e^{i (\text{qr3} - \text{qr1})} \big(e^{i (\text{qr2} - \text{qr3})} - 1\big)}{- (\text{qr3} - \text{qr1})(\text{qr2} - \text{qr3})}- \frac{e^{i (\text{qr2} - \text{qr1})} - 1}{- (\text{qr3} - \text{qr1})(\text{qr2} - \text{qr1})}.
$$

<!-- $$
\frac{e^{i (\text{qr3} - \text{qr1})} \big(e^{i (\text{qr2} - \text{qr3})} - 1\big)}{- (\text{qr3} - \text{qr1})(\text{qr2} - \text{qr3})}- \frac{e^{i (\text{qr2} - \text{qr1})} - 1}{- (\text{qr3} - \text{qr1})(\text{qr2} - \text{qr1})}.
$$ -->

Include the factor $e^{i \, \text{qr1}}$ back to the integral:

$$
e^{i \, \text{qr1}} \left(
\frac{e^{i (\text{qr3} - \text{qr1})} \big(e^{i (\text{qr2} - \text{qr3})} - 1\big)}{- (\text{qr3} - \text{qr1})(\text{qr2} - \text{qr3})}- \frac{e^{i (\text{qr2} - \text{qr1})} - 1}{- (\text{qr3} - \text{qr1})(\text{qr2} - \text{qr1})}
\right).
$$

## 6. A more symmetric form
The original expression
$$
e^{i \text{qr1}} \left(\frac{e^{i (\text{qr3} - \text{qr1})} (e^{i (\text{qr2} - \text{qr3})} - 1)}{- (\text{qr3} - \text{qr1})(\text{qr2} - \text{qr3})}- \frac{e^{i (\text{qr2} - \text{qr1})} - 1}{- (\text{qr3} - \text{qr1})(\text{qr2} - \text{qr1})}\right).
$$

Factor out $- (\text{qr3} - \text{qr1})(\text{qr2} - \text{qr3})(\text{qr2} - \text{qr1})$ as the denominator, yielding:
$$
\frac{e^{i \text{qr1}} \big[
e^{i (\text{qr3} - \text{qr1})} (e^{i (\text{qr2} - \text{qr3})} - 1) (\text{qr2} - \text{qr1})- (e^{i (\text{qr2} - \text{qr1})} - 1) (\text{qr2} - \text{qr3})
\big]}{- (\text{qr3} - \text{qr1})(\text{qr2} - \text{qr3})(\text{qr2} - \text{qr1})}.
$$

$$
=\frac{\big[
(e^{i (\text{qr2})} - e^{i (\text{qr3})}) (\text{qr2} - \text{qr1})- (e^{i (\text{qr2})} - e^{i (\text{qr1})}) (\text{qr2} - \text{qr3})
\big]}{- (\text{qr3} - \text{qr1})(\text{qr2} - \text{qr3})(\text{qr2} - \text{qr1})}.
$$

The numerator terms can now be reorganized
- The expression has a natural cyclic dependence on $\text{qr1}, \text{qr2}, \text{qr3}$.
- Using the symmetry of exponential terms, you can observe that each vertex contributes a term involving the phase difference with the other two vertices.

The symmetric form of the result is:

$$
\frac{e^{i \text{qr3}} (\text{qr1} - \text{qr2}) + e^{i \text{qr1}} (\text{qr2} - \text{qr3}) + e^{i \text{qr2}} (\text{qr3} - \text{qr1})}{(\text{qr1} - \text{qr2})(\text{qr2} - \text{qr3})(\text{qr3} - \text{qr1})}.
$$

## 7. The final expression
$$
\int_{\text{triangle}} e^{i (\mathbf{q} \cdot \mathbf{r})} \, dA = |\mathbf{n}| \cdot \frac{e^{i \text{qr3}} (\text{qr1} - \text{qr2}) + e^{i \text{qr1}} (\text{qr2} - \text{qr3}) + e^{i \text{qr2}} (\text{qr3} - \text{qr1})}{(\text{qr1} - \text{qr2})(\text{qr2} - \text{qr3})(\text{qr3} - \text{qr1})}.
$$

<!-- ## 8. Test the Scattering Amplitude for a Disk
To derive the scattering amplitude for a disk, we approximate the disk as a collection of pie-shaped triangles. By summing the contributions of individual triangles and transitioning to a continuous limit, we compute the scattering amplitude.

