Proof. According to linear system 7 and the non-singularity in 1, we have \[w^h = (G^h)^{-1} A^{-1} b^h, \quad w = A^{-1} b.\] By the assumptions on \(g(\boldsymbol{x})\), Taylor’s theorem gives \[|g(\boldsymbol{\beta}h) - 1| = |g(\boldsymbol{\beta}h)-g(0)| \leq C|\boldsymbol{\beta}h|^{2p+2} \leq C h^{2p+2}, \qquad 0\leq | \boldsymbol{\beta} |_1\leq p.\] Hence \(\|I-G^h\|_2\leq Ch^{2p+2}\) and \((G^h)^{-1}\) is uniformly bounded for all sufficiently small \(h\). Moreover, 2 implies \(\|b^h-b\|_2\leq Ch^{2p+2}\). Since \(b\) is fixed, this also gives the uniform bound \(\|b^h\|_2\leq \|b\|_2+\|b^h-b\|_2\leq C\) for all sufficiently small \(h\). Therefore \[\begin{align} \|w^h - w\|_2 &= \| (G^h)^{-1} A^{-1} b^h - A^{-1} b \|_2 = \| (G^h)^{-1} (I - G^h) A^{-1} b^h + A^{-1} (b^h - b) \|_2 \\ &\leq \|(G^h)^{-1}\|_2 \|I - G^h\|_2 \|A^{-1}\|_2 \|b^h\|_2 + \|A^{-1}\|_2 \|b^h - b\|_2 leq C h^{2p+2}, \end{align}\] which yields the stated componentwise bound. ◻

Proof. This is a special case of 18. ◻

Proof. Without loss of generality, suppose that \(\alpha_1\) is odd. Since \(g\) and \(\sigma\) are radial, then we have \[\begin{align} I[g\cdot\sigma\cdot\boldsymbol{x}^{\boldsymbol{\alpha}}] &=\int_{[-a,a]^{d-1}}\int_0^a g(|\boldsymbol{x}|)\sigma(|\boldsymbol{x}|) \left[x_1^{\alpha_1}+(-x_1)^{\alpha_1}\right] \prod_{j=2}^d x_j^{\alpha_j}\,\mathrm{d}x_1\cdots\mathrm{d}x_d =0. \end{align}\] Similarly, using \(c_{(i_1,i_2,\ldots,i_d)}=c_{(-i_1,i_2,\ldots,i_d)}\), we obtain \[\begin{align} T_h^0[g\cdot\sigma\cdot\boldsymbol{x}^{\boldsymbol{\alpha}}] &=h^d\sum_{i_2=-N}^N\cdots\sum_{i_d=-N}^N \sum_{i_1=1}^N c_{\boldsymbol{i}}\,g(|\boldsymbol{i}h|)\sigma(|\boldsymbol{i}h|) \left[(i_1h)^{\alpha_1}+(-i_1h)^{\alpha_1}\right] \prod_{j=2}^d(i_jh)^{\alpha_j} =0. \end{align}\] For every sign-orbit \(\mathcal{G}_{\boldsymbol{\beta}}\), the symmetry condition 4 gives \[\sum_{\boldsymbol{i}\in\mathcal{G}_{\boldsymbol{\beta}}}(\boldsymbol{i}h)^{\boldsymbol{\alpha}} =\frac{1}{2}\sum_{\boldsymbol{i}\in\mathcal{G}_{\boldsymbol{\beta}}} \left[(\boldsymbol{i}h)^{\boldsymbol{\alpha}} +((-i_1,i_2,\ldots,i_d)h)^{\boldsymbol{\alpha}}\right]=0.\] Consequently, the correction \(\mathcal{C}_h^p[g\cdot\boldsymbol{x}^{\boldsymbol{\alpha}}]\) is zero and we conclude that \(Q_h^p[g\cdot\sigma\cdot\boldsymbol{x}^{\boldsymbol{\alpha}}]=0\). ◻

Proof. Let \(P_{\phi}(\boldsymbol{x})\) be the Taylor polynomial of \(\phi(\boldsymbol{x})\) at \(\boldsymbol{x}=0\) of degree \(2p+1\) and define the localized surrogate as \[\tilde{\phi}(\boldsymbol{x}) := P_{\phi}(\boldsymbol{x}) g(\boldsymbol{x}), \quad P_{\phi}(\boldsymbol{x}) := \sum_{| \boldsymbol{\alpha} |_1 = 0}^{2p+1} \frac{\partial^{\boldsymbol{\alpha}} \phi(0)}{\boldsymbol{\alpha}!} \boldsymbol{x}^{\boldsymbol{\alpha}},\] where \(g(\boldsymbol{x}) \in C_c^{\infty}(\mathbb{R}^d)\) is compactly supported in domain \(V\) such that \[g(\boldsymbol{x})=g(|\boldsymbol{x}|),\qquad g(0)=1,\qquad \partial^{\boldsymbol{\alpha}}g(0)=0, \quad 1\leq | \boldsymbol{\alpha} |_1\leq 2p+1.\] Decompose the quadrature error of the singular integral as: \[\begin{align} I[\phi \cdot \sigma] - Q_h^p[\phi \cdot \sigma] &= (I - T^0_h) [(\phi - \tilde{\phi}) \cdot \sigma] + \mathcal{C}_{h}^p[\tilde{\phi}- \phi] + (I - Q_h^p)[\tilde{\phi}\cdot \sigma] \\ &:= E_1 + E_2 + E_3. \end{align}\]

For \(E_1\), denote \(f:= \phi - \tilde{\phi}\), then by direct calculation we have \[\label{eq:diff95phi95tphi} f(0) = 0, \quad \partial^{\boldsymbol{\alpha}} f(0) = 0, \quad \text{for } 1 \leq | \boldsymbol{\alpha} |_1 \leq 2p+1.\tag{12}\] Thus, using the fact that \(f \in C^{\infty}_{\text{c}}(V)\), we can write \[\begin{align} E_1 = I[f\cdot \sigma] - T^0_h[f \cdot \sigma] &= \int_{[-a, a]^d} f(\boldsymbol{x}) \frac{1}{|\boldsymbol{x}|^s} \mathrm{d}\boldsymbol{x} - \sum_{| \boldsymbol{i} |_\infty=1}^N h^d f(\boldsymbol{i}h) \frac{1}{|\boldsymbol{i}h|^s} \\ &= h^{d-s} \Big( \int_{\mathbb{R}^d} f(\boldsymbol{x}h) \frac{1}{|\boldsymbol{x}|^s} \mathrm{d}\boldsymbol{x} - \sum_{| \boldsymbol{i} |_\infty=1}^{\infty} f(\boldsymbol{i}h) \frac{1}{|\boldsymbol{i}|^s} \Big) = O(h^{2p+2+d-s}), \end{align}\] where we used 2 with \(\boldsymbol{\alpha}=\boldsymbol{0}\) and \(f(0)=0\) in the last step.

For \(E_2\), note that Taylor’s theorem with 12 gives \(|\tilde{\phi}(\boldsymbol{x})-\phi(\boldsymbol{x})|\leq C|\boldsymbol{x}|^{2p+2}\) on \(V\), hence on the fixed stencil \(\mathcal{X}_p\) with bounded weights, we have \[|E_2| \leq h^{d-s}\sum_{\boldsymbol{i}\in\mathcal{X}_p} |w_{\boldsymbol{i}}|\,|\tilde{\phi}(\boldsymbol{i}h)-\phi(\boldsymbol{i}h)| \leq C h^{2p+2+d-s}.\]

For \(E_3\), expand \(\tilde{\phi}=P_\phi g\) and use 7 to drop the odd monomials we obtain \[(I-Q_h^p)[\tilde{\phi}\cdot\sigma] = \sum_{| \boldsymbol{\alpha} |_1=0}^{p}\frac{\partial^{2\boldsymbol{\alpha}}\phi(0)}{(2\boldsymbol{\alpha})!}\, (I-Q_h^p)[\boldsymbol{x}^{2\boldsymbol{\alpha}}g\cdot\sigma].\] For each even monomial above, according to the moment conditions 3 we have \[I[\boldsymbol{x}^{2\boldsymbol{\alpha}}g\cdot\sigma] - T_h^0[\boldsymbol{x}^{2\boldsymbol{\alpha}}g\cdot\sigma] = h^{d-s}\sum_{\boldsymbol{i}\in\mathcal{X}_p} w_{\boldsymbol{i}}^h\,(\boldsymbol{i}h)^{2\boldsymbol{\alpha}}g(\boldsymbol{i}h).\] On the other hand, the SinCoTrap rule \(Q_h^p\) uses the converged weights \(w_{\boldsymbol{i}}\), so \[(I-Q_h^p)[\boldsymbol{x}^{2\boldsymbol{\alpha}}g\cdot\sigma] = \left(I-T_h^0\right)[\boldsymbol{x}^{2\boldsymbol{\alpha}}g\cdot\sigma] - h^{d-s}\sum_{\boldsymbol{i}\in\mathcal{X}_p} w_{\boldsymbol{i}}\,(\boldsymbol{i}h)^{2\boldsymbol{\alpha}}g(\boldsymbol{i}h).\] Substituting the moment identity gives \[(I-Q_h^p)[\boldsymbol{x}^{2\boldsymbol{\alpha}}g\cdot\sigma] = h^{d-s}\sum_{\boldsymbol{i}\in\mathcal{X}_p} (w_{\boldsymbol{i}}^h-w_{\boldsymbol{i}})\,(\boldsymbol{i}h)^{2\boldsymbol{\alpha}}g(\boldsymbol{i}h).\] Thus, by 3 we have \[\begin{align} |E_3| \leq C h^{d-s} \sum_{| \boldsymbol{\alpha} |_1 = 0}^{p} \sum_{\boldsymbol{i}\in \mathcal{X}_p} \left|(\boldsymbol{i}h)^{2\boldsymbol{\alpha}} g(\boldsymbol{i}h)\right| \left| w_{\boldsymbol{i}}^h - w_{\boldsymbol{i}} \right| \leq C h^{2p+2+d-s}. \end{align}\]

Combining the three bounds yields the claim. ◻

Proof. Using the smooth cutoff \(\eta\) defined in 1 , we split the integral as \[I[\varphi\cdot \sigma] = I[\varphi\cdot \sigma \cdot \eta] + I[\varphi\cdot \sigma \cdot (1 - \eta)].\] For the first term, since \(\varphi(\boldsymbol{x})\eta(\boldsymbol{x})\) is smooth and compactly supported in \(V\), 2 implies \[I[\varphi\cdot \sigma \cdot \eta] = T^0_h[\varphi\cdot \sigma \cdot \eta] + \mathcal{C}_{h}^p[\varphi\cdot \eta] + O(h^{2p+2+d-s}).\] For the second term, since \(\varphi(\boldsymbol{x}) \sigma(\boldsymbol{x}) (1 - \eta(\boldsymbol{x}))\) is \(0\) at \(x=0\), the punctured trapezoidal rule is equivalent to the standard trapezoidal rule. Moreover, since \(\varphi(\boldsymbol{x}) \sigma(\boldsymbol{x}) (1 - \eta(\boldsymbol{x}))\) is smooth and periodic with respect to the domain \(V\), the Euler-Maclaurin formula yields \[I[\varphi\cdot \sigma \cdot (1 - \eta)] = T^0_h[\varphi\cdot \sigma \cdot (1 - \eta)] + O(h^{q}),\] for any \(q>0\). Finally, note that \(\mathcal{C}_{h}^p[\varphi\cdot \eta] = \mathcal{C}_{h}^p[\varphi]\) for \(h>0\) sufficient small, we have \[\begin{align} I[\varphi\cdot \sigma] &= T^0_h[\varphi\cdot \sigma \cdot \eta] + \mathcal{C}_{h}^p[\varphi\cdot \eta] + O(h^{2p+2+d-s}) + T^0_h[\varphi\cdot \sigma \cdot (1 - \eta)] + O(h^{q}) \\ &= T^0_h[\varphi\cdot \sigma] + \mathcal{C}_{h}^p[\varphi] + O(h^{2p+2+d-s}), \end{align}\] which completes the proof. ◻

