Legendre polynomials from a convergent zonal kernel: generating function and exterior expansion

For $z\in [-1,1]$ put $w(\theta )=z+i\sqrt{1-z^2}\cos \theta$. Then $\mathrm{\mid }w(\theta ){\mathrm{\mid }}^2=z^2+(1-z^2){\cos }^2\theta \le z^2+1-z^2=1$, so $\mathrm{\mid }w(\theta )\mathrm{\mid }\le 1$, and the complex binomial theorem [L6] gives $w(\theta {)}^l={\sum }_{k=0}^l\left(\genfrac{}{}{0px}{}{l}{k}\right)z^{l-k}(i\sqrt{1-z^2}\cos \theta {)}^k$ for every $l\in {\mathbb{N}}_0$ and $\theta \in [0,\pi ]$.

The cosine integrals are ${\int }_0^{\pi }{\cos }^k\theta d\theta =0$ for odd $k$ and ${\int }_0^{\pi }{\cos }^{2j}\theta d\theta =2I_{2j}=\pi {\prod }_{i=1}^j\frac{2i-1}{2i}$ for $k=2j$. Indeed, For $a<c<b$: $f$ is integrable on $[a,b]$ if and only if it is integrable on $[a,c]$ and on $[c,b]$, and then ${\int }_a^{b}f={\int }_a^{c}f+{\int }_c^{b}f$; with the oriented form for arbitrary $a,b,c$ splits ${\int }_0^{\pi }={\int }_0^{\pi \mathrm{/}2}+{\int }_{\pi \mathrm{/}2}^{\pi }$; the substitution $\theta =\pi -t$ of [L4] on $[\pi \mathrm{/}2,\pi ]$, with $\cos (\pi -t)=-\cos t$ from [L5], turns the second integral into $(-1{)}^k{\int }_0^{\pi \mathrm{/}2}{\cos }^{k}tdt$; the cofunction identity $\cos t=\sin (\pi \mathrm{/}2-t)$ and the same substitution identify ${\int }_0^{\pi \mathrm{/}2}{\cos }^{k}tdt$ with $I_k$; and the values of $I_k$ are those of [L5].

For real $a>\mathrm{\mid }b\mathrm{\mid }\ge 0$ put $I(a,b)={\int }_0^{\pi }(a+b\cos \theta {)}^{-1}d\theta$. Then $(a^2-b^2)I(a,b{)}^2={\pi }^2$. Indeed, on $[0,\pi \mathrm{/}2]$ substitute $\theta =2\arctan t$, $t\in [0,1]$, which is legitimate by [L4] because $t\mapsto 2\arctan t$ has continuous derivative $2(1+t^2{)}^{-1}$ and $(a+b\cos \theta {)}^{-1}$ is continuous on $[0,\pi ]$; by [L5] and [L4], $\cos (2\arctan t)={\cos }^2(\arctan t)-{\sin }^2(\arctan t)=\frac{1-t^2}{1+t^2}$, because ${\cos }^{2}x=(1+{\tan }^{2}x{)}^{-1}$ at $\tan (\arctan t)=t$; hence ${\int }_0^{\pi \mathrm{/}2}(a+b\cos \theta {)}^{-1}d\theta ={\int }_0^1\frac{2dt}{(a+b)+(a-b)t^2}=\frac{2}{\sqrt{a^2-b^2}}\arctan c$ with $c=\sqrt{(a-b)\mathrm{/}(a+b)}>0$. On $[\pi \mathrm{/}2,\pi ]$ the reflection $\theta =\pi -t$ ([L5]) turns the integrand into $(a-b\cos t{)}^{-1}$ and the same substitution gives ${\int }_{\pi \mathrm{/}2}^{\pi }(a+b\cos \theta {)}^{-1}d\theta ={\int }_0^1\frac{2dt}{(a-b)+(a+b)t^2}=\frac{2}{\sqrt{a^2-b^2}}\arctan \frac{1}{c}$. Since $\tan (\pi \mathrm{/}2-x)=\cot x$ ([L5]) and $\cot (\arctan c)=1\mathrm{/}\tan (\arctan c)=1\mathrm{/}c$ ([L4]), with both angles in $(0,\pi \mathrm{/}2)$ where $\tan$ is injective ([L4]), $\arctan c+\arctan (1\mathrm{/}c)=\pi \mathrm{/}2$; adding the two halves by [L1] gives $I(a,b)=\pi \mathrm{/}\sqrt{a^2-b^2}$, hence $(a^2-b^2)I(a,b{)}^2={\pi }^2$.