To derive $F(q, \alpha)$, the scattering amplitude of a disk on the $xy$-plane with radius $R$, using the given triangle integration formula:

$$
\int_{\text{triangle}} e^{i (\mathbf{q} \cdot \mathbf{r})} \, dA = |\mathbf{n}| \cdot
\frac{
e^{i \text{qr3}} (\text{qr1} - \text{qr2}) +
e^{i \text{qr1}} (\text{qr2} - \text{qr3}) +
e^{i \text{qr2}} (\text{qr3} - \text{qr1})
}{
(\text{qr1} - \text{qr2})(\text{qr2} - \text{qr3})(\text{qr3} - \text{qr1})},
$$

we follow these steps:

### 8.1. Disk Geometry and Wave Vector Setup
The disk lies in the $xy$-plane, and the wave vector $\mathbf{q}$ forms an angle $\alpha$ with the $z$-axis. Let:
$$
\mathbf{q} = (q_x, q_y, q_z) = (q \sin\alpha \cos\phi_q, q \sin\alpha \sin\phi_q, q \cos\alpha),
$$
where $q = |\mathbf{q}|$ is the magnitude of the wave vector and $\phi_q$ is the azimuthal angle of $\mathbf{q}$.

The disk is approximated as $N$ pie-shaped triangles:
- The center of the disk is $\mathbf{r}_1 = (0, 0, 0)$,
- The vertices on the circumference are:
$$
\mathbf{r}_2 = (R \cos\theta, R \sin\theta, 0),
$$
$$
\mathbf{r}_3 = (R \cos(\theta + \Delta\theta), R \sin(\theta + \Delta\theta), 0),
$$
where $\Delta\theta = \frac{2\pi}{N}$.

### 8.2. Dot Products $\text{qr1}$, $\text{qr2}$, and $\text{qr3}$
The dot products are computed as:
1. $\text{qr1} = \mathbf{q} \cdot \mathbf{r}_1 = 0$,
2. $\text{qr2} = \mathbf{q} \cdot \mathbf{r}_2 = q_x R \cos\theta + q_y R \sin\theta$,
3. $\text{qr3} = \mathbf{q} \cdot \mathbf{r}_3 = q_x R \cos(\theta + \Delta\theta) + q_y R \sin(\theta + \Delta\theta)$.

In polar coordinates for $\mathbf{q}$, substituting $q_x = q \sin\alpha \cos\phi_q$ and $q_y = q \sin\alpha \sin\phi_q$, we find:
$$
\text{qr2} = q R \sin\alpha \cos(\theta - \phi_q),
$$
$$
\text{qr3} = q R \sin\alpha \cos(\theta + \Delta\theta - \phi_q).
$$

### 8.3. Triangle Area and Normal Vector
The area of each triangle:
$$
|\mathbf{n}| = \frac{1}{2} R^2 \Delta\theta.
$$

### 8.4. Contribution of a Single Triangle
Substituting $\text{qr1}$, $\text{qr2}$, and $\text{qr3}$ into the formula, the contribution from one triangle becomes:
$$
\int_{\text{triangle}} e^{i (\mathbf{q} \cdot \mathbf{r})} \, dA \approx \frac{1}{2} R^2 \Delta\theta \cdot
\frac{
e^{i \text{qr3}} (\text{qr1} - \text{qr2}) +
e^{i \text{qr1}} (\text{qr2} - \text{qr3}) +
e^{i \text{qr2}} (\text{qr3} - \text{qr1})
}{
(\text{qr1} - \text{qr2})(\text{qr2} - \text{qr3})(\text{qr3} - \text{qr1})}.
$$

### 8.5. Transition to Continuous Limit
In the limit as $N \to \infty$ ($\Delta\theta \to 0$), the disk's scattering amplitude is the sum of contributions from all triangles:
$$
F(q, \alpha) = \int_{\text{disk}} e^{i (\mathbf{q} \cdot \mathbf{r})} \, dA.
$$

Switching to polar coordinates:
- $\mathbf{r} = (r \cos\theta, r \sin\theta, 0)$,
- $\mathbf{q} \cdot \mathbf{r} = q r \sin\alpha \cos(\theta - \phi_q)$.