Proof. Define \(F(\boldsymbol{x}):=g(\boldsymbol{x})\sigma_B(\boldsymbol{x})\boldsymbol{x}^{\boldsymbol{\alpha}}\). The radiality of \(g\) and the central symmetry of \(\sigma_B\) imply that \(g(-\boldsymbol{x})=g(\boldsymbol{x})\) and \(\sigma_B(-\boldsymbol{x})=\sigma_B(\boldsymbol{x})\). Moreover, since \(| \boldsymbol{\alpha} |_1\) is odd, \[(-\boldsymbol{x})^{\boldsymbol{\alpha}} =(-1)^{| \boldsymbol{\alpha} |_1}\boldsymbol{x}^{\boldsymbol{\alpha}} =-\boldsymbol{x}^{\boldsymbol{\alpha}}.\] Hence \(F(-\boldsymbol{x})=-F(\boldsymbol{x})\). Because \(V\) is centrally symmetric, \[I[g\cdot\sigma_B\cdot\boldsymbol{x}^{\boldsymbol{\alpha}}] = \int_V F(\boldsymbol{x})\,\mathrm{d}\boldsymbol{x} = \frac{1}{2}\int_V\bigl(F(\boldsymbol{x})+F(-\boldsymbol{x})\bigr)\,\mathrm{d}\boldsymbol{x} = 0.\] For the punctured trapezoidal sum, the index set \(\{\boldsymbol{i}\in\mathbb{Z}^d:0<| \boldsymbol{i} |_\infty\le N\}\) is centrally symmetric and \(c_{-\boldsymbol{i}}=c_{\boldsymbol{i}}\), hence terms cancel in pairs: \[T_h^0[g\cdot\sigma_B\cdot\boldsymbol{x}^{\boldsymbol{\alpha}}] = \frac{h^d}{2}\sum_{0<| \boldsymbol{i} |_\infty\le N} c_{\boldsymbol{i}}\Bigl(F(\boldsymbol{i}h)+F(-\boldsymbol{i}h)\Bigr)=0.\] Likewise, central symmetry of \(\mathcal{X}_p\) and \(w_{-\boldsymbol{i}}=w_{\boldsymbol{i}}\) gives \[h^{d-s}\sum_{\boldsymbol{i}\in\mathcal{X}_p} w_{\boldsymbol{i}}\,g(\boldsymbol{i}h)\,(\boldsymbol{i}h)^{\boldsymbol{\alpha}} = \frac{h^{d-s}}{2}\sum_{\boldsymbol{i}\in\mathcal{X}_p} w_{\boldsymbol{i}}\Bigl(g(\boldsymbol{i}h)(\boldsymbol{i}h)^{\boldsymbol{\alpha}}+g(-\boldsymbol{i}h)(-\boldsymbol{i}h)^{\boldsymbol{\alpha}}\Bigr)=0.\] Therefore \(Q_h^p[g\cdot\sigma_B\cdot\boldsymbol{x}^{\boldsymbol{\alpha}}]=0\). ◻

Proof. The argument is similar to the proof of 2 after replacing \(\sigma(\boldsymbol{x})=|\boldsymbol{x}|^{-s}\) by \(\sigma_B(\boldsymbol{x})=|\boldsymbol{x}^\top B \boldsymbol{x}|^{-s/2}\). The only change is that the componentwise odd-moment cancellation (7) is replaced by the odd total-degree cancellation (8). ◻

Proof. The number of lattice points \(\#S_n\) is the difference of number of lattice points in nested cubes, i.e., \[\#\{\boldsymbol{k}\in\mathbb{Z}^d:| \boldsymbol{k} |_\infty\le n\}=(2n+1)^d, \quad \#\{\boldsymbol{k}\in\mathbb{Z}^d:| \boldsymbol{k} |_\infty\le n-1\}=(2n-1)^d.\] For the upper bound, apply the mean value theorem to \(f(x)=x^d\) on the interval \((2n-1,2n+1)\), then there exists \(\xi\in(2n-1,2n+1)\) such that \[(2n+1)^d-(2n-1)^d = 2 f'(\xi) = 2d\,\xi^{d-1} \leq 2d(2n+1)^{d-1} \leq 2d\,3^{d-1}\,n^{d-1}.\] ◻

Proof. Since \(|\boldsymbol{x}|>0\), for complex \(s\) we have \(|\boldsymbol{x}|^{-s}:=\exp(-s\log|\boldsymbol{x}|)\), where \(\log|\boldsymbol{x}|\) is real, so no choice of complex logarithm branch is involved. Differentiation with respect to the real variable \(x_j\) therefore gives \[\frac{\partial}{\partial x_j}|\boldsymbol{x}|^{-s} =-s x_j|\boldsymbol{x}|^{-s-2},\] which is the same formula as for real \(s\).

We complete the proof by induction on \(q\). For \(q=0\), \[\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} = \frac{1}{|\boldsymbol{x}|^{s-2| \boldsymbol{\alpha} |_1}} \left( \frac{\boldsymbol{x}}{|\boldsymbol{x}|} \right)^{2\boldsymbol{\alpha}}, \quad P_{0, \boldsymbol{\alpha},j} (\boldsymbol{x}) = \boldsymbol{x}^{2\boldsymbol{\alpha}}.\] For the inductive step, we assume the result holds for \(q\) and differentiate again: \[\begin{align} \frac{\partial^{q+1}}{\partial x_j^{q+1}} \left( \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} \right) &= \frac{\partial}{\partial x_j} \left( \frac{1}{|\boldsymbol{x}|^{s + q - 2| \boldsymbol{\alpha} |_1}} P_{q, \boldsymbol{\alpha},j}\left( \frac{\boldsymbol{x}}{|\boldsymbol{x}|} \right) \right) \\ &= \frac{-(s + q - 2| \boldsymbol{\alpha} |_1)}{|\boldsymbol{x}|^{s + q - 2| \boldsymbol{\alpha} |_1 + 1}} \, \frac{x_j}{|\boldsymbol{x}|} \, P_{q, \boldsymbol{\alpha},j}\left( \frac{\boldsymbol{x}}{|\boldsymbol{x}|} \right) \\ &\quad + \frac{1}{|\boldsymbol{x}|^{s + q - 2| \boldsymbol{\alpha} |_1}} \left[ P^{(j)}_{q, \boldsymbol{\alpha},j}\left(\frac{\boldsymbol{x}}{|\boldsymbol{x}|}\right) \left(\frac{1}{|\boldsymbol{x}|} - \frac{x_j^2}{|\boldsymbol{x}|^3} \right) - \sum_{l\neq j} P^{(l)}_{q, \boldsymbol{\alpha},j}\left(\frac{\boldsymbol{x}}{|\boldsymbol{x}|}\right) \frac{x_jx_l}{|\boldsymbol{x}|^3} \right] \\ &:= \frac{1}{|\boldsymbol{x}|^{s + q + 1 - 2| \boldsymbol{\alpha} |_1}} P_{q+1, \boldsymbol{\alpha},j}\left( \frac{\boldsymbol{x}}{|\boldsymbol{x}|} \right), \end{align}\] where \(P^{(l)}_{q,\boldsymbol{\alpha},j}(\boldsymbol{x})=\frac{\partial}{\partial x_l} P_{q,\boldsymbol{\alpha},j}(\boldsymbol{x})\) for \(l=1, \ldots,d\) are polynomials with \(\deg P^{(l)}_{q,\boldsymbol{\alpha},j} \leq \deg P_{q,\boldsymbol{\alpha},j} - 1\), and \[P_{q+1, \boldsymbol{\alpha},j}(\boldsymbol{x}) := -(s + q - 2| \boldsymbol{\alpha} |_1) x_j P_{q, \boldsymbol{\alpha},j}\left(\boldsymbol{x}\right) + P^{(j)}_{q, \boldsymbol{\alpha},j}\left(\boldsymbol{x}\right)\left(1 - x_j^2\right) - \sum_{l\neq j} P^{(l)}_{q, \boldsymbol{\alpha},j}\left(\boldsymbol{x}\right) x_j x_l\] is a polynomial with degree at most \(q + 2| \boldsymbol{\alpha} |_1 + 1\) in variable \(\boldsymbol{x}= (x_1, \ldots, x_d)\). ◻

Proof of 2. We begin by introducing a smooth cutoff function \(\eta(\boldsymbol{x}) \in C^{\infty}(\mathbb{R}^d)\) such that \[\label{eq:cutoff95rhs} \eta(\boldsymbol{x}) = \eta(|\boldsymbol{x}|), \quad \eta(\boldsymbol{x}) = \begin{cases} 0, & |\boldsymbol{x}| \leq \frac{1}{2}, \\ 1, & |\boldsymbol{x}| \geq 1. \end{cases}\tag{17}\] Then we rewrite the difference of the integral and summation as: \[\begin{align} & \quad \int_{\mathbb{R}^d} g(\boldsymbol{x}h) \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} \,\mathrm{d}\boldsymbol{x} - \sum_{| \boldsymbol{i} |_\infty=1}^{\infty} g(\boldsymbol{i}h) \frac{\boldsymbol{i}^{2\boldsymbol{\alpha}}}{|\boldsymbol{i}|^s} \\ &= \int_{|\boldsymbol{x}| \leq 1} g(\boldsymbol{x}h) \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} (1 - \eta(\boldsymbol{x})) \,\mathrm{d}\boldsymbol{x} + \left( \int_{\mathbb{R}^d} g(\boldsymbol{x}h) \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} \eta(\boldsymbol{x}) \,\mathrm{d}\boldsymbol{x} - \sum_{| \boldsymbol{i} |_\infty=1}^{\infty} g(\boldsymbol{i}h) \frac{\boldsymbol{i}^{2\boldsymbol{\alpha}}}{|\boldsymbol{i}|^s} \eta(\boldsymbol{i}) \right) \\ &:= {\rm I} + {\rm II}. \end{align}\]

8.0.0.1 Estimate of \({\rm I}\).

Note that \(\partial^{\boldsymbol{\beta}} g(0) = 0\) for \(1 \leq | \boldsymbol{\beta} |_1 \leq 2p+1\). Moreover, for multi-index \(| \boldsymbol{\beta} |_1=2p+2\) and constant \(\theta \in [0, 1]\), \[\int_{|\boldsymbol{x}| \leq 1} \Big| \frac{\partial^{\boldsymbol{\beta}} g(\theta \boldsymbol{x}h)}{\boldsymbol{\beta}!} (\boldsymbol{x}h)^{\boldsymbol{\beta}} \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} (1 - \eta(\boldsymbol{x})) \Big| \,\mathrm{d}\boldsymbol{x} \leq C h^{2p+2} \int_{|\boldsymbol{x}|\leq 1} |\boldsymbol{x}|^{2| \boldsymbol{\alpha} |_1 + 2p + 2 - \mathrm{Re}(s)} \,\mathrm{d}\boldsymbol{x} \leq C h^{2p+2},\] where \(0 < \mathrm{Re}(s) < d + 2| \boldsymbol{\alpha} |_1\) guarantees the convergence of the integral. Thus, the Taylor expansion of \(g(\boldsymbol{x} h)\) around \(\boldsymbol{x}=0\) gives \[{\rm I} = \int_{|\boldsymbol{x}| \leq 1} g(\boldsymbol{x}h) \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} (1 - \eta(\boldsymbol{x})) \,\mathrm{d}\boldsymbol{x} = g(0) \int_{|\boldsymbol{x}|\leq 1} \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} (1 - \eta(\boldsymbol{x})) \,\mathrm{d}\boldsymbol{x}+ O(h^{2p+2}).\]

8.0.0.2 Estimate of \({\rm II}\).

Define \[\label{eq:def95fh95rhs} f_h(\boldsymbol{x}) := g(\boldsymbol{x}h) \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} \eta(\boldsymbol{x}),\tag{18}\] then \(f_h(0)=0\) and \(f_h \in C_c^{\infty}(\mathbb{R}^d)\) for \(h > 0\). Apply the Poisson summation formula in 9 we obtain \[\begin{align} {\rm II} = \int_{\mathbb{R}^d} f_h(\boldsymbol{x}) \,\mathrm{d}\boldsymbol{x}- \sum_{| \boldsymbol{i} |_\infty=1}^{\infty} f_h(\boldsymbol{i}) = \widehat{f_h}(0) - \sum_{| \boldsymbol{i} |_\infty=0}^{\infty} f_h(\boldsymbol{i}) = \widehat{f_h}(0) - \sum_{| \boldsymbol{k} |_\infty=0}^{\infty} \widehat{f_h}(\boldsymbol{k}) = - \sum_{| \boldsymbol{k} |_\infty=1}^{\infty} \widehat{f_h}(\boldsymbol{k}). \end{align}\]

