First I try the plane of points equidistant from S and D which are 2 units apart. We divide the plane into thin circular zones concentric about the point in the plane closest to S and D. Let r be the radius of such a zone. The zone is dr thick and its area is 2πr dr. The distance from a zone to either S or D is √(1+r2). The integral is:
∫ 2πr sin(2ω√(1+r2))dr
It is immediately clear that this integral does not converge; the integrand does not even converge to 0! It diverges like Σ (−1)nn.
I do not see how to make this converge in any sense.

The expression
0r 2πx sin(2ω√(1+x2))dx/(πr2)
converges as r→∞.
The above amounts to averaging, not summing the “little arrows” however.

Here is a more sober development of the stuff below.

That wave theory bigot, jumping up and down in the back of the room, would probably try to interrupt at this point and say that we are merely doing Huygens’ wave computation and we forgot to divide by the distance that wave traveled. Indeed that seems to make the integral converge.
∫ (2πr/(1+r2)) sin(2ω√(1+r2))dr
looks like it converges. Note that the length, √(1+x2), appears squared because there are two equal straight path segments.
Alas “(1/(1+x^2))Sin[2 a Sqrt[1+x^2]]” stumps Mathematica as in indefinite integral.

To evaluate this definite integral I think we need to change coordinates. Let z = √(1+r2) whence r = √(1−z2) and
dr= ((√(1+r2))/r dz = z/(1−z2) dz.

Now ∫ (2πr/(1+r2)) sin(2ω√(1+r2))dr = ∫ (2π√(1−z2)/z2) sin(2ωz) z/(1−z2) dz = ∫ (1−z2)−1/2/z sin(2ωz) dz.
But “(1-x^2)^(-1/2)/x Sin[2 a x]” stumps Mathematica too.

### Perhaps contour integrals

Actually we need to compute ∫ (2πr/(1+r2)) exp(i 2ω √(1+r2))dr. That is fortunate for the magnitude of the integrand goes to 0 fast below the real axis, which is not the case for the sin function. We can thus do the normal contour integral thing. The contour is clockwise around the area under the real axis in the limit. There is one pole at r=−i in this area.
∫ (2πr/(1+r2)) exp(i 2ω √(1+r2))dr =
∫ (2πr/((1−ir)(1+ir))) exp(i 2ω √(1+r2))dr =
∫ (2πr/(1+ir)) exp(i 2ω √(1+r2))/(1−ir)dr =
−(1/2πi)(2π(−i)/(1+i(−i))) exp(i 2ω √(1+(−i)2))dr = 1/2 exp(0) = 1/2.
The sign of the whole expression is reversed because the contour is in the clockwise direction. A further factor of 2 arises because the original range of integration was [0,∞] whereas the contour integral produces the value for the integration range [-∞,∞].

I expected exp(2 i ω)/2 since the distance between S and D is 2. Perhaps I will do a numerical integration to see if my math is flawed.

The code below indicates there is an error in the contour integral.

```#include <stdio.h>
#include <math.h>
#define cis cos
// cis above should also be sin for other branch of answer.
double const omeg = 1.3;
double const pi = 3.14159265358979;
int main(){double s=0, x=0, dx = .01;
for(x=dx; x < 100000; x+=dx) s += x*cis(2*omeg * sqrt(1+x*x))/(1+x*x);
printf("Numerical: %e %e %e\n", cis(0), omeg, 2.*pi*s*dx);
return 0;}```
More later.
Ray Tracing
The Foundations of Photo-realistic Rendering: from Quantum Electrodynamics to Maxwell’s Equations
Physics, Topology, Logic and Computation: A Rosetta Stone Baez, Stay