Fix $\rho \in (0,1\mathrm{/}4)$, real $b$ with $\mathrm{\mid }b\mathrm{\mid }<\rho$, and complex $a$ with $\mathrm{\mid }a-1\mathrm{\mid }<\rho$; then $\mathrm{Re}a>3\mathrm{/}4>\mathrm{\mid }b\mathrm{\mid }$, so $a+b\cos \theta \mathrm{\ne }0$ on $[0,\pi ]$, and $I(a,b)={\int }_0^{\pi }(a+b\cos \theta {)}^{-1}d\theta$ is well defined as a $\mathbb{C}$-valued integral [L1]. The map $a\mapsto I(a,b)$ is holomorphic on $\mathrm{\mid }a-1\mathrm{\mid }<\rho$: for $h\mathrm{\ne }0$ small, the difference quotient equals $-{\int }_0^{\pi }((a+b\cos \theta )(a+h+b\cos \theta ){)}^{-1}d\theta$ by the elementary identity $\frac{1}{A+h}-\frac{1}{A}=-h\frac{1}{A(A+h)}$, and the integrands converge uniformly on $[0,\pi ]$ to $-(a+b\cos \theta {)}^{-2}$ because $\mathrm{\mid }a+b\cos \theta \mathrm{\mid }\ge \mathrm{Re}a-\mathrm{\mid }b\mathrm{\mid }>1\mathrm{/}2$ and, for $\mathrm{\mid }h\mathrm{\mid }\le \rho$, $\mathrm{\mid }a+h+b\cos \theta \mathrm{\mid }>1\mathrm{/}4$; [L2] applied to real and imaginary parts therefore gives complex differentiability at every such $a$, hence holomorphy [L3]. The same computation in the second variable shows that for each fixed $a$ with $\mathrm{\mid }a-1\mathrm{\mid }<\rho$ the map $b\mapsto I(a,b)$ is holomorphic on $\mathrm{\mid }b\mathrm{\mid }<\rho$.

Inserting step 1.1 into the defining integral and using step 1.2 with the linearity of [L1], only the terms with even $k=2j$ survive, and $i^{2j}=(-1{)}^j$ gives $P_l(z)={\sum }_{j=0}^{\lfloor l\mathrm{/}2\rfloor }(-1{)}^j\left(\genfrac{}{}{0px}{}{l}{2j}\right)W_j(1-z^2{)}^j{z}^{l-2j}$ with $W_j={\prod }_{i=1}^j\frac{2i-1}{2i}$ and $W_0=1$. Each summand is a real polynomial of degree $l$, so $P_l$ is a real polynomial of degree at most $l$; at $z=1$ every term with $j\ge 1$ carries the factor $(1-z^2{)}^j=0$, so $P_l(1)=W_0=1$.

Taking moduli in the defining integral, $\mathrm{\mid }{P}_l(z)\mathrm{\mid }={\pi }^{-1}\mid {\int }_0^{\pi }w(\theta {)}^{l}d\theta \mid \le {\pi }^{-1}{\int }_0^{\pi }\mathrm{\mid }w(\theta ){\mathrm{\mid }}^{l}d\theta \le 1$ by the norm inequality of [L1] and $\mathrm{\mid }w\mathrm{\mid }\le 1$ from step 1.1.

For each real $b$ with $\mathrm{\mid }b\mathrm{\mid }<\rho$ the function $F_b(a)=(a^2-b^2)I(a,b{)}^2-{\pi }^2$ is holomorphic on the disc $\mathrm{\mid }a-1\mathrm{\mid }<\rho$ (products and differences of holomorphic functions, [L3]); by step 1.3 it vanishes for every real $a\in (1-\rho ,1+\rho )$, since there $a>1-\rho >\mathrm{\mid }b\mathrm{\mid }$; that interval has accumulation points in the disc, so Identity theorem for holomorphic functions gives $F_b\equiv 0$ on the disc. For each fixed $a$ with $\mathrm{\mid }a-1\mathrm{\mid }<\rho$, the function $b\mapsto (a^2-b^2)I(a,b{)}^2-{\pi }^2$ is holomorphic on $\mathrm{\mid }b\mathrm{\mid }<\rho$ by step 1.4, and it vanishes for every real $b$ with $\mathrm{\mid }b\mathrm{\mid }<\rho$ by the first part; the identity theorem applied on the disc $\mathrm{\mid }b\mathrm{\mid }<\rho$ therefore gives $(a^2-b^2)I(a,b{)}^2={\pi }^2$ for all such $a$ and all complex $b$ with $\mathrm{\mid }b\mathrm{\mid }<\rho$.