Choose integer \[q = 2p+3+d+2| \boldsymbol{\alpha} |_1.\] For each \(\boldsymbol{k}\in \mathbb{Z}^d \setminus \{0\}\), let index \(j = j(\boldsymbol{k}) \in \text{argmax}_{\ell=1, \ldots, d} |k_{\ell}|\) and hence \(|k_j| = | \boldsymbol{k} |_\infty\). Then we have the following Fourier estimate for \(h>0\) sufficiently small, proved in 12 below: \[\widehat{f_h}(\boldsymbol{k}) = g(0)\frac{W(\boldsymbol{k})}{(2\pi k_j)^q} + R_h(\boldsymbol{k}), \qquad |R_h(\boldsymbol{k})|\le C h^{2p+2} \frac{1}{|k_j|^q}, \qquad |W(\boldsymbol{k})|\le C.\] Furthermore, for \(n\in \mathbb{N}_+\), let \(S_n := \{\boldsymbol{k}\in \mathbb{Z}^d : | \boldsymbol{k} |_\infty = n\}\), by 10 the number of elements in \(S_n\) satisfies \(\# S_n \leq C_d n^{d-1}\) for some constant \(C_d>0\) depending only on the dimension \(d\). Thus, whenever \(q > d\), we have \[\sum_{| \boldsymbol{k} |_\infty=1}^{\infty} \frac{1}{|k_j|^q} = \sum_{n=1}^{\infty} \sum_{\boldsymbol{k}\in S_n} \frac{1}{n^q} \leq C_d \sum_{n=1}^{\infty} \frac{1}{n^{q-d+1}} < \infty.\] Therefore, \[{\rm II} = - \sum_{| \boldsymbol{k} |_\infty=1}^{\infty} \widehat{f_h}(\boldsymbol{k}) = - g(0) \sum_{| \boldsymbol{k} |_\infty=1}^{\infty} \frac{W(\boldsymbol{k})}{(2\pi k_j)^q} + O(h^{2p+2}).\]

8.0.0.3 Conclusion.

Combining the estimates for \({\rm I}\) and \({\rm II}\) yields \[{\rm I} + {\rm II} = g(0)b_{\boldsymbol{\alpha}}(s) + O(h^{2p+2}), \quad b_{\boldsymbol{\alpha}}(s) := \int_{|\boldsymbol{x}|\le 1}\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}(1-\eta(\boldsymbol{x}))\,\mathrm{d}\boldsymbol{x} - \sum_{| \boldsymbol{k} |_\infty=1}^{\infty} \frac{W(\boldsymbol{k})}{(2\pi k_j)^q}.\] ◻

Proof. Since \(f_h\in C_c^{\infty}(\mathbb{R}^d)\), after integration by parts \(q\) times in the \(x_j\) variable we have \[\widehat{f_h}(\boldsymbol{k}) = \int_{\mathbb{R}^d} f_h(\boldsymbol{x})\, \mathrm{e}^{-2\pi \imath\, \boldsymbol{k}\cdot \boldsymbol{x}} \,\mathrm{d}\boldsymbol{x} = \left( \frac{1}{2\pi \imath k_j} \right)^q \int_{\mathbb{R}^d} \partial^q_j \left( g(\boldsymbol{x}h) \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} \eta(\boldsymbol{x}) \right) \mathrm{e}^{-2\pi \imath\, \boldsymbol{k}\cdot \boldsymbol{x}} \,\mathrm{d}\boldsymbol{x}.\]

8.0.0.4 Bound of \(W(\boldsymbol{k})\).

By definition of \(W(\boldsymbol{k})\), note that \(\eta(\boldsymbol{x})=0\) for \(|\boldsymbol{x}|\leq \frac{1}{2}\) and \(\eta(\boldsymbol{x})=1\) for \(|\boldsymbol{x}|\geq 1\), we have \[|W(\boldsymbol{k})| \leq \int_{\mathbb{R}^d}\left|\partial_j^q\!\left(\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}\eta(\boldsymbol{x})\right)\right|\,\mathrm{d}\boldsymbol{x} \leq \int_{\frac{1}{2} \leq |\boldsymbol{x}| \leq 1} \left|\partial_j^q\!\left(\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}\eta(\boldsymbol{x})\right)\right|\,\mathrm{d}\boldsymbol{x} + \int_{|\boldsymbol{x}| \geq 1} \left|\partial_j^q\!\left(\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}\right)\right|\,\mathrm{d}\boldsymbol{x}.\] On the annulus \(\{\frac{1}{2}\leq|\boldsymbol{x}|\leq1\}\), the integrand is smooth and hence integrable. For \(|\boldsymbol{x}|\geq1\), 11 and the boundedness of \(P_{q,\boldsymbol{\alpha},j}\) on the unit sphere imply \[\left|\partial_j^q\!\left(\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}\right)\right| \leq C|\boldsymbol{x}|^{-q-\mathrm{Re}(s)+2| \boldsymbol{\alpha} |_1}, \qquad |\boldsymbol{x}|\geq1.\] Here \(C\) is chosen independently of \(j\), since \(j\in\{1,\ldots,d\}\). Consequently, \[\int_{|\boldsymbol{x}|\geq1} \left|\partial_j^q\!\left(\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}\right)\right|\,\mathrm{d}\boldsymbol{x} \leq C\int_1^\infty r^{-q-\mathrm{Re}(s)+2| \boldsymbol{\alpha} |_1+d-1}\,\mathrm{d}r<\infty,\] because \(q+\mathrm{Re}(s)>d+2| \boldsymbol{\alpha} |_1\). Thus we have \(|W(\boldsymbol{k})|\leq C\).

8.0.0.5 Estimate of the remainder \(R_h(\boldsymbol{k})\).

Define \[\Delta_h(\boldsymbol{x}):=\bigl(g(h\boldsymbol{x})-g(0)\bigr)\eta(\boldsymbol{x})\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}.\] Then \[\widehat{f_h}(\boldsymbol{k}) - g(0)\frac{W(\boldsymbol{k})}{(2\pi k_j)^q} = \left(\frac{1}{2\pi \imath k_j}\right)^q \int_{\mathbb{R}^d} \partial_j^q \Delta_h(\boldsymbol{x})\,\mathrm{e}^{-2\pi \imath\boldsymbol{k}\cdot \boldsymbol{x}}\,\mathrm{d}\boldsymbol{x} =: R_h(\boldsymbol{k}),\] and hence \[|R_h(\boldsymbol{k})| \le (2\pi)^{-q}|k_j|^{-q}\,\|\partial_j^q\Delta_h\|_{L^1(\mathbb{R}^d)}.\] It remains to show \(\|\partial_j^q\Delta_h\|_{L^1(\mathbb{R}^d)}\le Ch^{2p+2}\). Since \(\eta(\boldsymbol{x})=0\) for \(|\boldsymbol{x}|\le 1/2\), we only integrate over \(|\boldsymbol{x}|\ge 1/2\). Furthermore, fix \(R>0\) such that \(\operatorname{supp}(g(\boldsymbol{x}))\subset\{\boldsymbol{x}:|\boldsymbol{x}| < R\}\), so \(\operatorname{supp}(g(h\boldsymbol{x}))\subset\{\boldsymbol{x}:|\boldsymbol{x}| < R/h\}\).

By Leibniz’ rule, \[\partial_j^q \Delta_h(\boldsymbol{x}) = \sum_{m=0}^{q}\binom{q}{m}\, \partial_j^{m}\bigl(g(h\boldsymbol{x})-g(0)\bigr)\, \partial_j^{\,q-m}\!\left(\eta(\boldsymbol{x})\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}\right).\] We prove \(\|\partial_j^q\Delta_h\|_{L^1(\mathbb{R}^d)}\le Ch^{2p+2}\) in three steps. We first give bounds for \(\partial_j^{m}(g(h\boldsymbol{x})-g(0))\) and \(\partial_j^{q-m}\!\left(\eta(\boldsymbol{x})\boldsymbol{x}^{2\boldsymbol{\alpha}}/|\boldsymbol{x}|^s\right)\), then combine these bounds to obtain the \(L^1\) estimate of \(\partial_j^q\Delta_h\).

8.0.0.6 Step 1: bounds for \(\partial_j^{m}(g(h\boldsymbol{x})-g(0))\).

Let \(m\in\{0,1,\ldots,q\}\). Since \(\partial^{\boldsymbol{\beta}}g(0)=0\) for \(1\le | \boldsymbol{\beta} |_1\le 2p+1\), Taylor’s theorem gives, for \(m=0\), we have \(|g(h\boldsymbol{x})-g(0)| \le C (h|\boldsymbol{x}|)^{2p+2}\); for \(m=1,\ldots,2p+1\), we have \[\left|\partial_j^m\bigl(g(h\boldsymbol{x})-g(0)\bigr)\right| = h^m |(\partial_j^m g)(h\boldsymbol{x}) - (\partial_j^m g)(0)| \le C h^m (h|\boldsymbol{x}|)^{2p+2-m} = C h^{2p+2}|\boldsymbol{x}|^{2p+2-m}.\] Further, for \(2p+2 \leq m \leq q\), the uniform boundedness of \(\partial_j^m g\) and the chain rule give \[\left|\partial_j^m\bigl(g(h\boldsymbol{x})-g(0)\bigr)\right| =h^m|(\partial_j^m g)(h\boldsymbol{x})|\leq Ch^m.\] Note that on the region \(\{|\boldsymbol{x}|\le R/h\}\), \(h|\boldsymbol{x}| \leq R\) implies \[h^m = h^{2p+2}(h|\boldsymbol{x}|)^{m-(2p+2)}|\boldsymbol{x}|^{2p+2-m} \le C_R h^{2p+2}|\boldsymbol{x}|^{2p+2-m}.\] Therefore, we conclude that for all \(m=0,1,\ldots,q\), \[\left|\partial_j^{m}\bigl(g(h\boldsymbol{x})-g(0)\bigr)\right| \le C h^{2p+2}|\boldsymbol{x}|^{2p+2-m}, \qquad \frac{1}{2}\le |\boldsymbol{x}|\le R/h.\]

8.0.0.7 Step 2: bounds for \(\partial_j^{q-m}\!\left(\eta(\boldsymbol{x})\boldsymbol{x}^{2\boldsymbol{\alpha}}/|\boldsymbol{x}|^s\right)\).

For \(|\boldsymbol{x}|\ge 1\), we have \(\eta(\boldsymbol{x})=1\) and thus 11, together with the boundedness of \(P_{q-m,\boldsymbol{\alpha},j}\) on the unit sphere, gives \[\left|\partial_j^{\,q-m}\!\left(\eta(\boldsymbol{x})\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}\right)\right| = \left|\partial_j^{\,q-m}\!\left(\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}\right)\right| \le C |\boldsymbol{x}|^{2| \boldsymbol{\alpha} |_1-\mathrm{Re}(s)-(q-m)},\qquad |\boldsymbol{x}|\ge 1,\] where \(C\) is chosen uniformly for \(m\in\{0,\ldots,q\}\) and \(j\in\{1,\ldots,d\}\). On the annulus \(\{1/2\le |\boldsymbol{x}|\le 1\}\), the function \(\eta(\boldsymbol{x})\boldsymbol{x}^{2\boldsymbol{\alpha}}/|\boldsymbol{x}|^s\) is smooth and the integral domain is bounded, hence \[\left|\partial_j^{\,q-m}\!\left(\eta(\boldsymbol{x})\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}\right)\right| \leq C \leq C |\boldsymbol{x}|^{2| \boldsymbol{\alpha} |_1-\mathrm{Re}(s)-(q-m)}, \qquad \frac{1}{2}\le |\boldsymbol{x}|\le 1.\] Therefore, we obtain the uniform bound for \(\partial_j^{q-m}\!\left(\eta(\boldsymbol{x})\boldsymbol{x}^{2\boldsymbol{\alpha}}/|\boldsymbol{x}|^s\right)\) over \(|\boldsymbol{x}|\ge 1/2\).

8.0.0.8 Step 3: \(L^1\) estimate of \(\partial_j^q\Delta_h\).

For \(h > 0\) sufficiently small, we have \(R/h > 1\). Using Steps 1, 2 and splitting \(\mathbb{R}^d\) into \(\{\frac{1}{2} \leq |\boldsymbol{x}|\leq R/h\}\) and \(\{|\boldsymbol{x}| \geq R/h\}\), we obtain \[\begin{align} \left\| \partial^q_j \Delta_h \right\|_{L^1(\mathbb{R}^d)} = \left( \int_{\frac{1}{2} \leq |\boldsymbol{x}| \leq R/h} + \int_{|\boldsymbol{x}| \geq R/h} \right) \left| \partial^q_j \Delta_h(\boldsymbol{x}) \right| \mathrm{d}\boldsymbol{x} := T_1 + T_2. \end{align}\] For the first part, we have \[\begin{align} T_{1} &= \int_{\frac{1}{2} \leq |\boldsymbol{x}| \leq R/h} \Big| \sum_{m=0}^q \binom{q}{m}\, \partial_j^{m}\bigl(g(h\boldsymbol{x})-g(0)\bigr)\, \partial_j^{\,q-m}\!\left(\eta(\boldsymbol{x})\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}\right) \Big| \, \mathrm{d}\boldsymbol{x}\\ &\le C h^{2p+2}\sum_{m=0}^{q}\binom{q}{m} \int_{\frac{1}{2} \le |\boldsymbol{x}|\le R/h} |\boldsymbol{x}|^{2p+2-m}\,|\boldsymbol{x}|^{2| \boldsymbol{\alpha} |_1-\mathrm{Re}(s)-(q-m)} \mathrm{d}\boldsymbol{x}\\ &\le C h^{2p+2} \int_{\frac{1}{2}}^{R/h} r^{2p+2+2| \boldsymbol{\alpha} |_1-\mathrm{Re}(s)-q+d-1}\,\mathrm{d}r \le C h^{2p+2}, \end{align}\] where the last inequality uses \(q\ge 2p+3+d+2| \boldsymbol{\alpha} |_1\), so the exponent of \(r\) is at most \(-2-\mathrm{Re}(s)<-2\).

