Does an an analytic solution exist to these integrals?

Question

For a quantum mechanical simulation I have these 3 functions of euclidian position (x,y) and constants (p,q,r,s,t,u):

$A1(x,y,p,q)=-\cos \left(\frac{\sqrt{3} \pi x (p+q)}{A}\right) \sin \left(\frac{\pi y (p-q)}{A}\right)+\cos \left(\frac{\sqrt{3} \pi q x}{A}\right) \sin \left(\frac{\pi y (2 p+q)}{A}\right)-\cos \left(\frac{\sqrt{3} \pi p x}{A}\right) \sin \left(\frac{\pi y (p+2 q)}{A}\right)$

$A2(x,y,r,s)=\sin \left(\frac{\sqrt{3} \pi x (r+s)}{A}\right) \sin \left(\frac{\pi y (r-s)}{A}\right)+\sin \left(\frac{\sqrt{3} \pi s x}{A}\right) \sin \left(\frac{\pi y (2 r+s)}{A}\right)-\sin \left(\frac{\sqrt{3} \pi r x}{A}\right) \sin \left(\frac{\pi y (r+2 s)}{A}\right)$

Where

$q=0,1,2,3... \quad p=q+1,q+2,q+3...$

$s=0,1,2,3..., \quad r=s+1,s+2,s+3...$

and

$E(x,y,t,u)=A2(x,y,t,u)+iA1(x,y,t,u)$

i.e.

$E(x,y,t,u)=\sin \left(\frac{\sqrt{3} \pi x (t+u)}{A}\right) \sin \left(\frac{\pi y (t-u)}{A}\right)+\sin \left(\frac{\sqrt{3} \pi u x}{A}\right) \sin \left(\frac{\pi y (2 t+u)}{A}\right)-\sin \left(\frac{\sqrt{3} \pi t x}{A}\right) \sin \left(\frac{\pi y (t+2 u)}{A}\right)+i \Biggl(-\cos \left(\frac{\sqrt{3} \pi x (t+u)}{A}\right) \sin \left(\frac{\pi y (t-u)}{A}\right)+\cos \left(\frac{\sqrt{3} \pi u x}{A}\right) \sin \left(\frac{\pi y (2 t+u)}{A}\right)-\cos \left(\frac{\sqrt{3} \pi t x}{A}\right) \sin \left(\frac{\pi y (t+2 u)}{A}\right)\Biggr)$

Where

$u=0,\frac{1}{3},\frac{2}{3},\frac{3}{3}...\quad t=q+1,q+2,q+3...$

I wish to calculate the integrals of products of pairs of different combinations of these functions in equilateral triangle shaped regions of the plane, i.e. I want to find expressions for

$\int\int A1(x,y,p,q)A1(x,y,r,s) dxdy$

$\int\int A1(x,y,p,q)A2(x,y,r,s) dxdy$

$\int\int A2(x,y,p,q)A2(x,y,r,s) dxdy$

$\int\int A1(x,y,p,q)E(x,y,t,u) dxdy$

$\int\int A2(x,y,p,q)E(x,y,t,u) dxdy$

$\int \int E(x,y,t,u)E(x,y,v,w) dxdy$

Where the integral is over the surface of an equilateral triangle. I have tried transforming the problem, like this: https://math.stackexchange.com/a/955188/441529
But have not yet managed to find an analytic solution to the integral.

Answer 1

If I am right, all the terms are of the form

$$\int_D\text{ soc}(ax)\text{ soc}(by)\text{ soc}(cx)\text{ soc}(dy)\,dx\,dy$$

where $\text{soc}$ denotes a sine or a cosine, and the integration domain is a triangle.

By repeated application of the product-to-sum formulas, you can turn that to a sum of

$$\int_D \text{ soc}(a'x+b'y)\text{ soc}(c'x+d'y)\,dx\,dy,$$

then

$$\int_D \text{ soc}(a''x+b''y)\,dx\,dy.$$

You can decompose the domain as an algebraic sum of trapezoids and evaluate

$$\int_{x=x_0}^{x_1}\int_{y=0}^{px+q}\text{ soc}(a''x+b''y)\,dx\,dy.$$

The integration on $y$ will yield terms in $\text{soc}(a'''x+b''')$, so, yes, there is an analytic solution.

