## 4 Error Analysis  
The following theorem establishes the truncation error of the \(\alpha\)th-order GBFA quadrature associated with the \(\alpha\)th-order FSGIM \({}^{\varepsilon}\mathbf{Q}_{n}^{\alpha}\) in closed form. **Theorem 1**. : _Suppose that \(f\in C^{n+1}(\mathbf{\Omega}_{1})\) is approximated by the GBFA interpolant (7). Assume also that the integrals_  
\[\mathcal{I}_{1}^{(y)}\hat{G}_{0,n}^{\lambda}[t(1-y^{1/\alpha})], \tag{22}\]  
_are computed exactly \(\forall\,t\in\mathbf{\Omega}_{1}\). Then \(\exists\,\xi=\xi(t)\in\mathbf{\Omega}_{1}^{\circ}\) such that the truncation error, \({}^{\alpha}\mathcal{I}_{n}^{\lambda}(t,\xi)\), in the RLFI approximation (15) is given by_  
\[{}^{\alpha}\mathcal{I}_{n}^{\lambda}(t,\xi)=\frac{t^{\alpha}f^{(n+1)}(\xi)}{( n+1)!\Gamma(\alpha+1)\hat{K}_{n+1}^{\lambda}}\mathcal{I}_{1}^{(y)}\hat{G}_{n+1}^ {\lambda}[t(1-y^{1/\alpha})], \tag{23}\]  
_where \(\hat{K}_{n}^{\lambda}\) is the leading coefficient of the \(n\)th-degree, \(\lambda\)-indexed SG polynomial, defined as_  
\[\hat{K}_{n}^{\lambda}=2^{2n-1}\frac{\Gamma(n+\lambda)\Gamma(2\lambda+1)}{ \Gamma(n+2\lambda)\Gamma(\lambda+1)},\quad n\in\mathbb{Z}_{0}^{+}. \tag{24}\]  
Proof. : The Lagrange interpolation error associated with the GBFA interpolation (7) is given by  
\[f(t)=I_{n}f(t)+\frac{f^{(n+1)}(\xi)}{(n+1)!\hat{K}_{n+1}^{\lambda}}\hat{G}_{n+ 1}^{\lambda}(t), \tag{25}\]  
where \(\hat{K}_{n}^{\lambda}\) is given by Equation (24), which can be directly derived from [34] (Equation (4.12)) by setting \(L=1\). Applying the RLFI operator \(\mathcal{I}_{t}^{\alpha}\) on both sides of Equation (25) results in the truncation error  
\[{}^{\alpha}\mathcal{I}_{n}^{\lambda}(t,\xi)=\frac{f^{(n+1)}(\xi)}{(n+1)!\hat{ K}_{n+1}^{\lambda}}\mathcal{I}_{t}^{\alpha}\hat{G}_{n+1}^{\lambda}. \tag{26}\]  
The proof is accomplished from Equation (26) by applying the change of variables (13) on the RLFI of \(\hat{G}_{n+1}^{\lambda}\). The following theorem provides an upper bound for the truncation error (23). **Theorem 2**. : _Let \(\left\|f^{(n)}\right\|_{L^{\infty}(\mathbf{\Omega}_{1})}=\mathcal{A}_{n}\), and suppose that the assumptions of Theorem 1 hold. Then the truncation error \({}^{\alpha}\mathcal{I}_{n}^{\lambda}(t,\xi)\) satisfies the asymptotic bound_  
\[\left|{}^{\alpha}\mathcal{I}_{n}^{\lambda}(t,\xi)\right|\lesssim\mathcal{A}_{n +1}\,\vartheta_{\alpha,\lambda}\Big{(}\frac{e}{4}\Big{)}^{n}n^{-\frac{3}{2}-n+ \lambda}\,\mathrm{Y}_{\sigma^{\lambda}}(n)\quad\forall_{t}\,n, \tag{27}\]  
_where_  
\[\mathrm{Y}_{\sigma^{\lambda}}(n)=\left\{ \begin{array}{ll}1,&\lambda\in \mathbb{R}_{0}^{+},\\ \sigma^{\lambda}n^{-\lambda},&\lambda\in\mathbb{R}_{-1/2}^{-1/2},\end{array} \right.\]\(\sigma^{\lambda}>1\) is a constant dependent on \(\lambda\), and_  
\[\theta_{\alpha,\lambda}=\tfrac{1}{4\pi}\epsilon^{\alpha+1}\alpha^{- \frac{1}{2}-\alpha}(1+2\lambda)^{-\frac{1}{2}-2\lambda}\left[1+\frac{1}{1620(1+ \lambda)^{5}}\right]\times\] \[\left[\frac{(1+\lambda)\operatorname{csch}\!\left(\tfrac{1}{1+2 \lambda}\right)}{1+2\lambda}\right]^{\frac{1}{2}+\lambda}\left[\alpha\sinh \!\left(\frac{1}{\alpha}\right)\right]^{-\alpha/2}\!\left[(1+\lambda)\sinh \!\left(\frac{1}{1+\lambda}\right)\right]^{\frac{1+\lambda}{2}}. \tag{28}\]  
Proof. : Observe first that \(t^{\alpha}\leq 1\,\forall t\in\mathbf{\Omega}_{1}\). Notice also that  
\[\frac{1}{(n+1)!\Gamma(\alpha+1)\bar{K}_{n+1}^{\lambda}}=\frac{2^{-2n-1}(2 \lambda+1)\Gamma(\lambda+2)\Gamma(n+2\lambda+1)}{(\lambda+1)\Gamma(\alpha+1) \Gamma(2\lambda+2)\Gamma(n+2)\Gamma(n+\lambda+1)}, \tag{29}\]  
by definition. The asymptotic inequality (27) results immediately from (23) after applying the sharp inequalities of the Gamma function [36] ([InEquation (96)]) on Equation (29) and using [37] ([Lemma 5.1]), which gives the uniform norm of Gegenbauer polynomials and their associated shifted forms.