For $\mathrm{\mid }q\mathrm{\mid }<1$ the series ${\sum }_l{q}^{l}w(\theta {)}^l$ has the majorant ${\sum }_l\mathrm{\mid }q{\mathrm{\mid }}^l<\mathrm{\infty }$ and converges uniformly on $[0,\pi ]$ to $(1-qw(\theta ){)}^{-1}$; by [L2] it may be integrated term by term, so $G_z(q):={\sum }_{l\ge 0}{q}^l{P}_l(z)$ satisfies $G_z(q)={\pi }^{-1}{\int }_0^{\pi }(1-qw(\theta ){)}^{-1}d\theta$, and the majorant $\mathrm{\mid }q{\mathrm{\mid }}^l$ also gives absolute convergence and uniform convergence on $[-1,1]$ of ${\sum }_l{q}^l{P}_l(z)$ because $\mathrm{\mid }{P}_l\mathrm{\mid }\le 1$ by step 2.2.

$G_z(q)={\sum }_l{q}^l{P}_l(z)$ is a complex power series whose coefficients are bounded (step 2.2), so its radius of convergence is at least $1$; by [L3] it is holomorphic on $\mathrm{\mid }q\mathrm{\mid }<1$, and $G_z(0)=P_0(z)=1$ because $P_0(z)={\pi }^{-1}{\int }_0^{\pi }1d\theta =1$. The finite-Riemann and analytic constructions up to this point use no Choice.

Fix real $q$ with $\mathrm{\mid }q\mathrm{\mid }<\rho$ and set $a=1-qz$, $b=-iq\sqrt{1-z^2}$; then $\mathrm{\mid }a-1\mathrm{\mid }=\mathrm{\mid }qz\mathrm{\mid }\le \mathrm{\mid }q\mathrm{\mid }<\rho$ and $\mathrm{\mid }b\mathrm{\mid }\le \mathrm{\mid }q\mathrm{\mid }<\rho$, and $a^2-b^2=(1-qz{)}^2+q^2(1-z^2)=1-2qz+q^2$. Since $a+b\cos \theta =1-q(z+i\sqrt{1-z^2}\cos \theta )=1-qw(\theta )$, step 3.1 gives $\pi G_z(q)={\int }_0^{\pi }(1-qw(\theta ){)}^{-1}d\theta =I(a,b)$; hence step 2.3 yields $(1-2qz+q^2)G_z(q{)}^2=1$ for every real $q$ with $\mathrm{\mid }q\mathrm{\mid }<\rho$.

Define $\mathrm{\Psi }(q)=(1-2qz+q^2)G_z(q{)}^2-1$ on $\mathrm{\mid }q\mathrm{\mid }<1$; by step 4.1 and [L3] the function $\mathrm{\Psi }$ is holomorphic there, and by step 4.2 it vanishes on the interval $(-\rho ,\rho )$, which has accumulation points in the disc. The identity theorem [L3] gives $\mathrm{\Psi }\equiv 0$, that is $(1-2qz+q^2)G_z(q{)}^2=1$ for every $\mathrm{\mid }q\mathrm{\mid }<1$. The polynomial $1-2qz+q^2$ has the two roots $q=z\pm i\sqrt{1-z^2}$ of modulus $\sqrt{z^2+(1-z^2)}=1$, and equals $1$ at $q=0$, so it is nonzero and holomorphic on the connected disc $\mathrm{\mid }q\mathrm{\mid }<1$, and $G_z$ satisfies $G_z(0)=1$ and $G_z^2=(1-2qz+q^2{)}^{-1}$ there; if $S$ is any holomorphic function on the disc with $S(0)=1$ and $S^2=(1-2qz+q^2{)}^{-1}$, then $(G_z-S)(G_z+S)\equiv 0$ while $G_z+S$ is continuous with value $2$ at $0$, hence nonzero near $0$, so $G_z=S$ near $0$ and then on the whole connected disc by the identity theorem. Thus $G_z$ is exactly the analytic square root of $(1-2qz+q^2{)}^{-1}$ equal to $1$ at $q=0$.