For \(T_2\), note that \(g(h\boldsymbol{x})=0\) when \(|\boldsymbol{x}|\ge R/h\), hence \(\Delta_h(\boldsymbol{x})=-g(0)\eta(\boldsymbol{x})\boldsymbol{x}^{2\boldsymbol{\alpha}}/|\boldsymbol{x}|^s\) there, and since \(\eta(\boldsymbol{x})=1\) for \(|\boldsymbol{x}|\ge 1\) and \(R/h > 1\) for \(h>0\) sufficiently small, \[T_2 = |g(0)| \int_{|\boldsymbol{x}|\ge R/h}\left|\partial_j^q\left(\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}\right)\right|\mathrm{d}\boldsymbol{x} \le C \int_{R/h}^{\infty} r^{-q-\mathrm{Re}(s)+2| \boldsymbol{\alpha} |_1+d-1}\,\mathrm{d}r \le C h^{2p+2}.\] Therefore \(\|\partial_j^q\Delta_h\|_{L^1(\mathbb{R}^d)}\le C h^{2p+2}\), which completes the proof. ◻

Proof of 1. Recall that \[A_{jk}=\sum_{\boldsymbol{i}\in\mathcal{G}_{\boldsymbol{\beta}_k}}\boldsymbol{i}^{2\boldsymbol{\alpha}_j}.\] Since the exponents are even, \(\boldsymbol{i}^{2\boldsymbol{\alpha}_j}\) is constant on the sign-orbit \(\mathcal{G}_{\boldsymbol{\beta}_k}\), hence \[A_{jk}=c(\boldsymbol{\beta}_k)\,\boldsymbol{\beta}_k^{2\boldsymbol{\alpha}_j},\qquad c(\boldsymbol{\beta}):=\#\mathcal{G}_{\boldsymbol{\beta}}=2^{r(\boldsymbol{\beta})},\,\, r(\boldsymbol{\beta}):=\#\{\ell:\beta_\ell\neq 0\}.\] Therefore \(A=MD\), where \(D=\mathrm{diag}(c(\boldsymbol{\beta}_k))\) has strictly positive diagonal entries and \[M_{jk}=\boldsymbol{\beta}_k^{2\boldsymbol{\alpha}_j}.\] Since \(D\) is invertible, it suffices to show that \(M\) is nonsingular.

For \(m\in\mathbb{N}\), define the even 1D polynomials \[q_m(x) := \prod_{r=0}^{m-1} (x^2-r^2) = x^2(x^2-1^2)\cdots(x^2-(m-1)^2), \qquad q_0(x):=1.\] and for \(\boldsymbol{\gamma}\in\mathbb{N}^d\) define the tensor-product polynomials \[Q_{\boldsymbol{\gamma}}(\boldsymbol{x}):=\prod_{\ell=1}^d q_{\gamma_\ell}(x_\ell) = \prod_{\ell=1}^d \prod_{r=0}^{\gamma_\ell-1} (x_\ell^2-r^2).\] For \(\boldsymbol{\beta}\in\mathbb{N}^d\), we have \(Q_{\boldsymbol{\gamma}}(\boldsymbol{\beta})=0\) whenever \(\beta_\ell<\gamma_\ell\) for some \(\ell\) (since then the factor with \(r=\beta_\ell\) appears and vanishes). In particular, \[Q_{\boldsymbol{\gamma}}(\boldsymbol{\beta})=0 \quad \text{unless } \boldsymbol{\gamma}\le \boldsymbol{\beta},\] where \(\boldsymbol{\gamma}\le \boldsymbol{\beta}\) means componentwise inequality. Moreover, for \(| \boldsymbol{\beta} |_1 > 0\), \[Q_{\boldsymbol{\beta}}(\boldsymbol{\beta}) =\prod_{\ell=1}^d\prod_{r=0}^{\beta_\ell-1}(\beta_\ell^2-r^2)\neq 0, \quad \text{and} \quad Q_{\boldsymbol{0}}(\boldsymbol{0}) = 1.\] Order the index set \(\{\boldsymbol{\gamma}\in\mathbb{N}^d:0\le | \boldsymbol{\gamma} |_1\le p\}\) by any linear extension of the partial order \(\boldsymbol{\gamma}\le\boldsymbol{\beta}\) (componentwise). Then the evaluation matrix \[E_{\boldsymbol{\gamma},\boldsymbol{\beta}}:=Q_{\boldsymbol{\gamma}}(\boldsymbol{\beta})\] is upper triangular with nonzero diagonal, hence \(E\) is invertible.

Now we express \(Q_{\boldsymbol{\gamma}}(\boldsymbol{x})\) in the monomial basis. Since \(q_m(x)\) is an even polynomial of degree \(2m\) with leading coefficient \(1\), it admits an expansion \[q_m(x)=\sum_{r=0}^{m}l_{m,r}\,x^{2r},\qquad l_{m,m}=1.\] Consequently, for every \(\boldsymbol{\gamma}\in\mathbb{N}^d\) we can expand the tensor product as \[Q_{\boldsymbol{\gamma}}(\boldsymbol{x}) =\prod_{\ell=1}^d\Bigl(\sum_{r=0}^{\gamma_\ell}l_{\gamma_\ell,r}\,x_\ell^{2r}\Bigr) =\sum_{\boldsymbol{\delta}\le \boldsymbol{\gamma}} L_{\boldsymbol{\gamma},\boldsymbol{\delta}}\,\boldsymbol{x}^{2\boldsymbol{\delta}}, \qquad L_{\boldsymbol{\gamma},\boldsymbol{\delta}}:=\prod_{\ell=1}^d l_{\gamma_\ell,\delta_\ell}.\] Specifically, we have \(L_{\boldsymbol{\gamma},\boldsymbol{\gamma}}=1\). Furthermore, we extend this notation by setting \(L_{\boldsymbol{\gamma},\boldsymbol{\delta}}=0\) whenever \(\delta_\ell>\gamma_\ell\) for at least one component \(\ell\). Under the same ordering for \(\{\boldsymbol{\gamma}\in\mathbb{N}^d:0\le | \boldsymbol{\gamma} |_1\le p\}\) as above, the matrix \(L=[L_{\boldsymbol{\gamma},\boldsymbol{\delta}}]\) is lower triangular with ones on the diagonal, hence invertible.

Evaluating the expansion at \(\boldsymbol{x}=\boldsymbol{\beta}\) yields \[Q_{\boldsymbol{\gamma}}(\boldsymbol{\beta})=\sum_{\boldsymbol{\delta}\le \boldsymbol{\gamma}} L_{\boldsymbol{\gamma},\boldsymbol{\delta}}\,\boldsymbol{\beta}^{2\boldsymbol{\delta}}.\] Stacking these identities for all \(\boldsymbol{\gamma},\boldsymbol{\beta}\) gives the matrix relation \[E = L M.\] Since \(E\) and \(L\) are invertible, then \(M\) is invertible. Therefore \(A=MD\) is invertible as well. ◻

Proof. By dominated convergence, \(f\) is continuous on \(\Omega\). Let \(T\) be any triangle whose closure \(\overline{T}\) is contained in \(\Omega\). By assumption (2), there exists \(M\in L^1(U)\) such that \(|F(z,\boldsymbol{x})|\leq M(\boldsymbol{x})\) for \(z\in\overline{T}\). Hence Fubini’s theorem and Cauchy’s theorem give \[\int_{\partial T}f(z)\,\mathrm{d}z =\int_U\left(\int_{\partial T}F(z,\boldsymbol{x})\,\mathrm{d}z\right)\mathrm{d}\boldsymbol{x} =0.\] Thus \(f\) is holomorphic by Morera’s theorem. ◻

Proof of 4. Recall that in the proof of 2 we obtained \[\begin{align} b_{\boldsymbol{\alpha}}(s) &= \lim_{N\rightarrow \infty} \left( \int_{\mathbb{R}^d} g(\boldsymbol{x}\frac{a}{N}) \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} \mathrm{d}\boldsymbol{x} - \sum_{| \boldsymbol{i} |_\infty=1}^{\infty} g(\boldsymbol{i}\frac{a}{N}) \frac{\boldsymbol{i}^{2\boldsymbol{\alpha}}}{|\boldsymbol{i}|^s} \right) \\ &= \int_{|\boldsymbol{x}|\le 1}\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}(1-\eta(\boldsymbol{x}))\,\mathrm{d}\boldsymbol{x} - \sum_{| \boldsymbol{k} |_\infty=1}^{\infty}\frac{W(\boldsymbol{k};s)}{(2\pi k_j)^q} =: b_{\boldsymbol{\alpha}}^{(1)}(s) - b_{\boldsymbol{\alpha}}^{(2)}(s), \qquad s\in D_1, \end{align}\] where \(j=j(\boldsymbol{k})\in \text{argmax}_{\ell=1, \ldots, d} |k_{\ell}|\) and \(q = 2p+3+d+2| \boldsymbol{\alpha} |_1\) are chosen as in the proof of 2. We show that both \(b_{\boldsymbol{\alpha}}^{(1)}\) and \(b_{\boldsymbol{\alpha}}^{(2)}\) are holomorphic on \(D_1 = \left\{ s \in \mathbb{C}\, : \, 0 < \mathrm{Re}(s) < d+ 2| \boldsymbol{\alpha} |_1 \right\}\).

For the first term \(b_{\boldsymbol{\alpha}}^{(1)}(s)\), we split the integral as \[b_{\boldsymbol{\alpha}}^{(1)}(s) = \int_{\frac{1}{2}\le |\boldsymbol{x}|\le 1}\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}(1-\eta(\boldsymbol{x}))\,\mathrm{d}\boldsymbol{x} + \int_{|\boldsymbol{x}|\le \frac{1}{2}}\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}\,\mathrm{d}\boldsymbol{x} =: J_1(s)+J_2(s).\] For \(J_1\), the integrand is holomorphic in \(s\) for each \(\boldsymbol{x}\) in the integration domain, and is uniformly bounded on compact subsets of \(D_1\). Thus, by 13, \(J_1\) is holomorphic on \(D_1\). For \(J_2\), using polar coordinates \(\boldsymbol{x}=r\omega\) with \(\omega\in\mathbb{S}^{d-1}\), \[\begin{align} J_2(s) = \int_{\mathbb{S}^{d-1}}\omega^{2\boldsymbol{\alpha}}\,\mathrm{d}S(\omega)\int_0^{1/2} r^{2| \boldsymbol{\alpha} |_1-s+d-1}\,\mathrm{d}r = \left(\int_{\mathbb{S}^{d-1}}\omega^{2\boldsymbol{\alpha}}\,\mathrm{d}S(\omega)\right) \frac{2^{\,s-d-2| \boldsymbol{\alpha} |_1}}{d+2| \boldsymbol{\alpha} |_1-s}. \end{align}\] Since \(D_1\) excludes \(s=d+2| \boldsymbol{\alpha} |_1\), the right-hand side is holomorphic on \(D_1\). Hence \(b_{\boldsymbol{\alpha}}^{(1)}\) is holomorphic on \(D_1\).

