Proposition: Explicit spectral integration enclosure

Let $U\ge 1$, $\alpha >0$, and use composite Simpson quadrature with an even number $n$ of equal intervals on $[0,U]$. For $j=0,\dots ,4$, define

Let $P_0=1$ and

If $P_j(u)={\sum }_{\ell }{p}_{j,\ell }{u}^{\ell }$, put

Then the finite-interval quadrature error for $b(\alpha )$ is at most

The omitted tail is bounded by

Validated nodal intervals and an interval for $2/\pi$, combined with these errors, give a rigorous enclosure for $b(\alpha )$.

An enclosure at ${\alpha }_0$ extends to a whole regularization cell by adding $48{\sup }_{\alpha \text{ in cell}}|{\alpha }^2-{{\alpha }_0}^2|$.

Proof. The normalized sphere form factor is an average of $e^{-iuZ}$ with $|Z|\le 1$, so $|F^{(j)}|\le 1$. Leibniz's rule gives $|(F^2{)}^{(j)}|\le 2^j$. Since $\mathrm{sinc}(vu)={{\int }_0}^1\cos (vtu)\mathrm{d}t$, $|S^{(j)}|\le s_j$. The Gaussian derivative is $P_j{e}^{-{\alpha }^2{u}^2}$, bounded by $e_j$. The product derivative bound is therefore $M_4$. The standard Simpson remainder, obtained by integrating the fourth-order interpolation remainder panel by panel, gives the stated finite-interval error.

For the first tail bound, use $|F|\le 1$, $|S|\le 16$, and

For $u\ge 1$, $|F(u)|\le 6/u^2$. Integrating $16\cdot 36/u^4$ gives the second bound after multiplying by $2/\pi$. The cell extension is Theorem thm:small.