Thus:
$$
F(q, \alpha) = \int_0^{2\pi} \int_0^R e^{i q r \sin\alpha \cos(\theta - \phi_q)} \, r \, dr \, d\theta.
$$

### 8.6. Evaluate the Radial Integral
The radial integral is:
$$
\int_0^R r e^{i q r \sin\alpha \cos(\theta - \phi_q)} \, dr.
$$

Using the result for $\int_0^R r e^{i a r} \, dr$:
$$
\int_0^R r e^{i a r} \, dr = \frac{e^{i a R} - 1}{a^2} - \frac{i R e^{i a R}}{a},
$$
where $a = q \sin\alpha \cos(\theta - \phi_q)$.

### 8.7. Evaluate the Angular Integral
The angular integral becomes:
$$
\int_0^{2\pi} e^{i q r \sin\alpha \cos(\theta - \phi_q)} \, d\theta = 2\pi J_0(q r \sin\alpha),
$$
where $J_0(x)$ is the Bessel function of the first kind.

### 8.8. Final Expression
Substituting back into the full expression:
$$
F(q, \alpha) = 2\pi \int_0^R r J_0(q r \sin\alpha) \, dr.
$$

The integral of $r J_0(x)$ gives:
$$
\int_0^R r J_0(q r \sin\alpha) \, dr = \frac{R^2 J_1(q R \sin\alpha)}{q R \sin\alpha}.
$$

Thus, the scattering amplitude for the disk is:
$$
F(q, \alpha) = 2\pi R^2 \frac{J_1(q R \sin\alpha)}{q R \sin\alpha}.
$$

**Fournet (1949) Expression of Cylinder:**
$$
F(q, \alpha) = 2 (\Delta \rho) V \cdot
\frac{\sin \left(\tfrac12 qL\cos\alpha \right)}
{\tfrac12 qL \cos \alpha}
\cdot
\frac{J_1 \left(q R \sin \alpha\right)}{q R \sin \alpha}.
$$

Both results coincide when $L \rightarrow 0$. -->

---

# II. Tetrahedron Volume Element
To evaluate the integral of $\exp(i(\mathbf{q} \cdot \mathbf{r}))$ over the tetrahedron defined by vertices $\mathbf{r}_1, \mathbf{r}_2, \mathbf{r}_3, \mathbf{r}_4$ in 3D space, we follow a similar approach to the triangle case.

## 1. **Setup and Parametrization of the Tetrahedron**
The integral is:
$$
\int_{\text{tetrahedron}} e^{i (\mathbf{q} \cdot \mathbf{r})} \, dV,
$$
where $\mathbf{r}$ lies within the tetrahedron. Using barycentric coordinates, any point $\mathbf{r}$ inside the tetrahedron can be expressed as:
$$
\mathbf{r}(u, v, w) = (1-u-v-w)\mathbf{r}_1 + u\mathbf{r}_2 + v\mathbf{r}_3 + w\mathbf{r}_4,
$$
where $u, v, w \geq 0$ and $u+v+w \leq 1$.

The differential volume element $dV$ is proportional to the determinant of the vectors forming the tetrahedron:
$$
dV = |\mathbf{n}| \, du \, dv \, dw,
$$
where $|\mathbf{n}| = 6V$, and $V$ is the volume of the tetrahedron:
$$
V = \frac{1}{6} \left| \det\left([\mathbf{r}_2 - \mathbf{r}_1, \mathbf{r}_3 - \mathbf{r}_1, \mathbf{r}_4 - \mathbf{r}_1]\right) \right|.
$$

