Simulation

Proof. Take ψ(r) = r^α and ϕ(r)=(e_k·r)²/2=r_k²/2. Then, ∇ψ(r)·e_k = α_kr^α−e_k, ∇ϕ(r)=r_ke_k, and Δϕ(r) = 1. Using Green’s first identity yields

Lemma 1 For ω∈ℝ^d∖ {0} and α∈ℕ^d,

f_B^α (ω) = j g_B^α (ω) +

∑

⎛
⎜
⎜
⎜
⎝

−j ω_i

⎪⎪
⎪⎪

⎞
⎟
⎟
⎟
⎠

α_i f_B^α−e_i (ω). (1)

Proof. Use Green’s first identity with ψ(r) = r^α, and ϕ(r)=−e^−jω·r. Then, x·∇ψ(r) = ∑_i x_iα_ir^α−e_i, ∇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=e_i, and Lemma 1, we obtain f_B^e_i (ω) = j g_B^e_i (ω) +ω_i/||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_B^α(ω) = j g_B^α(ω) + ∑_i ( −j ω_i/||ω ||² )α_i f_B^α−e_i(ω) . Since |α−e_i|=n, we substitute f_B^α−e_i using the induction hypothesis and, after simplifications, we obtain

with d_i(x)=x_i/|x| for x≠ 0 and d_i(0)=0. By permutation of the sums and noting that ∑_i d_i(x)=1 for x≠ 0 and ∑_i d_i(0)=0, we get

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) − δ_m = 0. (2)

Proof. Integrate ∫₀¹ −j(a+2bλ)λ^me^{−jλ(a+λ b)}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) =√πe^{j a² /4 b}/(2 √j b) [ erf((a+2 b)√j/(2√b))−erf(a√j/(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/√π∫₀¹e^−z²t²d t.

Proof. From Proposition 1 with m=0 and b=0, we get h⁽⁰⁾(a,0) = e^{−j a/2}sinc(a/(2π)). In the case b≠ 0, we define t=λ+a/2b such that λ (a+b λ) = a²/4b−bt². By Definition (22), we get

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) = z^s

∫

λ^s−1e^−λ zd λ.

Proof. Note that e^{−jλ(a+λ b)}=e^{−jλ a}∑_n=0^∞(−jλ² b)ⁿ/n!. By virtue of Fubini’s theorem, we get

A.3 Proof of Proposition 2

Proof. We rewrite g_B^α using Equation (21) with F(r)=r^αe^{−j ω·r}ω/||ω ||² for ω≠0 and F(r)=r^α+e_ke_k/(1+α_k) 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 r^α and r^α+e_k.

Appendix A Simulation

A.1 Proof of Theorem 1

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

A.3 Proof of Proposition 2