Appendix A  Simulation

A.1  Proof of Theorem 1

First, we consider the case ω=0.

Proof. Take ψ(r) = rα and ϕ(r)=(ek·r)2/2=rk2/2. Then, ψ(rek = αkrαek, ϕ(r)=rkek, and Δϕ(r) = 1. Using Green’s first identity yields

(1+αk)f Bα (0) =
 ∂ B
rα+ekek·nd σ=(1+αk)g Bα (0).

For the case ω0, we need an intermediate result.

Lemma 1   For ω∈ℝd∖ {0} and α∈ℕd,
f Bα (ω) = j g Bα (ω) +
 ∑ i

−j ωi
 ⎪⎪ ⎪⎪ ω ⎪⎪ ⎪⎪ 2

αi f Bαei (ω).     (1)

Proof. Use Green’s first identity with ψ(r) = rα, and ϕ(r)=−e−jω·r. Then, x·ψ(r) = ∑i xiαirαei, ϕ(r)=je−jω·rω, and Δϕ(r) = ||ω ||2e−jω·r.

Let us continue the proof of Theorem 1 by induction on n=|α|. For n=0, α=0 and the result holds true according to Lemma 1. When considering n=1, α=ei, and Lemma 1, we obtain f Bei (ω) = j g Bei (ω) +ωi/||ω ||2g B0 (ω). This is true for all i, hence the result holds true for n=1. Now, we suppose the result holds true at order n and we consider α such that |α|=n+1. From Lemma 1, we have that f Bα(ω) = j g Bα(ω) + ∑i ( −j ωi/||ω ||2i f Bαei(ω) . Since |αei|=n, we substitute f Bαei using the induction hypothesis and, after simplifications, we obtain

f Bα(ω)=
g Bα(ω) +j
 ∑ i

 α ∑ m=0

−j ω
 ⎪⎪ ⎪⎪ ω ⎪⎪ ⎪⎪ 2

 α− m
×|αm|! di(αm) Cαmg Bm(ω)

with di(x)=xi/|x| for x0 and di(0)=0. By permutation of the sums and noting that ∑i di(x)=1 for x0 and ∑i di(0)=0, we get

f Bα(ω) =j
 α ∑ m=0

−j ω
 ⎪⎪ ⎪⎪ ω ⎪⎪ ⎪⎪ 2

 α− m
|αm|! Cαmg Bm(ω).

This is valid for all α such that |α|=n+1. Hence, we just proved the result for n+1 assuming it holds true for n.

A.2  Characterization and Computations of a Family of 1-D Integrals

Proposition 1   For m∈ℕ, h(m) follows the recursion rule
 2j b h(m+1)(a,b)+j a h(m)(a,b) − m h(m−1)(a,b)+e−j (a+b) − δm = 0.     (2)

Proof. Integrate ∫01 −j(a+2bλ)λme−jλ(ab)d λ by parts and identify h(m+1), h(m) and h(m−1) if m>0.

Corollary 2   For small values of a and b, one can rely on the backward iteration starting from a higher order M>m to get accurate results
• h~(M+1)(a,b)=h~(M)(a,b)=0
• h~(m)(a,b) = (2j b h~(m+2)(a,b) +j a h~(m+1)(a,b) +e−j (a+b))/(m+1).
Proposition 3   For b nonzero and m≥ 1, the forward iteration is used
• h(0)(a,b) =√πej a2 /4 b/(2 √j b) [ erf((a+2 b)√j/(2√b))−erf(aj/(2√b))]
• h(m+1)(a,b)= (m h(m−1)(a,b)−j a h(m)(a,b)+e−j (a+b))/(2j b)
with erf(z) = 2z/√π01ez2t2d t.

Proof. From Proposition 1 with m=0 and b=0, we get h(0)(a,0) = e−j a/2sinc(a/(2π)). In the case b≠ 0, we define t=λ+a/2b such that λ (a+b λ) = a2/4bbt2. By Definition (22), we get

e
−j
 a2 4b

 a 2b
+1

 a 2b
e−j b t2t,

after the change of variable. Splitting this integral, we get

e
j
 a2 4b

 a 2b
+1

0
e−j bt2t −
 a 2b

0
e−j bt2t

.

The result follows from normalizing the integration intervals.

Proposition 4   For b small, the truncated Taylor series in b=0 provides accurate results
h(m)(a,b) =
 ∞ ∑ n=0
 (−j b)nγ(m+2n+1,j a) n!(j a)m+2n+1
,     (3)
where the lower incomplete gamma function is defined as
γ(s,z) = zs
 1 0
λs−1e−λ zd λ.

Proof. Note that e−jλ(ab)=e−jλ an=0(−jλ2 b)n/n!. By virtue of Fubini’s theorem, we get

h(m)(a,b) =
 ∞ ∑ n=0
(−j b)n h(m+2n)(a,0)/n!.

Identify h(m+2n)(a,0) to γ(m+2n+1,j a)/(j a)m+2n+1.

A.3  Proof of Proposition 2

Proof. We rewrite g Bα using Equation (21) with F(r)=rαe−j ω·rω/||ω ||2 for ω0 and F(r)=rα+ekek/(1+αk) for ω=0. The piecewise parameterization of the contour (Table 4.2) is then used, and by virtue of the multinomial theorem, we expand the terms rα and rα+ek.