## 2. **Integration**
Substitute the parametrization of \(\mathbf{r}\) into $\exp(i (\mathbf{q} \cdot \mathbf{r}))$:
$$
\mathbf{q} \cdot \mathbf{r}(u, v, w) = (1-u-v-w)(\mathbf{q} \cdot \mathbf{r}_1) + u(\mathbf{q} \cdot \mathbf{r}_2) + v(\mathbf{q} \cdot \mathbf{r}_3) + w(\mathbf{q} \cdot \mathbf{r}_4).
$$
Define:
$$
\text{qr1} = \mathbf{q} \cdot \mathbf{r}_1, \quad \text{qr2} = \mathbf{q} \cdot \mathbf{r}_2, \quad \text{qr3} = \mathbf{q} \cdot \mathbf{r}_3, \quad \text{qr4} = \mathbf{q} \cdot \mathbf{r}_4.
$$
Then:
$$
\mathbf{q} \cdot \mathbf{r}(u, v, w) = \text{qr1} + u(\text{qr2} - \text{qr1}) + v(\text{qr3} - \text{qr1}) + w(\text{qr4} - \text{qr1}).
$$

The integral becomes:
$$
\int_{\text{tetrahedron}} e^{i (\mathbf{q} \cdot \mathbf{r})} \, dV = |\mathbf{n}| \int_0^1 \int_0^{1-u} \int_0^{1-u-v} e^{i (\text{qr1} + u(\text{qr2} - \text{qr1}) + v(\text{qr3} - \text{qr1}) + w(\text{qr4} - \text{qr1}))} \, dw \, dv \, du.
$$

## 3. **Simplify the Exponential**
Factorize the exponential:
$$
e^{i \text{qr1}} \int_0^1 \int_0^{1-u} \int_0^{1-u-v} e^{i \big(u(\text{qr2} - \text{qr1}) + v(\text{qr3} - \text{qr1}) + w(\text{qr4} - \text{qr1})\big)} \, dw \, dv \, du.
$$

Perform the integration over \(w\), then \(v\), and finally \(u\), following the sequence of nested integrals.

## 4. **Evaluate the $w$-Integral**
The \(w\)-dependent term is:
$$
\int_0^{1-u-v} e^{i w (\text{qr4} - \text{qr1})} \, dw = \frac{e^{i (1-u-v)(\text{qr4} - \text{qr1})} - 1}{i (\text{qr4} - \text{qr1})}.
$$

## 5. **Evaluate the $v$-Integral**
The result of the \(v\)-integration involves substituting the \(w\)-integral:
$$
\int_0^{1-u} e^{i v (\text{qr3} - \text{qr1})} \cdot \frac{e^{i (1-u-v)(\text{qr4} - \text{qr1})} - 1}{i (\text{qr4} - \text{qr1})} \, dv.
$$
Solve this using the same technique, applying the formula for the integral of an exponential.

## 6. **Evaluate the $u$-Integral**
The \(u\)-integral similarly involves the result of the \(v\)-integration:
$$
\int_0^1 e^{i u (\text{qr2} - \text{qr1})} \cdot \text{(result from \(v\)-integration)} \, du.
$$

## 7. **Final Expression**
After evaluating all integrals, the final result is:
$$
\int_{\text{tetrahedron}} e^{i (\mathbf{q} \cdot \mathbf{r})} \, dV = |\mathbf{n}| \cdot \frac{\text{(cyclic sum of terms involving \(e^{i \text{qr}}\))}}{\prod_{1 \leq i < j \leq 4} (\text{qr}_i - \text{qr}_j)}.
$$

The integral of $\exp(i (\mathbf{q} \cdot \mathbf{r}))$ over the tetrahedron defined by the vertices $\mathbf{r}_1, \mathbf{r}_2, \mathbf{r}_3, \mathbf{r}_4$ can be expressed as:

$$
\int_{\text{tetrahedron}} e^{i (\mathbf{q} \cdot \mathbf{r})} \, dV = |\mathbf{n}| \cdot \frac{\text{Numerator}}{\text{Denominator}},
$$

where the **Denominator** is:
$$
\text{Denominator} = (\text{qr1} - \text{qr2})(\text{qr1} - \text{qr3})(\text{qr1} - \text{qr4})(\text{qr2} - \text{qr3})(\text{qr2} - \text{qr4})(\text{qr3} - \text{qr4}),
$$