For the second term \(b_{\boldsymbol{\alpha}}^{(2)}\), fix an arbitrary compact set \(K\subset D_1\) and set \(\sigma_K:=\min_{s\in K}\mathrm{Re}(s)>0\). For each fixed \(\boldsymbol{k}\neq0\), \[\begin{align} W(\boldsymbol{k};s) &= (-\imath)^q \int_{\mathbb{R}^d} \partial_j^q\!\left(\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}\eta(\boldsymbol{x})\right) \mathrm{e}^{-2\pi \imath\, \boldsymbol{k}\cdot \boldsymbol{x}}\,\mathrm{d}\boldsymbol{x}\\ &= (-\imath)^q \int_{\frac{1}{2}\le |\boldsymbol{x}|\le 1}\partial_j^q\!\left(\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}\eta(\boldsymbol{x})\right) \mathrm{e}^{-2\pi \imath\,\boldsymbol{k}\cdot \boldsymbol{x}}\,\mathrm{d}\boldsymbol{x} + (-\imath)^q \int_{|\boldsymbol{x}|\geq 1}\partial_j^q\!\left(\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}\right) \mathrm{e}^{-2\pi \imath\,\boldsymbol{k}\cdot \boldsymbol{x}}\,\mathrm{d}\boldsymbol{x}. \end{align}\] The integrand is holomorphic in \(s\) for each fixed \(\boldsymbol{x}\). On the annulus \(\{1/2\le|\boldsymbol{x}|\le1\}\), it is bounded uniformly for \(s\in K\). On \(\{|\boldsymbol{x}|\ge1\}\), 11 gives, uniformly for \(s\in K\), \[\left|\partial_j^q\!\left(\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}\right)\right| \le C_K|\boldsymbol{x}|^{-q-\mathrm{Re}(s)+2| \boldsymbol{\alpha} |_1} \le C_K|\boldsymbol{x}|^{-q-\sigma_K+2| \boldsymbol{\alpha} |_1}.\] This majorant is integrable because \(q+\sigma_K>d+2| \boldsymbol{\alpha} |_1\). Since \(K\subset D_1\) was arbitrary, 13 shows that \(W(\boldsymbol{k};\cdot)\) is holomorphic on \(D_1\). Moreover, the estimates for the integrands in \(W(\boldsymbol{k};s)\) are uniform in \(\boldsymbol{k}\). Thus, \[\sup_{s\in K}\left|\frac{W(\boldsymbol{k};s)}{(2\pi k_j)^q}\right| \leq \frac{C_K}{|k_j|^q}, \qquad \boldsymbol{k}\in\mathbb{Z}^d\setminus\{0\},\] where \(C_K\) is a constant independent of \(\boldsymbol{k}\). Since \(|k_j|=| \boldsymbol{k} |_\infty\), 10 and \(q>d\) imply \[\sum_{| \boldsymbol{k} |_\infty=1}^{\infty}|k_j|^{-q} \leq C_d\sum_{n=1}^{\infty}n^{d-1-q}<\infty.\] The Weierstrass M-test therefore gives uniform convergence of the series defining \(b_{\boldsymbol{\alpha}}^{(2)}\) on every compact \(K\subset D_1\). By the Weierstrass theorem, \(b_{\boldsymbol{\alpha}}^{(2)}\) is holomorphic on \(D_1\). ◻

Proof. We first show \(\{I_N\}\) is locally bounded on domain \(D\) via a shell decomposition and uniform Taylor estimates. This local boundedness, together with pointwise convergence on \(\mathrm{Re}(s) > d+2| \boldsymbol{\alpha} |_1\), implies the limit \(I(s)\) exists and is holomorphic on \(D\) by Vitali-Porter theorem [25].

10.0.0.1 Shell decomposition and local Taylor expansion.

For each \(N\), decompose the integral into shells and shift each shell to the cube \(\{| \boldsymbol{x} |_\infty\le 1/2\}\): \[\begin{align} \int_{\frac{1}{2} \leq | \boldsymbol{x} |_\infty \leq N + \frac{1}{2}} g\!\left(\boldsymbol{x}\frac{a}{N}\right)\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}\,\mathrm{d}\boldsymbol{x} &= \sum_{n=1}^N \int_{n-\frac{1}{2} \leq | \boldsymbol{x} |_\infty \leq n + \frac{1}{2}} g\!\left(\boldsymbol{x}\frac{a}{N}\right)\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}\,\mathrm{d}\boldsymbol{x}\\ &= \sum_{n=1}^N \sum_{| \boldsymbol{i} |_\infty=n} \int_{| \boldsymbol{x} |_\infty\le \frac{1}{2}} g\!\left((\boldsymbol{x}+\boldsymbol{i})\frac{a}{N}\right)\frac{(\boldsymbol{x}+\boldsymbol{i})^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}+\boldsymbol{i}|^s}\,\mathrm{d}\boldsymbol{x}. \end{align}\] Hence we can write \[I_N(s) = \sum_{n=1}^N \delta_{n,N}(s),\] where \[\delta_{n,N}(s) := \sum_{| \boldsymbol{i} |_\infty=n} \int_{| \boldsymbol{x} |_\infty\le \frac{1}{2}} \left[ g\!\left((\boldsymbol{x}+\boldsymbol{i})\frac{a}{N}\right)\frac{(\boldsymbol{x}+\boldsymbol{i})^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}+\boldsymbol{i}|^s} - g\!\left(\boldsymbol{i}\frac{a}{N}\right)\frac{\boldsymbol{i}^{2\boldsymbol{\alpha}}}{|\boldsymbol{i}|^s} \right]\mathrm{d}\boldsymbol{x}.\] For any fixed \(N\), define \[f_{N}(\boldsymbol{x};s) := g\!\left(\boldsymbol{x}\frac{a}{N}\right) \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}.\] Taylor’s theorem with integral remainder gives, for \(| \boldsymbol{x} |_\infty\le 1/2\), \[\label{eq:taylor95bs95regular} f_{N}(\boldsymbol{x}+ \boldsymbol{i};s) - f_{N}(\boldsymbol{i};s) = \sum_{j=1}^d \partial_{x_j} f_{N}(\boldsymbol{i};s)\,x_j + \sum_{j,k=1}^d x_jx_k\int_0^1(1-t) \partial_{x_jx_k}^2f_N(\boldsymbol{i}+t\boldsymbol{x};s)\,\mathrm{d}t.\tag{19}\]

10.0.0.2 Local boundedness of \(\{I_N\}\) on \(D\).

Now we show that \(\{I_N\}\) is uniformly bounded on any compact subset of \(D\). Fix an arbitrary compact set \(K\subset D\) and set \[\sigma_K:=\min_{s\in K}\mathrm{Re}(s)>d+2| \boldsymbol{\alpha} |_1-2.\] According to 11, for \(j, k=1, \ldots, d\), we have \[\frac{\partial f_N}{\partial x_j}(\boldsymbol{x};s) = (\partial_j g)\!\left(\boldsymbol{x}\frac{a}{N}\right) \frac{a}{N} \cdot \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} + g\!\left(\boldsymbol{x}\frac{a}{N}\right) \cdot \frac{1}{|\boldsymbol{x}|^{1+s-2| \boldsymbol{\alpha} |_1}} P_{1, \boldsymbol{\alpha},j}\!\left(\frac{\boldsymbol{x}}{|\boldsymbol{x}|} \right),\] and \[\begin{align} \frac{\partial^2 f_N}{\partial x_j\partial x_k}(\boldsymbol{x};s) &= (\partial_{jk}^2 g)\!\left(\boldsymbol{x}\frac{a}{N}\right) (\frac{a}{N})^2 \cdot \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} \\ &\quad + \frac{a}{N} \frac{1}{|\boldsymbol{x}|^{1+s-2| \boldsymbol{\alpha} |_1}} \left[ (\partial_j g)\!\left(\boldsymbol{x}\frac{a}{N}\right) P_{1, \boldsymbol{\alpha},k}\!\left(\frac{\boldsymbol{x}}{|\boldsymbol{x}|}\right) + (\partial_k g)\!\left(\boldsymbol{x}\frac{a}{N}\right) P_{1, \boldsymbol{\alpha},j}\!\left(\frac{\boldsymbol{x}}{|\boldsymbol{x}|}\right) \right] \\ &\quad - g\!\left(\boldsymbol{x}\frac{a}{N}\right) \cdot \frac{1+s-2| \boldsymbol{\alpha} |_1}{|\boldsymbol{x}|^{2+s-2| \boldsymbol{\alpha} |_1}} \frac{x_k}{|\boldsymbol{x}|} \cdot P_{1, \boldsymbol{\alpha},j}\!\left(\frac{\boldsymbol{x}}{|\boldsymbol{x}|}\right) \\ &\quad + g\!\left(\boldsymbol{x}\frac{a}{N}\right) \cdot \frac{1}{|\boldsymbol{x}|^{2+s-2| \boldsymbol{\alpha} |_1}} \left[ (\partial_kP_{1, \boldsymbol{\alpha},j})\!\left(\frac{\boldsymbol{x}}{|\boldsymbol{x}|}\right) (1 - \frac{x_k^2}{|\boldsymbol{x}|^2}) - \sum_{\ell\neq k} (\partial_\ell P_{1, \boldsymbol{\alpha},j})\!\left(\frac{\boldsymbol{x}}{|\boldsymbol{x}|}\right) \frac{x_{\ell} x_k}{|\boldsymbol{x}|^2} \right], \end{align}\] where \(P_{1, \boldsymbol{\alpha},j}(\boldsymbol{x})\) is a polynomial of \(\boldsymbol{x}\).

Given \(| \boldsymbol{i} |_\infty \leq N\), let \(n=| \boldsymbol{i} |_\infty\), then for \(| \boldsymbol{x} |_\infty\le 1/2\), using \(n \leq N\) it follows that \[\max_{j,k}\sup_{| \boldsymbol{x} |_\infty\le 1/2} \left|\partial^2_{x_jx_k} f_N(\boldsymbol{x}+\boldsymbol{i};s)\right| \le C_K\, n^{2| \boldsymbol{\alpha} |_1-\mathrm{Re}(s)-2} \le C_K\, n^{2| \boldsymbol{\alpha} |_1-\sigma_K-2}.\] Since \(x_j\) is odd and the domain is centrally symmetric, the first order term in 19 integrates to \(0\) over \(\{| \boldsymbol{x} |_\infty\le 1/2\}\). Consequently, the Hessian estimate above gives \[\left|\int_{| \boldsymbol{x} |_\infty\le \frac{1}{2}} \left[f_N(\boldsymbol{i}+\boldsymbol{x};s)-f_N(\boldsymbol{i};s)\right]\,\mathrm{d}\boldsymbol{x}\right| \le C_K\, n^{2| \boldsymbol{\alpha} |_1-\sigma_K-2}.\] Summing over the shell \(\{\boldsymbol{i}\in\mathbb{Z}^d:| \boldsymbol{i} |_\infty=n\}\) and using \(\#\{\boldsymbol{i}:| \boldsymbol{i} |_\infty=n\}\le C_d n^{d-1}\) by 10 gives \[\label{eq:delta95nN95bound} \sup_{s\in K}|\delta_{n,N}(s)| \leq C_K\, n^{d+2| \boldsymbol{\alpha} |_1-\sigma_K-3}, \qquad 1\leq n\leq N.\tag{20}\] Combining \(I_N(s)=\sum_{n=1}^N\delta_{n,N}(s)\) with 20 gives \[\sup_{N\in\mathbb{N}_+}\sup_{s\in K}|I_N(s)| \leq \sum_{n=1}^{\infty} \sup_{N\geq n}\sup_{s\in K}|\delta_{n,N}(s)| \leq C_K\sum_{n=1}^{\infty} n^{d+2| \boldsymbol{\alpha} |_1-\sigma_K-3}<\infty,\] where the last series converges since \(\sigma_K>d+2| \boldsymbol{\alpha} |_1-2\).

10.0.0.3 Holomorphicity of the limit.

For each fixed \(N\), the map \(s\mapsto I_N(s)\) is holomorphic on \(D\) by 13. If \(\mathrm{Re}(s)>d+2| \boldsymbol{\alpha} |_1\), then \(|\boldsymbol{x}|^{2| \boldsymbol{\alpha} |_1-\mathrm{Re}(s)}\) is integrable on \(\{| \boldsymbol{x} |_\infty\geq1/2\}\) and summable over \(\mathbb{Z}^d\setminus\{0\}\). Since \(\operatorname{supp}(g)\subset[-a,a]^d\), we have \(g(a\boldsymbol{i}/N)=0\) whenever \(| \boldsymbol{i} |_\infty>N\). Note that \(g\) is bounded and \(g(a\boldsymbol{x}/N)\to g(0)\) pointwise, dominated convergence with respect to Lebesgue and counting measure gives \[\lim_{N\to\infty}I_N(s) =g(0)\left[ \int_{| \boldsymbol{x} |_\infty\geq1/2}\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}\,\mathrm{d}\boldsymbol{x} -\sum_{| \boldsymbol{i} |_\infty=1}^{\infty} \frac{\boldsymbol{i}^{2\boldsymbol{\alpha}}}{|\boldsymbol{i}|^s}\right].\] Thus \(I_N\) converges pointwise on the nonempty open subset \(\{\mathrm{Re}(s)>d+2| \boldsymbol{\alpha} |_1\}\) of \(D\). Since \(\{I_N\}\) is locally bounded on \(D\), Vitali-Porter theorem implies that \(I_N\to I\) uniformly on compact subsets of \(D\), and the limit \(I(s)\) is holomorphic on \(D\). ◻