Answer 2

Notice functions $A_1, A_2$ and $E$ are linear combinations of traveling waves $e^{ip\cdot r} = e^{i(p_1 x + p_2 y)}$. So do integrals of their product. All integrals you mentioned are linear combinations of the form $$\int_\Delta e^{i p\cdot r} dxdy$$ where $\Delta$ is your triangle.

Instead of triangle, consider generalization of this sort of integral over any simple polygon.

Let $\Omega$ be any simple polygonal region with vertices $v_1 = (x_1,y_1), v_2 = (x_2,y_2), \ldots, v_{n} = (x_n,y_n)$ ordered in counter-clockwise orientation. For convenience, let $v_{n+1} = v_1$. Let $p = (p_1,p_2)$ be any non-zero vector such that all $p \cdot v_k \ne p \cdot v_{k+1}$.

Embed the plane $\mathbb{R}^2$ into $\mathbb{R}^3$, lift the polygon $\Omega$ to a polygonal prism $\Omega \times [0,1]$ in $\mathbb{R}^3$ and apply divergence theorem, we have

$$\begin{align} \int_\Omega e^{ip\cdot r} dxdy &= \int_{\Omega \times [0,1]} e^{ip\cdot r}dxdydz = -\frac{i}{|p|^2} \int_{\Omega \times [0,1]} \nabla \cdot (p e^{ip\cdot r}) dxdydz\\ &= -\frac{i}{|p|^2} \int_{\partial(\Omega \times [0,1])} e^{ip\cdot r} p \cdot dS = -\frac{i}{|p|^2} \int_{\partial\Omega} e^{ip\cdot r} p\cdot(ds \times \hat{z})\\ &= -\frac{i}{|p|^2} \int_{\partial\Omega} e^{ip\cdot r} \hat{z}\cdot (p \times ds) \end{align} $$

where $dS$ is the surface element of $\partial(\Omega \times [0,1])$ and $ds$ is the line element for $\partial\Omega$.

Break $\partial\Omega$ into $n$ line segments $v_iv_{i+1}$, integrate over each line segment and sum, we obtain $$ \int_\Omega e^{ip\cdot r} dxdy = -\frac{1}{|p|^2}\sum_{k=1}^n (e^{ip\cdot v_{k+1}} - e^{ip\cdot v_k})\frac{ \hat{z}\cdot( p \times (v_{k+1} - v_k))}{p\cdot(v_{k+1}-v_k)} \tag{*1} $$ Expanding all factors out, RHS becomes following mess: $$ -\frac{1}{p_1^2+p_2^2}\sum_{k=1}^n \big(e^{i(p_1x_{k+1} + p_2y_{k+1})} - e^{i(p_1x_{k} + p_2y_{k})}\big) \frac{p_1(y_{k+1}-y_k) - p_2(x_{k+1}-x_k)}{p_1(x_{k+1}-x_k) + p_2(y_{k+1}-y_k)} $$ In general, the integral of $e^{ip\cdot r}$ over polygon $\Omega$ will be a linear combination of the value of $e^{ip\cdot r}$ at the vertices $v_k$ weighted by rational function of coordinates of $p$ and $v_k$.

The formula $(*1)$ is valid only when all $p\cdot v_k \ne p \cdot v_{k+1}$. In the case when this fails, one could use the fact the integral depends on $p$ continuously and extract the desired coefficient in front of $e^{ip\cdot v_k}$ by taking appropriate limit in $(*1)$. Introduce two set of new variables $$r_k = \frac12(v_{k+1} + v_{k})\quad\text{ and }\quad \epsilon_k = \frac12(v_{k+1} - v_k)$$ We obtain following expression which works as long as $p \ne 0$. $$\int_\Omega e^{ip\cdot r} dxdy = \frac{2}{i|p|^2}\sum_{k=1}^n e^{ip\cdot r_k} {\rm sinc}(p\cdot \epsilon_k) ( \hat{z} \cdot p \times \epsilon_k ),\quad {\rm sinc}(x) = \begin{cases} \frac{\sin x}{x}, & x \ne 0\\ 1, & x = 0 \end{cases} $$

I worked out above formulas but this stuff has been known for a long time. A literature search return a recent article Form factor (Fourier shape transform) of polygon and polyhedron by Joachim Wuttke. Look at that and references there for more details.