and the **Numerator** is the cyclic sum of contributions from all vertices:
$$
\text{Numerator} = e^{i \text{qr1}} \big[ (\text{qr2} - \text{qr3})(\text{qr3} - \text{qr4})(\text{qr4} - \text{qr2}) \big]
$$
$$+ e^{i \text{qr2}} \big[ (\text{qr3} - \text{qr4})(\text{qr4} - \text{qr1})(\text{qr1} - \text{qr3}) \big]
$$
$$+ e^{i \text{qr3}} \big[ (\text{qr4} - \text{qr1})(\text{qr1} - \text{qr2})(\text{qr2} - \text{qr4}) \big]
$$
$$+ e^{i \text{qr4}} \big[ (\text{qr1} - \text{qr2})(\text{qr2} - \text{qr3})(\text{qr3} - \text{qr1}) \big].
$$

This result is symmetric under any permutation of the vertices, as expected for a tetrahedron. Each term in the numerator captures the cyclic contributions of the vertices to the integral, similar to the result for the triangle but generalized for four vertices.

We can write a general formula for the integral of $\exp(i(\mathbf{q} \cdot \mathbf{r}))$ over a $D$-dimensional simplex defined by vertices $\mathbf{r}_1, \dots, \mathbf{r}_{D+1}$ in 3D space. Let’s derive the general pattern for $D = 0, 1, 2, 3$:

---
# **General Form**
For a $D$-dimensional simplex:
$$
\int_{\text{simplex}} e^{i (\mathbf{q} \cdot \mathbf{r})} \, dV_D = |\mathbf{n}| \cdot \frac{\text{Numerator}}{\text{Denominator}},
$$
where:

- **Denominator**: The product of pairwise differences of the $\text{qr}_i = \mathbf{q} \cdot \mathbf{r}_i$:
$$
\text{Denominator} = \prod_{1 \leq i < j \leq D+1} (\text{qr}_i - \text{qr}_j).
$$

- **Numerator**: A cyclic sum over all vertices:
$$
\text{Numerator} = \sum_{k=1}^{D+1} e^{i \text{qr}_k} \prod_{\substack{j=1 \\ j \neq k}}^{D+1} (\text{qr}_j - \text{qr}_\ell)_{j, \ell \neq k}.
$$

- **Magnitude**: $|\mathbf{n}|$ accounts for the volume scaling:
- For $D = 0$: $|\mathbf{n}| = 1$.
- For $D = 1$: $|\mathbf{n}| = \| \mathbf{r}_2 - \mathbf{r}_1 \|$.
- For $D = 2$: $|\mathbf{n}| = 2 \times \text{Area of triangle} = \| (\mathbf{r}_2 - \mathbf{r}_1) \times (\mathbf{r}_3 - \mathbf{r}_1) \|$.
- For $D = 3$: $|\mathbf{n}| = 6 \times \text{Volume of tetrahedron} = |\det([\mathbf{r}_2 - \mathbf{r}_1, \mathbf{r}_3 - \mathbf{r}_1, \mathbf{r}_4 - \mathbf{r}_1])|$.

---

### **Explicit Results for $D = 0, 1, 2, 3$**

#### **0D: Point**
For $D = 0$, a single vertex $\mathbf{r}_1$:
$$
\int_{\text{point}} e^{i (\mathbf{q} \cdot \mathbf{r})} \, dV_0 = e^{i \text{qr}_1}.
$$

#### **1D: Line Segment**
For $D = 1$, a line segment defined by $\mathbf{r}_1, \mathbf{r}_2$:
$$
\int_{\text{line}} e^{i (\mathbf{q} \cdot \mathbf{r})} \, dV_1 = |\mathbf{n}| \cdot \frac{e^{i \text{qr}_2} - e^{i \text{qr}_1}}{\text{qr}_1 - \text{qr}_2},
$$
where $|\mathbf{n}| = \| \mathbf{r}_2 - \mathbf{r}_1 \|$.