Proof. We begin by splitting the cube into the ball plus the remaining corner region: \[\int_{| \boldsymbol{x} |_\infty \leq t} g(\boldsymbol{x}h) \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} \,\mathrm{d}\boldsymbol{x} = \int_{|\boldsymbol{x}| \leq t} g(\boldsymbol{x}h) \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} \,\mathrm{d}\boldsymbol{x} + \int_{|\boldsymbol{x}| \geq t, \, | \boldsymbol{x} |_\infty \leq t} g(\boldsymbol{x}h) \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} \,\mathrm{d}\boldsymbol{x},\] where the second term is an entire function in \(s\) by 13. For the first term, using polar coordinates \(x=r\omega\) with \(\omega \in \mathbb{S}^{d-1}\) and radiality \(g(\boldsymbol{x})=g(|\boldsymbol{x}|)\) we have \[\begin{align} \int_{|\boldsymbol{x}|\le t} g(\boldsymbol{x}h)\,\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}\,\mathrm{d}\boldsymbol{x} = \int_{\mathbb{S}^{d-1}} \omega^{2\boldsymbol{\alpha}}\,\mathrm{d}S(\omega) \int_{0}^{t} g(rh)\,r^{d+2| \boldsymbol{\alpha} |_1-1-s}\,\mathrm{d}r. \end{align}\] Thus, the integral converges and is analytic for \(\mathrm{Re}(s) < d+2| \boldsymbol{\alpha} |_1\). We then rewrite the radial integral as \[\begin{align} \label{eq:integral95gr} \int_{0}^{t} g(rh) r^{d+2| \boldsymbol{\alpha} |_1-1-s} \,\mathrm{d}r = \int_{0}^{t} \bigl(g(rh) - g(0)\bigr) r^{d+2| \boldsymbol{\alpha} |_1-1-s} \,\mathrm{d}r + g(0) \frac{t^{d+2| \boldsymbol{\alpha} |_1 - s}}{d+2| \boldsymbol{\alpha} |_1 - s}. \end{align}\tag{21}\] The second term in 21 is meromorphic in \(s\) with a simple pole at \(s=d+2| \boldsymbol{\alpha} |_1\), and hence is holomorphic on \(\Omega_{\boldsymbol{\alpha}}\). For the first term, Taylor’s theorem gives \[| g(rh) - g(0) | \leq C (rh)^{2p+2}.\] If \(K\subset\Omega_{\boldsymbol{\alpha}}\) is compact and \(\sigma_K:=\max_{s\in K}\mathrm{Re}(s)\), then the integrand in the first term of 21 is bounded near \(r=0\) by \(C_K r^{d+2| \boldsymbol{\alpha} |_1+2p+1-\sigma_K}\), which is integrable because \(\sigma_K<d+2| \boldsymbol{\alpha} |_1+1\). Thus 13 shows that the first term is holomorphic on \(\Omega_{\boldsymbol{\alpha}}\) as well. Consequently, the right-hand side of ?? defines a holomorphic function on \(\Omega_{\boldsymbol{\alpha}}\) that agrees with the original integral on \(D_1\). Its restriction to \(D_2\) is therefore the stated analytic continuation.

Next we verify the interchange of orders of limit with \(h\) and analytic continuation operator. By applying Lebesgue’s dominated convergence theorem to the right-hand side of ?? , we have \[\begin{align} & \quad \lim_{h\rightarrow 0^+} \mathcal{A}_{D_1}^{D_2} \left[ \int_{| \boldsymbol{x} |_\infty \leq t} g(\boldsymbol{x}h) \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} \,\mathrm{d}\boldsymbol{x}\right](s) \\ &= \int_{|\boldsymbol{x}| \geq t, \, | \boldsymbol{x} |_\infty \leq t} g(0) \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} \,\mathrm{d}\boldsymbol{x} + \int_{\mathbb{S}^{d-1}} \omega^{2\boldsymbol{\alpha}} \mathrm{d}S(\omega) \left[ g(0)\frac{t^{d+2| \boldsymbol{\alpha} |_1 - s}}{d+2| \boldsymbol{\alpha} |_1 - s} \right]. \end{align}\] On the other hand, for fixed \(s\in D_1\), the dominated convergence theorem gives \[\label{eq:lim95box95integral} \lim_{h\rightarrow 0^+} \int_{| \boldsymbol{x} |_\infty \leq t} g(\boldsymbol{x}h) \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} \,\mathrm{d}\boldsymbol{x} = \int_{| \boldsymbol{x} |_\infty \leq t} g(0) \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} \,\mathrm{d}\boldsymbol{x}.\tag{22}\] Applying the same ball–corner decomposition, the analytic continuation of 22 to \(D_2\) is \[\int_{|\boldsymbol{x}| \geq t, \, | \boldsymbol{x} |_\infty \leq t} g(0) \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} \,\mathrm{d}\boldsymbol{x} + \int_{\mathbb{S}^{d-1}} \omega^{2\boldsymbol{\alpha}} \mathrm{d}S(\omega) \left[ g(0)\frac{t^{d+2| \boldsymbol{\alpha} |_1 - s}}{d+2| \boldsymbol{\alpha} |_1 - s} \right].\] Therefore, by comparing these two expressions we complete the proof of the lemma. ◻

Proof of 5. Since \(\operatorname{supp}(g)\subset[-a,a]^d\), we restrict the integral in \(u_N^{D_1}(s)\) to \(\{| \boldsymbol{x} |_\infty \leq N + \frac{1}{2}\}\) as \[u_N^{D_1}(s) = \int_{\mathbb{R}^d} g(\boldsymbol{x}\frac{a}{N}) \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} \,\mathrm{d}\boldsymbol{x} = \int_{| \boldsymbol{x} |_\infty \leq N + \frac{1}{2}} g(\boldsymbol{x}\frac{a}{N}) \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} \,\mathrm{d}\boldsymbol{x}.\]

To perform analytic continuation from \(D_1\) to \(D_2\), we decompose the difference \(\bigl[u_N^{D_1} - v_N^{D_1}\bigr](s)\) into regular and singular parts as, \[\begin{align} u_N^{D_1} - v_N^{D_1} &= \int_{| \boldsymbol{x} |_\infty \leq N + \frac{1}{2}} g(\boldsymbol{x}\frac{a}{N}) \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} \,\mathrm{d}\boldsymbol{x} - \sum_{| \boldsymbol{i} |_\infty=1}^{N} g(\boldsymbol{i}\frac{a}{N}) \frac{\boldsymbol{i}^{2\boldsymbol{\alpha}}}{|\boldsymbol{i}|^s} \\ &= \left( \int_{\frac{1}{2} \leq | \boldsymbol{x} |_\infty \leq N + \frac{1}{2}} g(\boldsymbol{x}\frac{a}{N}) \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} \,\mathrm{d}\boldsymbol{x}- \sum_{| \boldsymbol{i} |_\infty=1}^{N} g(\boldsymbol{i}\frac{a}{N}) \frac{\boldsymbol{i}^{2\boldsymbol{\alpha}}}{|\boldsymbol{i}|^s} \right) + \int_{| \boldsymbol{x} |_\infty \leq \frac{1}{2}} g(\boldsymbol{x}\frac{a}{N}) \frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s} \,\mathrm{d}\boldsymbol{x}\\ &:= I_N(s) + J_N^{D_1}(s). \end{align}\] For the singular part, \(J_N^{D_1}(s)\) is initially defined for \(s\in D_1\). After applying 15 with \(h=a/N\) and \(t=1/2\), \(J_N^{D_1}(s)\) and \(\lim_{N\to\infty} J_N^{D_1}(s)\) possess analytic continuation over \(D_1 \cup D_2\) and we further have \[\lim_{N\rightarrow\infty} \mathcal{A}_{D_1}^{D_2} \left[ J^{D_1}_N \right] (s) = \mathcal{A}_{D_1}^{D_2} \left[ \lim_{N\rightarrow\infty} J_N^{D_1} \right] (s).\] For the regular part, \(I_N(s)\) is analytic on \(D_1 \cup D_2\), and by 14, the limit \(I(s):=\lim_{N\to\infty}I_N(s)\) is analytic on \(D=\{s\in\mathbb{C}:\mathrm{Re}(s)>d+2| \boldsymbol{\alpha} |_1-2\}\). Introduce the overlap strip between \(D_1\) and \(D\) as \[D_0 := \left\{ s\in\mathbb{C}:\;d+2| \boldsymbol{\alpha} |_1-1<\mathrm{Re}(s)<d+2| \boldsymbol{\alpha} |_1\right\} \subset (D \cap D_1).\] Then the analyticity of \(I_N(s)\) and \(I(s)\) over \(D_0 \cup D_2\) gives \[\lim_{N\rightarrow\infty} \mathcal{A}_{D_0}^{D_2} \left[ I^{D_0}_N \right] (s) = \lim_{N\rightarrow\infty} I^{D_2}_N(s) = I^{D_2}(s) = \mathcal{A}_{D_0}^{D_2} \left[I^{D_0}\right] (s) = \mathcal{A}_{D_0}^{D_2} \left[ \lim_{N\rightarrow\infty}I_N^{D_0} \right] (s).\]

The relative positions of the four domains used in this argument are summarized in 5.

Combine the two results, since \(D_0 \subset D_1\), \(I_N^{D_0}(s)+J_N^{D_0}(s)\) and \(\lim_{N\to\infty} I_N^{D_0}(s)+J_N^{D_0}(s)\) possess analytic continuation over \(D_0 \cup D_2\). Finally, by the identity theorem of analytic function on connected open sets, we have \[\begin{align} \mathcal{A}_{D_1}^{D_2} \left[ \lim_{N\rightarrow \infty} \left( u_N^{D_1} - v_N^{D_1} \right) \right] (s) &= \mathcal{A}_{D_0}^{D_2} \left[ \lim_{N\rightarrow \infty} \left( I_N^{D_0} + J_N^{D_0} \right) \right] (s) = \lim_{N\rightarrow\infty} \mathcal{A}_{D_0}^{D_2} \left[ I^{D_0}_N + J^{D_0}_N \right] (s) \\ &= \lim_{N\rightarrow \infty} \left( \mathcal{A}_{D_1}^{D_2} \left[u_N^{D_1} - v_N^{D_1}\right](s) \right). \end{align}\] ◻

Proof of 6. Since \(g\) is compactly supported in \([-a,a]^d\), by changing variables \(\boldsymbol{y}=(a/N)\boldsymbol{x}\) we have \[u_N^{D_1}(s) = \int_{| \boldsymbol{x} |_\infty\le N} g\!\left(\boldsymbol{x}\frac{a}{N}\right)\frac{\boldsymbol{x}^{2\boldsymbol{\alpha}}}{|\boldsymbol{x}|^s}\,\mathrm{d}\boldsymbol{x} = \left(\frac{N}{a}\right)^{d+2| \boldsymbol{\alpha} |_1-s} \int_{| \boldsymbol{y} |_\infty\le a} g(\boldsymbol{y})\frac{\boldsymbol{y}^{2\boldsymbol{\alpha}}}{|\boldsymbol{y}|^s}\,\mathrm{d}\boldsymbol{y}.\] The prefactor is entire in \(s\), and by 15 the analytic continuation of the \(\boldsymbol{y}\)-integral to \(D_3\) exists and does not depend on \(N\). Since \(s\in D_3\) implies \(\mathrm{Re}(s)>d+2| \boldsymbol{\alpha} |_1\), we obtain \[\lim_{N\to\infty}\mathcal{A}_{D_1}^{D_3}[u_N^{D_1}](s) = \lim_{N\rightarrow\infty} \left(\frac{N}{a}\right)^{d+2| \boldsymbol{\alpha} |_1-s} \mathcal{A}_{D_1}^{D_3} \left[\int_{| \boldsymbol{y} |_\infty \leq a} g(\boldsymbol{y}) \frac{\boldsymbol{y}^{2\boldsymbol{\alpha}}}{|\boldsymbol{y}|^s} \mathrm{d}\boldsymbol{y}\right](s) = 0.\] For \(v_N^{D_1}(s)=\sum_{| \boldsymbol{i} |_\infty=1}^{N} g(\boldsymbol{i} a/N)\,\boldsymbol{i}^{2\boldsymbol{\alpha}}/|\boldsymbol{i}|^s\), analytic continuation does not change it because it is a finite sum. Moreover, for \(s\in D_3\) the series \(\sum_{| \boldsymbol{i} |_\infty=1}^{\infty} |\boldsymbol{i}|^{2| \boldsymbol{\alpha} |_1-\mathrm{Re}(s)}\) converges, so dominated convergence yields \[\lim_{N\to\infty}\mathcal{A}_{D_1}^{D_3}[v_N^{D_1}](s) = \lim_{N\rightarrow\infty} \sum_{| \boldsymbol{i} |_\infty=1}^{N} g(\boldsymbol{i}\frac{a}{N}) \frac{\boldsymbol{i}^{2\boldsymbol{\alpha}}}{|\boldsymbol{i}|^s} = \lim_{N\rightarrow\infty} \sum_{| \boldsymbol{i} |_\infty=1}^{\infty} g(\boldsymbol{i}\frac{a}{N}) \frac{\boldsymbol{i}^{2\boldsymbol{\alpha}}}{|\boldsymbol{i}|^s} = \sum_{| \boldsymbol{i} |_\infty=1}^{\infty}\frac{\boldsymbol{i}^{2\boldsymbol{\alpha}}}{|\boldsymbol{i}|^s},\] since \(g(\boldsymbol{i}a/N)\to g(0)=1\) for each fixed \(\boldsymbol{i}\). Combining the two limits gives \[b_{\boldsymbol{\alpha}}^{D_3}(s) = \lim_{N\to\infty}\left(\mathcal{A}_{D_1}^{D_3}[u_N^{D_1}](s)-\mathcal{A}_{D_1}^{D_3}[v_N^{D_1}](s)\right) = -\sum_{| \boldsymbol{i} |_\infty=1}^{\infty}\frac{\boldsymbol{i}^{2\boldsymbol{\alpha}}}{|\boldsymbol{i}|^s}, \qquad s\in D_3,\] as claimed. ◻