For $x\mathrm{\ne }0$, $y\in {\mathbb{R}}^3$ with $\mathrm{\mid }x\mathrm{\mid }>\mathrm{\mid }y\mathrm{\mid }$, write $r=\mathrm{\mid }x\mathrm{\mid }$, $n=x\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$, $q=\mathrm{\mid }y\mathrm{\mid }\mathrm{/}r<1$; if $y\mathrm{\ne }0$ and $\widehat{y}=y\mathrm{/}\mathrm{\mid }y\mathrm{\mid }$, then $\mathrm{\mid }x-y{\mathrm{\mid }}^{-1}=r^{-1}(1-2q(n\cdot \widehat{y})+q^2{)}^{-1\mathrm{/}2}=r^{-1}{\sum }_{l\ge 0}{q}^l{P}_l(n\cdot \widehat{y})={\sum }_{l\ge 0}\mathrm{\mid }y{\mathrm{\mid }}^l\mathrm{\mid }x{\mathrm{\mid }}^{-l-1}{P}_l(n\cdot y\mathrm{/}\mathrm{\mid }y\mathrm{\mid })$ by the generating identity of step 5.1, since $n\cdot \widehat{y}\in [-1,1]$. For real $0\le q<1$, $G_z(q)$ is real because all coefficients are real, is continuous and never zero by step 5.1, and has value $1$ at zero. By the intermediate value theorem [L3] it cannot become negative; hence it is the positive reciprocal square root used by the positive real kernel. The remainder satisfies $\mid {\sum }_{l>N}{q}^l{P}_l(n\cdot \widehat{y})\mid \le {\sum }_{l>N}{q}^l=\frac{q^{N+1}}{1-q}$ by step 2.2, which after multiplication by $r^{-1}$ gives the stated bound, uniformly on $\mathrm{\mid }y\mathrm{\mid }\le a$ whenever $\mathrm{\mid }x\mathrm{\mid }\ge r>a$; for $y=0$ the formula reduces to $\mathrm{\mid }x{\mathrm{\mid }}^{-1}$ with all higher terms zero, consistently with $c_l=\mathrm{\mid }y{\mathrm{\mid }}^l\mathrm{\mid }x{\mathrm{\mid }}^{-l-1}{P}_l(\cdot )$ vanishing for $l\ge 1$ at $y=0$.

Fix $x\mathrm{\ne }0$ and $y\in {\mathbb{R}}^3$ with $\mathrm{\mid }x\mathrm{\mid }>\mathrm{\mid }y\mathrm{\mid }$. For $y=0$ define $c_0(x,0)=\mathrm{\mid }x{\mathrm{\mid }}^{-1}$ and $c_l(x,0)=0$ for $l\ge 1$; the parameter kernel $\mathrm{\mid }x-t0{\mathrm{\mid }}^{-1}$ is constant and these are its Taylor coefficients, with no division by $\mathrm{\mid }y\mathrm{\mid }$. For $y\mathrm{\ne }0$, step 6.1 applied to $ty$ gives the expansion for $t\ge 0$ with $t\mathrm{\mid }y\mathrm{\mid }<\mathrm{\mid }x\mathrm{\mid }$. The polynomial formula of step 2.1 gives $P_l(-z)=(-1{)}^l{P}_l(z)$ because each monomial $z^{l-2j}$ has parity $l$, while $(1-z^2{)}^j$ is even. Therefore for $t<0$ the same expansion is ${\sum }_{l\ge 0}{t}^l{c}_l(x,y)$ as well, where $c_l(x,y)=\mathrm{\mid }y{\mathrm{\mid }}^l\mathrm{\mid }x{\mathrm{\mid }}^{-l-1}{P}_l(x\cdot y\mathrm{/}(\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }y\mathrm{\mid }))$. Hence $\mathrm{\mid }x-ty{\mathrm{\mid }}^{-1}={\sum }_{l\ge 0}{t}^l{c}_l(x,y)$ on the open interval $\mathrm{\mid }t\mathrm{\mid }<\mathrm{\mid }x\mathrm{\mid }\mathrm{/}\mathrm{\mid }y\mathrm{\mid }$. By [L8], $c_l(x,y)=\frac{1}{l!}{\mathrm{\partial }}_t^l{\mid }_{t=0}\mathrm{\mid }x-ty{\mathrm{\mid }}^{-1}$. Expanding the polynomial formula in step 2.1 shows that $\mathrm{\mid }y{\mathrm{\mid }}^l{P}_l(n\cdot y\mathrm{/}\mathrm{\mid }y\mathrm{\mid })$ is a homogeneous polynomial of degree $l$ in $y$: its $j$-term is a constant times $[\mathrm{\mid }y{\mathrm{\mid }}^2-(n\cdot y{)}^2{]}^j(n\cdot y{)}^{l-2j}$. Thus the coefficients extend polynomially to $y=0$ with the values already assigned.