#### **2D: Triangle**
For $D = 2$, a triangle defined by $\mathbf{r}_1, \mathbf{r}_2, \mathbf{r}_3$:

$$
\int_{\text{triangle}} e^{i (\mathbf{q} \cdot \mathbf{r})} \, dV_2 = |\mathbf{n}| \cdot \frac{e^{i \text{qr}_3} (\text{qr}_1 - \text{qr}_2) + e^{i \text{qr}_1} (\text{qr}_2 - \text{qr}_3) + e^{i \text{qr}_2} (\text{qr}_3 - \text{qr}_1)}{(\text{qr}_1 - \text{qr}_2)(\text{qr}_2 - \text{qr}_3)(\text{qr}_3 - \text{qr}_1)},
$$
where $|\mathbf{n}| = 2 \times \text{Area of the triangle} = \| (\mathbf{r}_2 - \mathbf{r}_1) \times (\mathbf{r}_3 - \mathbf{r}_1) \|$.

#### **3D: Tetrahedron**
For $D = 3$, a tetrahedron defined by $\mathbf{r}_1, \mathbf{r}_2, \mathbf{r}_3, \mathbf{r}_4$:
$$
\int_{\text{tetrahedron}} e^{i (\mathbf{q} \cdot \mathbf{r})} \, dV_3 = |\mathbf{n}| \cdot \frac{\text{Numerator}}{\text{Denominator}},
$$
where:
$$
\text{Denominator} = (\text{qr}_1 - \text{qr}_2)(\text{qr}_1 - \text{qr}_3)(\text{qr}_1 - \text{qr}_4)(\text{qr}_2 - \text{qr}_3)(\text{qr}_2 - \text{qr}_4)(\text{qr}_3 - \text{qr}_4),
$$

and:
$$
\text{Numerator} = e^{i \text{qr}_1} (\text{qr}_2 - \text{qr}_3)(\text{qr}_3 - \text{qr}_4)(\text{qr}_4 - \text{qr}_2)+ e^{i \text{qr}_2} (\text{qr}_3 - \text{qr}_4)(\text{qr}_4 - \text{qr}_1)(\text{qr}_1 - \text{qr}_3)
$$

$$+ e^{i \text{qr}_3} (\text{qr}_4 - \text{qr}_1)(\text{qr}_1 - \text{qr}_2)(\text{qr}_2 - \text{qr}_4)+ e^{i \text{qr}_4} (\text{qr}_1 - \text{qr}_2)(\text{qr}_2 - \text{qr}_3)(\text{qr}_3 - \text{qr}_1).
$$

Here, $|\mathbf{n}| = 6 \times \text{Volume of the tetrahedron}$.
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# Degenerate cases

## 2-D Triangle

$$
J_2=e^{i\text{qr1}} \int_0^1 \int_0^{1-u} e^{i \big(u(\text{qr2} - \text{qr1}) + v(\text{qr3} - \text{qr1})\big)} \, dv \, du.
$$

$$
= e^{i\text{qr1}} \int_0^1 e^{iu\Delta_{21}} \int_0^{1-u} e^{i v\Delta_{31}} \, dv \, du.
$$

$$
= e^{i\text{qr1}} \int_0^1 e^{iu\Delta_{21}} \frac{e^{i(1-u)\Delta_{31}}-1}{i\Delta_{31}} \, du.
$$

$$
= e^{i\text{qr1}}\big[
\frac{e^{i \Delta_{31}}}{i \Delta_{31}}\int_0^1 e^{iu(\Delta_{21}-\Delta_{31})} \, du-\frac{1}{i \Delta_{31}}\int_0^1 e^{iu\Delta_{21}} \, du.\big]
$$

$$
= e^{i\text{qr1}}\big[
\frac{e^{i \Delta_{31}}}{i \Delta_{31}}
\frac{e^{i(\Delta_{21}-\Delta_{31})}-1}{i(\Delta_{21}-\Delta_{31})}-\frac{1}{i \Delta_{31}}\frac{e^{i\Delta_{21}}-1}{i\Delta_{21}}\big]
$$

$$
= e^{i\text{qr1}}\big[
\frac{e^{i \Delta_{31}}}{i \Delta_{31}}
\frac{e^{i\Delta_{23}}-1}{i\Delta_{23}}-\frac{1}{i \Delta_{31}}\frac{e^{i\Delta_{21}}-1}{i\Delta_{21}}\big]
$$

$$
= \frac{e^{\text{qr2}}-e^{\text{qr3}}}{i \Delta_{31}i\Delta_{23}}
-\frac{e^{\text{qr2}}-e^{\text{qr1}}}{i \Delta_{31}i\Delta_{21}}
$$

$$
= \frac{\Delta_{12}e^{\text{qr3}}+\Delta_{23}e^{\text{qr1}}+\Delta_{31}e^{\text{qr2}}}{\Delta_{12}i\Delta_{23}\Delta_{31}}
$$