Proof. By Taylor’s formula and 2, the generating function is \[\sum_{\boldsymbol{\alpha}\in\mathbb{N}^d} H_{\boldsymbol{\alpha}}(\boldsymbol{x};A) \frac{\boldsymbol{t}^{\boldsymbol{\alpha}}}{\boldsymbol{\alpha}!} = \mathrm{e}^{\boldsymbol{x}^\top A\boldsymbol{x}}\mathrm{e}^{-(\boldsymbol{x}-\boldsymbol{t})^\top A(\boldsymbol{x}-\boldsymbol{t})} = \mathrm{e}^{2\boldsymbol{t}^\top A\boldsymbol{x}-\boldsymbol{t}^\top A\boldsymbol{t}}.\] Differentiating with respect to \(t_j\) gives \[\sum_{\boldsymbol{\alpha}\in\mathbb{N}^d} H_{\boldsymbol{\alpha}+\boldsymbol{e}_j}(\boldsymbol{x};A) \frac{\boldsymbol{t}^{\boldsymbol{\alpha}}}{\boldsymbol{\alpha}!} = \left(2(A\boldsymbol{x})_j-2(A\boldsymbol{t})_j\right) \sum_{\boldsymbol{\alpha}\in\mathbb{N}^d} H_{\boldsymbol{\alpha}}(\boldsymbol{x};A) \frac{\boldsymbol{t}^{\boldsymbol{\alpha}}}{\boldsymbol{\alpha}!}.\] Comparing the coefficient of \(\boldsymbol{t}^{\boldsymbol{\alpha}}/\boldsymbol{\alpha}!\) yields \[H_{\boldsymbol{\alpha}+\boldsymbol{e}_j}(\boldsymbol{x};A) =2(A\boldsymbol{x})_jH_{\boldsymbol{\alpha}}(\boldsymbol{x};A) -\sum_{k=1}^d2A_{jk}\alpha_kH_{\boldsymbol{\alpha}-\boldsymbol{e}_k}(\boldsymbol{x};A).\] On the other hand, differentiating the same generating function with respect to \(x_j\) gives \[\sum_{\boldsymbol{\alpha}\in\mathbb{N}^d} \partial_{x_j}H_{\boldsymbol{\alpha}}(\boldsymbol{x};A) \frac{\boldsymbol{t}^{\boldsymbol{\alpha}}}{\boldsymbol{\alpha}!} =2(A\boldsymbol{t})_j \sum_{\boldsymbol{\alpha}\in\mathbb{N}^d} H_{\boldsymbol{\alpha}}(\boldsymbol{x};A) \frac{\boldsymbol{t}^{\boldsymbol{\alpha}}}{\boldsymbol{\alpha}!}.\] Comparing coefficients again gives \(\partial_{x_j}H_{\boldsymbol{\alpha}}(\boldsymbol{x};A) =\sum_{k=1}^d2A_{jk}\alpha_kH_{\boldsymbol{\alpha}-\boldsymbol{e}_k}(\boldsymbol{x};A)\), which is exactly ?? . ◻

Proof of 1. Firstly, note that for \(\lambda > 0\) and \(\mathrm{Re}(s) > 0\) one has \[\frac{1}{\lambda^s} = \frac{1}{\Gamma(s)} \int_{0}^{\infty} \mathrm{e}^{-\lambda x} x^{s-1} \,\mathrm{d}x.\] Then for \(\mathrm{Re}(s) > d+2| \boldsymbol{\alpha} |_1\), we have \[Z_{2\boldsymbol{\alpha}}(s) = \sum_{\boldsymbol{n}\in \mathbb{Z}^{d} \setminus \{\boldsymbol{0}\}} \frac{\boldsymbol{n}^{2\boldsymbol{\alpha}}}{|\boldsymbol{n}|^s} = \sum_{| \boldsymbol{n} |_\infty=1}^{\infty} \boldsymbol{n}^{2\boldsymbol{\alpha}} \frac{\pi^{s/2}}{\Gamma(s/2)} \int_{0}^{\infty} \mathrm{e}^{-|\boldsymbol{n}|^2\pi t} \, t^{\frac{s}{2} -1} \,\mathrm{d}t.\] Then we split the integral at \(t=1\) and change variables \(t\mapsto 1/t\) on \((0,1)\) as \[\begin{align} \frac{\Gamma(s/2)}{\pi^{s/2}} Z_{2\boldsymbol{\alpha}}(s) &= \sum_{| \boldsymbol{n} |_\infty=1}^{\infty} \boldsymbol{n}^{2\boldsymbol{\alpha}} \left( \int_{1}^{\infty} \mathrm{e}^{-|\boldsymbol{n}|^2\pi t} \, t^{\frac{s}{2} -1} \,\mathrm{d}t + \int_{0}^{1} \mathrm{e}^{-|\boldsymbol{n}|^2\pi t} \, t^{\frac{s}{2} -1} \,\mathrm{d}t \right) \\ &= \sum_{| \boldsymbol{n} |_\infty=1}^{\infty} \boldsymbol{n}^{2\boldsymbol{\alpha}} \int_{1}^{\infty} \mathrm{e}^{-|\boldsymbol{n}|^2\pi t} \, t^{\frac{s}{2} -1} \,\mathrm{d}t + \int_{1}^{\infty} \sum_{| \boldsymbol{n} |_\infty=1}^{\infty} \boldsymbol{n}^{2\boldsymbol{\alpha}} \mathrm{e}^{-\frac{\pi}{t} |\boldsymbol{n}|^2} \left(\frac{1}{t}\right)^{\frac{s}{2}+1} \,\mathrm{d}t \end{align}\] with the interchange of summation and integral justified by the absolute convergence when \(\mathrm{Re}(s) > d+2| \boldsymbol{\alpha} |_1\). Further, for the second term, by introducing the Kronecker \(\delta\) notation we have \[\begin{align} \int_{1}^{\infty} \sum_{| \boldsymbol{n} |_\infty=1}^{\infty} \boldsymbol{n}^{2\boldsymbol{\alpha}} \mathrm{e}^{-\frac{\pi}{t} |\boldsymbol{n}|^2} \left(\frac{1}{t}\right)^{\frac{s}{2}+1} \,\mathrm{d}t &= \int_{1}^{\infty} \sum_{\boldsymbol{n}\in \mathbb{Z}^d} \boldsymbol{n}^{2\boldsymbol{\alpha}} \mathrm{e}^{-\frac{\pi}{t} |\boldsymbol{n}|^2} \left(\frac{1}{t}\right)^{\frac{s}{2}+1} \,\mathrm{d}t - \delta_{0, \boldsymbol{\alpha}} \int_{1}^{\infty} \left(\frac{1}{t}\right)^{\frac{s}{2}+1} \,\mathrm{d}t \\ &= \int_{1}^{\infty} \sum_{\boldsymbol{n}\in \mathbb{Z}^d} \boldsymbol{n}^{2\boldsymbol{\alpha}} \mathrm{e}^{-\frac{\pi}{t} |\boldsymbol{n}|^2} \left(\frac{1}{t}\right)^{\frac{s}{2}+1} \,\mathrm{d}t - \delta_{0, \boldsymbol{\alpha}} \frac{2}{s} . \end{align}\] Define \[f(\boldsymbol{x}) := \boldsymbol{x}^{2\boldsymbol{\alpha}} \mathrm{e}^{-\frac{\pi}{t}|\boldsymbol{x}|^2}, \qquad \boldsymbol{x}\in \mathbb{R}^d.\] Then we compute the Fourier transform of \(f\) as \[\begin{align} \mathcal{F}\left[f\right](\boldsymbol{y}) &= \mathcal{F}\left[\boldsymbol{x}^{2\boldsymbol{\alpha}} \mathrm{e}^{-\frac{\pi}{t}|\boldsymbol{x}|^2}\right] (\boldsymbol{y}) = \left(\frac{\imath}{2\pi}\right)^{2| \boldsymbol{\alpha} |_1} \left( \partial^{2\boldsymbol{\alpha}}_{\boldsymbol{y}} \mathcal{F}\left[\mathrm{e}^{-\frac{\pi}{t}|\boldsymbol{x}|^2}\right](\boldsymbol{y}) \right) \\ &= \left(\frac{\imath}{2\pi}\right)^{2| \boldsymbol{\alpha} |_1} \partial^{2\boldsymbol{\alpha}}_{\boldsymbol{y}} \left( t^{d/2} \mathrm{e}^{-\pi t|\boldsymbol{y}|^2} \right) \\ &= \left(\frac{\imath}{2\pi}\right)^{2| \boldsymbol{\alpha} |_1} t^{d/2} \cdot (-1)^{2| \boldsymbol{\alpha} |_1} (\pi t)^{| \boldsymbol{\alpha} |_1} H_{2\boldsymbol{\alpha}}(\sqrt{\pi t} \, \boldsymbol{y}) \, \mathrm{e}^{-\pi t |\boldsymbol{y}|^2} \\ &= \left(\frac{-1}{4\pi}\right)^{| \boldsymbol{\alpha} |_1} t^{\frac{d}{2}+| \boldsymbol{\alpha} |_1} \mathrm{e}^{-\pi t |\boldsymbol{y}|^2} H_{2\boldsymbol{\alpha}}(\sqrt{\pi t} \, \boldsymbol{y}), \end{align}\] where we have used the multivariate Hermite polynomials for the derivatives according to 4. Therefore, apply the Poisson summation formula introduced in 9 to \(f(\boldsymbol{x})\), we have \[\sum_{\boldsymbol{n}\in \mathbb{Z}^d} \boldsymbol{n}^{2\boldsymbol{\alpha}} \mathrm{e}^{-\frac{\pi}{t} |\boldsymbol{n}|^2} = \sum_{\boldsymbol{k}\in \mathbb{Z}^d} \left(\frac{-1}{4\pi}\right)^{| \boldsymbol{\alpha} |_1} t^{\frac{d}{2}+| \boldsymbol{\alpha} |_1} \mathrm{e}^{-\pi t |\boldsymbol{k}|^2} H_{2\boldsymbol{\alpha}}(\sqrt{\pi t} \, \boldsymbol{k}).\] Thus, for \(\mathrm{Re}(s) > d+2| \boldsymbol{\alpha} |_1\), \[\begin{align} & \quad \int_{1}^{\infty} \sum_{\boldsymbol{n}\in \mathbb{Z}^d} \boldsymbol{n}^{2\boldsymbol{\alpha}} \mathrm{e}^{-\frac{\pi}{t} |\boldsymbol{n}|^2} \left(\frac{1}{t}\right)^{\frac{s}{2}+1} \mathrm{d}t = \int_{1}^{\infty} \sum_{\boldsymbol{k}\in \mathbb{Z}^d} \left(\frac{-1}{4\pi}\right)^{| \boldsymbol{\alpha} |_1} \mathrm{e}^{-\pi t |\boldsymbol{k}|^2} H_{2\boldsymbol{\alpha}}(\sqrt{\pi t} \, \boldsymbol{k}) \, t^{\frac{d}{2}+| \boldsymbol{\alpha} |_1-\frac{s}{2} - 1} \mathrm{d}t \\ &= \left(\frac{-1}{4\pi}\right)^{| \boldsymbol{\alpha} |_1} \left( \sum_{| \boldsymbol{k} |_\infty=1}^{\infty} \int_{1}^{\infty} \mathrm{e}^{-\pi t |\boldsymbol{k}|^2} H_{2\boldsymbol{\alpha}}(\sqrt{\pi t} \, \boldsymbol{k}) \, t^{\frac{d}{2}+| \boldsymbol{\alpha} |_1-\frac{s}{2} - 1} \mathrm{d}t + H_{2\boldsymbol{\alpha}}(0) \int_{1}^{\infty} t^{\frac{d}{2}+| \boldsymbol{\alpha} |_1-\frac{s}{2} - 1} \mathrm{d}t \right) \\ &= \left(\frac{-1}{4\pi}\right)^{| \boldsymbol{\alpha} |_1} \left( \sum_{| \boldsymbol{k} |_\infty=1}^{\infty} \int_{1}^{\infty} \mathrm{e}^{-\pi t |\boldsymbol{k}|^2} H_{2\boldsymbol{\alpha}}(\sqrt{\pi t} \, \boldsymbol{k}) \, t^{\frac{d}{2}+| \boldsymbol{\alpha} |_1-\frac{s}{2} - 1} \mathrm{d}t - \frac{2H_{2\boldsymbol{\alpha}}(0)}{d+2| \boldsymbol{\alpha} |_1-s} \right). \end{align}\] Hence, we obtain the functional equation as \[\begin{align} \frac{\Gamma(s/2)}{\pi^{s/2}} Z_{2\boldsymbol{\alpha}}(s) &= \left(-\frac{1}{4\pi}\right)^{| \boldsymbol{\alpha} |_1} \frac{2H_{2\boldsymbol{\alpha}}(0)}{s - d - 2| \boldsymbol{\alpha} |_1} - \frac{2\delta_{0, \boldsymbol{\alpha}}}{s} + \sum_{\boldsymbol{n}\in \mathbb{Z}^{d} \setminus \{\boldsymbol{0}\}} \boldsymbol{n}^{2\boldsymbol{\alpha}} \int_{1}^{\infty} \mathrm{e}^{-|\boldsymbol{n}|^2\pi t} \, t^{\frac{s}{2} -1} \mathrm{d}t \\ &\quad + \left(-\frac{1}{4\pi}\right)^{| \boldsymbol{\alpha} |_1} \sum_{\boldsymbol{k}\in \mathbb{Z}^{d} \setminus \{\boldsymbol{0}\}} \int_{1}^{\infty} \mathrm{e}^{-|\boldsymbol{k}|^2 \pi t} H_{2\boldsymbol{\alpha}}(\sqrt{\pi t} \, \boldsymbol{k}) \, t^{\frac{d-s}{2}+| \boldsymbol{\alpha} |_1 - 1} \mathrm{d}t. \end{align}\]