For fixed $x\mathrm{\ne }0$ and small real $t$ the function $(x,t)\mapsto \mathrm{\mid }x-ty{\mathrm{\mid }}^{-1}$ is smooth and harmonic in $x$ off $x=ty$ by [L7]; since mixed partials commute ([L7]) the operator ${\mathrm{\Delta }}_x$ commutes with ${\mathrm{\partial }}_t^l$, so ${\mathrm{\partial }}_t^l\mathrm{\mid }x-ty{\mathrm{\mid }}^{-1}$ is harmonic in $x$ on the open set where $x\mathrm{\ne }ty$; restricting to $t=0$ and using step 7.1, every $c_l(\cdot ,y)$ is smooth and harmonic on ${\mathbb{R}}^3\setminus \{0\}$. Replacing $x$ by $y$ in the same computation, ${\mathrm{\Delta }}_y\mathrm{\mid }x-ty{\mathrm{\mid }}^{-1}=t^2{\mathrm{\Delta }}_x\mathrm{\mid }x-ty{\mathrm{\mid }}^{-1}=0$, shows that the degree-$l$ solid polynomial $y\mapsto c_l(x,y)$ is harmonic in $y$ for each fixed $x\mathrm{\ne }0$.

Fix a nonempty compact $K$ with ${\inf }_K\mathrm{\mid }x\mathrm{\mid }>a$; choose $\delta >0$ and an open bounded neighbourhood $K_{+}$ of $K$ with ${\inf }_{K_{+}}\mathrm{\mid }x\mathrm{\mid }>a$ and every closed $\delta$-ball centred at a point of $K$ contained in $K_{+}$. For $\mathrm{\mid }y\mathrm{\mid }\le a$ and $x\in K_{+}$ the tail $v_N(x,y)=\mathrm{\mid }x-y{\mathrm{\mid }}^{-1}-{\sum }_{l=0}^N{c}_l(x,y)$ is harmonic in $x$ by [L7] and step 8.1, and equals ${\sum }_{l>N}{c}_l(x,y)$ by step 6.1 and satisfies $\mathrm{\mid }{v}_N(x,y)\mathrm{\mid }\le a^{N+1}\left({\inf }_{K_{+}}\mathrm{\mid }x\mathrm{\mid }{)}^{-N-2}\right(1-a\mathrm{/}{\inf }_{K_{+}}\mathrm{\mid }x\mathrm{\mid }{)}^{-1}\to 0$ uniformly on $K_{+}\times \overline{B_a(0)}$ by step 6.1. Hence for every multi-index $\alpha$ the interior derivative estimates of [L7] give $\mathrm{\mid }{D}^{\alpha }{v}_N(x,y)\mathrm{\mid }\le C_{3,\alpha }{\delta }^{-3-\mathrm{\mid }\alpha \mathrm{\mid }}\mathrm{\mid }{B}_{\delta }\mathrm{\mid }{\sup }_{K_{+}}\mathrm{\mid }{v}_N(\cdot ,y)\mathrm{\mid }\to 0$ uniformly on $K$, uniformly for $\mathrm{\mid }y\mathrm{\mid }\le a$. Each sequence $N\mapsto D^{\alpha }{S}_N$ of derivatives of the partial sums is therefore uniformly Cauchy, and its uniform limit is identified with $D^{\alpha }$ of the kernel along short coordinate segments by Newton–Leibniz needs only continuity on $[a,b]$, differentiability on $(a,b)$, and a Riemann-integrable extension of the interior derivative; thus every fixed spatial derivative series of the kernel converges uniformly on $K$, uniformly for $\mathrm{\mid }y\mathrm{\mid }\le a$. Countable Choice is used only through the harmonic-estimate interface [L7]; the empty $K$ case is vacuous.