---
For $\Delta_{21}=\Delta_{31}=0$
$$
J_2 = e^{i\text{qr1}} \int_0^1 \int_0^{1-u} \, dv \, du.
$$
$$
= \frac{e^{i\text{qr1}}}{2}
$$

---
For $\Delta_{21}=0$
$$
J_2= e^{i\text{qr1}} \int_0^1 \int_0^{1-u} e^{i v\Delta_{31}} \, dv \, du.
$$
$$
= e^{i\text{qr1}} \int_0^1 \frac{e^{i(1-u)\Delta_{31}}-1}{i\Delta_{31}} \, du.
$$
$$
= e^{i\text{qr1}}\big[
\frac{e^{i \Delta_{31}}}{i \Delta_{31}}
\frac{e^{i\Delta_{23}}-1}{i\Delta_{23}}-\frac{1}{i \Delta_{31}}\big]
$$
$$
= e^{i\text{qr1}}\big[
\frac{1-e^{i \Delta_{31}}-i \Delta_{31}}{i^2 \Delta_{31}^2}
\big]
$$
$$
=\frac{e^{i\text{qr3}}-e^{i\text{qr1}}-ie^{i\text{qr1}}\Delta_{31}}{\Delta_{31}^2}
$$

## 3-D Tetrahedron
$$
J_{3} = |\mathbf{n}| \int_0^1 \int_0^{1-u} \int_0^{1-u-v} e^{i (\text{qr1} + u(\text{qr2} - \text{qr1}) + v(\text{qr3} - \text{qr1}) + w(\text{qr4} - \text{qr1}))} \, dw \, dv \, du.
$$

Let $s=qr2-qr1$, $t=qr3-qr1$, $r=qr4-qr1$. Considering the following integral
$$
J_{3} =Je^{i\text{qr1}} = e^{i\text{qr1}}\int_0^1 e^{i u s} \left( \int_0^{1-u} e^{i v t} \left( \int_0^{1-u-v} e^{i w r} \, dw \right) dv \right) du
$$

evaluates to:

$$
J = \frac{i}{r (r - t) t} \left( \frac{e^{i t} r}{s - t} + \frac{e^{i r} t}{r - s} + \frac{e^{i s} r (r - t) t}{s (s - t) (r - s)} + \frac{-r + t}{s} \right).
$$

**Case 1. $r=0$**
The integral reduces to:

$$
J = \int_0^1 e^{i u s} \left( \int_0^{1-u} e^{i v t} \left( \int_0^{1-u-v} 1 \, dw \right) dv \right) du,
$$

which evaluates to:

$$
J = \frac{-i s^2 (-1 + e^{i t} - i t) + i (-1 + e^{i s}) t^2 + s t^2}{s^2 (s - t) t^2}.
$$

**Case 2. $r=0$ and $t=0$**
The integral simplifies to:

$$
J = \int_0^1 e^{i u s} \left( \int_0^{1-u} \left( \int_0^{1-u-v} 1 \, dw \right) dv \right) du,
$$

which evaluates to:

$$
J = \frac{i}{2} \cdot \frac{-2 + 2 e^{i s} + s(-2i + s)}{s^3}.
$$

**Case 3. $r=0$ and $s=t$, $s\neq 0$**
The integral simplifies to:

$$
J = \int_0^1 e^{i u s} \left( \int_0^{1-u} e^{i v s} \left( \int_0^{1-u-v} 1 \, dw \right) dv \right) du,
$$
Same as case 2.
$$
J = \frac{i}{2} \cdot \frac{-2 + 2 e^{i s} + s(-2i + s)}{s^3}.
$$

Case 4. $s=0$ and $t=0$ and $r=0$
$$
J = \int_0^1 \left( \int_0^{1-u} e^{i v s} \left( \int_0^{1-u-v} 1 \, dw \right) dv \right) du,
$$

$$
J = \frac{1}{6}.
$$