Absolute convergence and local parameter domination for Gaussian basis integrals

Repeated product-rule differentiation shows that every fixed spatial or primitive-parameter derivative of a Gaussian factor is a finite sum of polynomial-Gaussian terms. On the compact parameter sets in the Statement, choose $0<a_0\le a\le a_1<\mathrm{\infty }$ and $\mathrm{\parallel }A\mathrm{\parallel }\le R$. The coefficients of these finitely many polynomials are bounded uniformly, since they are polynomials in the bounded parameters. Moreover, $\mathrm{\parallel }x-A{\mathrm{\parallel }}^2\ge \frac{1}{2}\mathrm{\parallel }x{\mathrm{\parallel }}^2-\mathrm{\parallel }A{\mathrm{\parallel }}^2$ and $e^{-a\mathrm{\parallel }x-A{\mathrm{\parallel }}^2}\le e^{-a_0\mathrm{\parallel }x-A{\mathrm{\parallel }}^2}\le e^{a_0{R}^2}{e}^{-a_0\mathrm{\parallel }x{\mathrm{\parallel }}^2\mathrm{/}2}$. Each term is therefore bounded by $K(1+\mathrm{\parallel }x\mathrm{\parallel }{)}^N{e}^{-a_0\mathrm{\parallel }x{\mathrm{\parallel }}^2\mathrm{/}2}$ for finite $K,N$ uniform on those parameter sets. This majorant is integrable by [F2] and bounded, so each pair product is in $L^1\cap L^{\mathrm{\infty }}$; products involving $\mathrm{\Delta }\psi$ have the same form. For fixed parameters use singleton compact sets.

The overlap and kinetic integrands are polynomial-Gaussian products and are absolutely integrable by step 1.1. For a nuclear center $C$, the pair product is bounded on $\mathrm{\parallel }x-C\mathrm{\parallel }<1$. To check local kernel integrability without a radial-coordinate formula, partition $0<\mathrm{\parallel }z\mathrm{\parallel }<1$ into $A_k=\{2^{-(k+1)}\le \mathrm{\parallel }z\mathrm{\parallel }<2^{-k}\}$: on $A_k$, $\mathrm{\parallel }z{\mathrm{\parallel }}^{-1}\le 2^{k+1}$ and $A_k$ lies in a cube of volume $8\cdot 2^{-3k}$, so ${\int }_{A_k}\mathrm{\parallel }z{\mathrm{\parallel }}^{-1}dz\le 16\cdot 4^{-k}$. The sum is $64\mathrm{/}3$. Outside this unit ball the kernel is at most one, so the remaining integral is bounded by the pair product's $L^1$ norm. This proves absolute convergence of the point-nucleus attraction.

Let $f,g$ be two pair products from finite contractions; by step 1.1 they belong to $L^1\cap L^{\mathrm{\infty }}$. On $\mathrm{\parallel }x-y\mathrm{\parallel }<1$, Tonelli [F3] and $z=y-x$ give the bound $\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}{\int }_{\mathrm{\parallel }z\mathrm{\parallel }<1}\mathrm{\parallel }z{\mathrm{\parallel }}^{-1}dz\le (64\mathrm{/}3)\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$ by the shell calculation in step 2.1. On the complement, $\mathrm{\parallel }x-y{\mathrm{\parallel }}^{-1}\le 1$, giving bound $\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1$. Hence the two-electron integrand is absolutely integrable, and [F4] supplies its product and iterated integral meanings.

All the allowed derivatives act on Gaussian factors, so at every fixed finite order step 1.1 applies; the kernels, singular centers $C$, and diagonal $x=y$ remain fixed. Products of those uniform Gaussian envelopes, multiplied by the unchanged kernel when present, are common integrable majorants by steps 2.1 and 3.1. For difference quotients, enlarge the compact parameter bounds slightly within the positive-exponent domain so they contain all sufficiently short parameter segments. Apply [L8] to each real Gaussian term on such a segment; its quotient is bounded by the envelope for the corresponding next derivative. The same near-nucleus and near-diagonal estimates make these quotient bounds integrable. Applying [F5] passes the derivative through the integral; repeating at each fixed finite order proves the mixed-derivative assertion. Fixed complex coefficients and finite contraction sums are handled by finite linearity and the triangle inequality.