For a density $\rho$ vanishing outside $\overline{B_a(0)}$ and $\epsilon >0$ there is $N$ with $\mid \mathrm{\mid }x-y{\mathrm{\mid }}^{-1}-{\sum }_{l\le N}{c}_l(x,y)\mid \le \epsilon$ for all $x\in K$ and $\mathrm{\mid }y\mathrm{\mid }\le a$ by step 9.1, so $\mid \int \rho (y)[\mathrm{\mid }x-y{\mathrm{\mid }}^{-1}-{\sum }_{l\le N}{c}_l(x,y)]dy\mid \le \epsilon \mathrm{\parallel }\rho {\mathrm{\parallel }}_1$; the partial-sum integrals equal ${\sum }_{l\le N}\int \rho (y)c_l(x,y)dy$ by the complex linearity of the Lebesgue integral [L8], and hence $\int \rho (y)\mathrm{\mid }x-y{\mathrm{\mid }}^{-1}dy={\sum }_{l\ge 0}\int \rho (y)c_l(x,y)dy$ with the remainder bounded by $\mathrm{\parallel }\rho {\mathrm{\parallel }}_1{\sup }_{\mathrm{\mid }y\mathrm{\mid }\le a}\mid \mathrm{\mid }x-y{\mathrm{\mid }}^{-1}-{\sum }_{l\le N}{c}_l(x,y)\mid$ and, on $\mathrm{\mid }x\mathrm{\mid }\ge r>a$, by $\mathrm{\parallel }\rho {\mathrm{\parallel }}_1\mathrm{\mid }a{\mathrm{\mid }}^{N+1}{r}^{-N-2}(1-r^{-1}a{)}^{-1}$ (step 6.1, with Monotonicity and nonnegative homogeneity of the nonnegative integral for the $L^1$ factor). Moreover, on a small open coordinate interval $J$ with $\bar{J}\subset K_{+}$ every derivative $D_x^{\alpha }\mathrm{\mid }x-y{\mathrm{\mid }}^{-1}$ is bounded on $J\times \overline{B_a(0)}$ by a constant $C_{\alpha }$, since $\mathrm{\mid }x-y\mathrm{\mid }\ge {\inf }_{K_{+}}\mathrm{\mid }x\mathrm{\mid }-a>0$ there; the majorant $C_{\alpha }\mathrm{\mid }\rho (y)\mathrm{\mid }$ is integrable, so Differentiation under the integral sign applies (with empty exceptional set) and identifies every spatial derivative of $x\mapsto \int \rho (y)\mathrm{\mid }x-y{\mathrm{\mid }}^{-1}dy$ on $J$ with $\int \rho (y)D_x^{\alpha }\mathrm{\mid }x-y{\mathrm{\mid }}^{-1}dy$, i.e. with the corresponding termwise-differentiated multipole series.

Collecting the results: $P_l$ is a real polynomial of degree at most $l$ with $P_l(1)=1$ and $\mathrm{\mid }{P}_l\mathrm{\mid }\le 1$ on $[-1,1]$ (steps 2.1, 2.2); for $\mathrm{\mid }q\mathrm{\mid }<1$ the series ${\sum }_l{q}^l{P}_l(z)$ converges absolutely and uniformly on $[-1,1]$ and equals the analytic inverse square root of $1-2qz+q^2$ by steps 3.1, 4.1 and 5.1; the exterior expansion and its geometric remainder hold uniformly on $\mathrm{\mid }y\mathrm{\mid }\le a<r\le \mathrm{\mid }x\mathrm{\mid }$ by step 6.1, with $y=0$ handled by the zero higher terms; the integrated $L^1$-source expansion converges with error at most $\mathrm{\parallel }\rho {\mathrm{\parallel }}_1$ times the geometric tail, and every fixed spatial derivative series converges uniformly on observation compacta, uniformly over $\mathrm{\mid }y\mathrm{\mid }\le a$, with derivatives of the integral identified directly by differentiation under the integral sign (steps 9.1, 10.1); the coefficient terms are smooth and harmonic in $x$ away from the origin and are solid harmonics in $y$ (step 8.1). The finite-Riemann polynomial and generating assertions use no Choice (step 4.1); Countable Choice enters only through the Lebesgue and harmonic-estimate interfaces [L7, L8].