We now verify that the functional equation yields the meromorphic continuation of \(Z_{2\boldsymbol{\alpha}}(s)\). The two series appearing there are entire. Take the first series as an example, rewrite it with the upper incomplete gamma functions as \[\sum_{| \boldsymbol{n} |_\infty=1}^{\infty} \boldsymbol{n}^{2\boldsymbol{\alpha}} \int_{1}^{\infty} \mathrm{e}^{-|\boldsymbol{n}|^2\pi t} \, t^{\frac{s}{2} -1} \,\mathrm{d}t = \frac{1}{\pi^{s/2}} \sum_{| \boldsymbol{n} |_\infty=1}^{\infty} \left(\frac{\boldsymbol{n}}{|\boldsymbol{n}|}\right)^{2\boldsymbol{\alpha}} \frac{1}{|\boldsymbol{n}|^{s-2| \boldsymbol{\alpha} |_1}} \Gamma(\frac{s}{2}, |\boldsymbol{n}|^2 \pi).\] According to the tail estimate in 18, let \(N\) be the smallest integer such that \(\pi N^2 \geq \max \{\mathrm{Re}(s)/2, (d + 2| \boldsymbol{\alpha} |_1)/2 \}\), then we have \[\sum_{| \boldsymbol{n} |_\infty=N+1}^{\infty} \left(\frac{\boldsymbol{n}}{|\boldsymbol{n}|}\right)^{2\boldsymbol{\alpha}} \frac{1}{|\boldsymbol{n}|^{s-2| \boldsymbol{\alpha} |_1}} \Gamma(\frac{s}{2}, |\boldsymbol{n}|^2 \pi) \leq C N^{d+2| \boldsymbol{\alpha} |_1} \mathrm{e}^{-\pi N^2}.\] Hence, the series is uniformly convergent in any given compact subset of the complex plane. By the Weierstrass theorem, this series is holomorphic with respect to \(s\) over the complex plane. The second series is treated similarly: expanding the Hermite polynomial \(H_{2\boldsymbol{\alpha}}\) reduces it to a finite linear combination of series of the same type, each of which is entire by the same argument.

Therefore the right-hand side of the functional equation is meromorphic, with possible poles only at \(s=0\) and \(s=d+2| \boldsymbol{\alpha} |_1\). Since the two series in the functional equation are entire, for \(s\) near \(0\) we have \[\frac{\Gamma(s/2)}{\pi^{s/2}}Z_{2\boldsymbol{\alpha}}(s) = -\frac{2\delta_{0,\boldsymbol{\alpha}}}{s}+O(1).\] Furthermore, since \(\lim_{s\to 0} |\Gamma(s)| = +\infty\), we have \[\lim_{s\to 0} Z_{2\boldsymbol{\alpha}}(s) = \lim_{s\to 0} -\frac{\pi^{s/2}}{\Gamma(s/2)} \frac{\delta_{0,\boldsymbol{\alpha}}}{s/2} = \lim_{s\to 0} -\delta_{0,\boldsymbol{\alpha}}\frac{\pi^{s/2}}{\Gamma(s/2+1)} = -\delta_{0, \boldsymbol{\alpha}}.\] Thus \(s=0\) is not a pole of \(Z_{2\boldsymbol{\alpha}}(s)\). Finally, we conclude that \(Z_{2\boldsymbol{\alpha}}(s)\) has a meromorphic continuation to the whole complex plane \(\mathbb{C}\) with a simple pole at \(s = d + 2| \boldsymbol{\alpha} |_1\). ◻

Proof of 3. The proof is similar to that of 1 and we omit most details for brevity. The key step is to compute a different Fourier transform as \[\begin{align} \mathcal{F}\left[\boldsymbol{x}^{\boldsymbol{\alpha}} \mathrm{e}^{-\frac{\pi}{t} \boldsymbol{x}^\top B \boldsymbol{x}}\right] (\boldsymbol{y}) = \left(\frac{-1}{4\pi}\right)^{\frac{| \boldsymbol{\alpha} |_1}{2}} t^{\frac{d + | \boldsymbol{\alpha} |_1}{2}} |B|^{-1/2} \mathrm{e}^{-\pi t \boldsymbol{y}^\top B^{-1} \boldsymbol{y}} H_{\boldsymbol{\alpha}}(\sqrt{\pi t} \,\boldsymbol{y}; B^{-1}), \end{align}\] where \(H_{\boldsymbol{\alpha}}(\cdot; B^{-1})\) is the multivariate Hermite polynomial associated with matrix \(B^{-1}\). Note that here we have used the assumption that \(| \boldsymbol{\alpha} |_1\) is even to ensure the Fourier-transform prefactor is real-valued. ◻

Proof. Since the integrand is positive for \(t\ge x>0\), we have \(\Gamma(a,x)>0\). Make the change of variables \(t=x(1+u)\) to obtain \[\Gamma(a,x) = \int_x^\infty t^{a-1}\mathrm{e}^{-t}\,\mathrm{d}t = x^a\mathrm{e}^{-x}\int_0^\infty (1+u)^{a-1}\mathrm{e}^{-ux}\,\mathrm{d}u.\] When \(a \leq 1\), \(|(1+u)^{a-1}| \leq 1\) for \(u \geq 0\), thus we have \[\Gamma(a, x) \leq x^a \mathrm{e}^{-x} \int_0^{\infty} \mathrm{e}^{-ux} \,\mathrm{d}u = x^a \mathrm{e}^{-x} \cdot \frac{1}{x} \leq x^a \mathrm{e}^{-x}, \qquad x \geq 1.\] When \(a > 1\), by \(1 + u \leq e^u\) we have \[\Gamma(a, x) \leq x^a \mathrm{e}^{-x} \int_0^{\infty} \mathrm{e}^{-u(x-a+1)} \mathrm{d}u = x^a \mathrm{e}^{-x} \cdot \frac{1}{x - a + 1} \leq x^a \mathrm{e}^{-x}, \qquad x \geq a.\] Combining the discussions above we obtain the desired estimate. ◻

Proof. For \(n\in\mathbb{N}_+\), let \(S_n:=\{\boldsymbol{k}\in\mathbb{Z}^d:| \boldsymbol{k} |_\infty=n\}\). If \(\boldsymbol{k}\in S_n\), then \(n^2\le |\boldsymbol{k}|^2\le dn^2\), hence by the Rayleigh quotient bounds \[\lambda_{\min}n^2 \leq \lambda_{\min}|\boldsymbol{k}|^2 \leq \boldsymbol{k}^\top B \boldsymbol{k}\leq \lambda_{\max}|\boldsymbol{k}|^2 \leq d \lambda_{\max}n^2.\] Since \(\Gamma(a, x)\) is decreased with \(x\) increasing when \(x> 0\), we have \[\Gamma(a, \pi \boldsymbol{k}^\top B \boldsymbol{k}) \leq \Gamma(a, \pi \lambda_{\min}n^2).\] Moreover, \(x\mapsto x^{b/2}\) is decreasing for \(b\le 0\) and increasing for \(b>0\), so for \(\boldsymbol{k}\in S_n\), \[(\boldsymbol{k}^\top B\boldsymbol{k})^{b/2}\le C_{B,b}\,n^{b}.\] Note that 10 gives \(\#S_n \leq 2d\,3^{d-1}n^{d-1}\), so \[\begin{align} \sum_{| \boldsymbol{k} |_\infty=N+1}^{\infty} (\boldsymbol{k}^\top B \boldsymbol{k})^{b/2}\, \Gamma(a,\pi\,\boldsymbol{k}^\top B \boldsymbol{k}) &= \sum_{n=N+1}^{\infty}\;\sum_{\boldsymbol{k}\in S_n} (\boldsymbol{k}^\top B \boldsymbol{k})^{b/2}\, \Gamma(a,\pi\,\boldsymbol{k}^\top B \boldsymbol{k}) \\ &\le 2d\,3^{d-1} C_{B,b}\sum_{n=N+1}^{\infty} n^{b+d-1}\,\Gamma(a,\pi \lambda_{\min}n^2). \end{align}\] By 17 and the assumption \(\pi \lambda_{\min} n^2\ge \max\{a,1\}\) for all \(n\ge N\), we have \[\Gamma(a,\pi \lambda_{\min}n^2)\le (\pi \lambda_{\min}n^2)^a\,\mathrm{e}^{-\pi \lambda_{\min}n^2}.\] Hence, with \(c:=2a+b+d-1\), \[\sum_{n=N+1}^{\infty} n^{b+d-1}\,\Gamma(a,\pi \lambda_{\min}n^2) \le (\pi \lambda_{\min})^a\sum_{n=N+1}^{\infty} \mathrm{e}^{-\pi \lambda_{\min}n^2}\,n^{c}.\]

Let \(f(x):=\mathrm{e}^{-\pi \lambda_{\min}x^2}x^c\). Then \(f'(x)=\mathrm{e}^{-\pi \lambda_{\min} x^2}x^{c-1}(c-2\pi \lambda_{\min}x^2)\). Thus, we have \(f'(x)\leq 0\) for \(x > \sqrt{\frac{|c|}{2\pi \lambda_{\min}}}\). The condition \(\pi \lambda_{\min}N^2\ge (c+1)/2\) implies \(f\) is decreasing on \([N,\infty)\), so \[\sum_{n=N+1}^{\infty} \mathrm{e}^{-\pi \lambda_{\min}n^2} n^{c} = \sum_{n=N+1}^{\infty} f(n) \leq \sum_{n=N+1} \int_{n-1}^{n} f(x) \mathrm{d}x = \int_N^{\infty} \mathrm{e}^{-\pi \lambda_{\min}x^2}x^c \,\mathrm{d}x.\] Further, by changing variables \(t=\pi \lambda_{\min}x^2\), \[\begin{align} \int_N^{\infty} \mathrm{e}^{-\pi \lambda_{\min}x^2}x^c \,\mathrm{d}x &= \frac{1}{2} \left(\frac{1}{\pi \lambda_{\min}}\right)^{\frac{c+1}{2}} \Gamma\!\left(\frac{c+1}{2}, \pi \lambda_{\min}N^2\right) \\ & \leq \frac{1}{2} \left(\frac{1}{\pi \lambda_{\min}}\right)^{\frac{c+1}{2}} (\pi \lambda_{\min}N^2)^{\frac{c+1}{2}} \mathrm{e}^{-\pi \lambda_{\min}N^2} = \frac{1}{2} N^{c+1} \mathrm{e}^{-\pi \lambda_{\min}N^2}, \end{align}\] where we have used 17 with the assumption \(\pi \lambda_{\min}N^2 \geq \max\{\frac{c+1}{2}, 1\}\). Therefore, we conclude that \[\sum_{| \boldsymbol{k} |_\infty=N+1}^{\infty} (\boldsymbol{k}^\top B \boldsymbol{k})^{b/2} \, \Gamma(a, \pi \, \boldsymbol{k}^\top B \boldsymbol{k}) \leq d\, 3^{d-1} \, C_{B, b} \cdot (\pi \lambda_{\min})^a \cdot N^{2a+b+d} \mathrm{e}^{-\pi \lambda_{\min}N^2},\] as claimed. ◻