## Appendix A Description of Relaxation in Time and Frequency
### Truncation Error in Large-\(\omega\) Expansion  
In this appendix, an upper bound for the truncation error of the large-\(\omega\) expansion (21) is derived. The following two unproven statements of [2] turn out to be wrong: (i) the most accurate results are obtained by truncating the summation before the smallest term; and (ii) the truncation error is less than twice the first neglected term. We specialize again to the cosine transform \(Q_{\beta}(\omega)\). If we choose \(\beta=4/3\) the oscillatory factor \(\sin(k\beta\pi/2)\) in Equation (21) is zero for \(k=3\). If the statements of [2] were correct, then we could stopthe summation at \(k=2\) with a truncation error of zero for all values of \(\omega\). This is obviously wrong. A correct truncation criterion can only be based on the amplitudes \(B_{k}\); it must disregard the oscillating prefactor \(\sin(k\beta\pi/2)\). But even after omitting oscillatory factors the two statements are unfounded. In [2] they are underlaid by a reference to a specific page in a book on numerical analysis [42]. However, that page only says \"the error committed is usually less than twice the first neglected term\", followed by a reference to a specific page in a 1907 book on Celestial Mechanics [43]. Going back to this source, we find a rigorous theorem, which however holds only under very restrictive conditions not fulfilled here. Therefore we have to restart from scratch. We will simplify an argument of [PERSON] [34], and generalize it to cover not only the convergent case \(\beta\leq 1\) but also the asymptotic expansion for \(\beta>1\). Substituting \(s=\omega t\), the Fourier integral (4) takes the form  
\[F_{\beta}(\omega)=\omega^{-1}\int_{0}^{\infty}\!\mathrm{d}s\,\exp\left(is- \omega^{-\beta}s^{\beta}\right) \tag{89}\]  
We consider the cosine transform \(Q_{\beta}(\omega)=\text{Re}\;F_{\beta}(\omega)\), which we rewrite as \(Q_{\beta}(\omega)=G(\omega^{-\beta})/\omega\), introducing the functions  
\[G(x):=\text{Re}\,\int_{0}^{\infty}\!\mathrm{d}s\,\gamma(s,x,0) \tag{90}\]  
and  
\[\gamma(s,x,a):=s^{a}\exp\left(is-xs^{\beta}\right) \tag{91}\]  
The Taylor expansion of \(G(x)\), including the Lagrange remainder, reads  
\[G(x)=\sum_{k=0}^{n-1}G^{(k)}(0)\frac{x^{k}}{k!}+G^{(n)}(\xi)\frac{x^{n}}{n!} \tag{92}\]  
with \(0\leq\xi\leq x\) and  
\[G^{(k)}(\xi)=\left(-1\right)^{k}\text{Re}\,\int_{0}^{\infty}\!\mathrm{d}s\, \gamma(s,\xi,k\beta) \tag{93}\]  
Now, we choose an integration path \(C\) in the complex plane consisting of two line segments, \(s\) and \(s\mathrm{e}^{i\phi}\), and two arcs, \(r\mathrm{e}^{i\varphi}\) and \(R\mathrm{e}^{i\varphi}\), with \(0<r\leq s\leq R<\infty\) and \(0\leq\varphi\leq\phi\leq\pi/2\) as shown in Figure 4. The integral of \(\gamma\) along this path is zero:  
\[\int_{C}\mathrm{d}z\,\gamma(z,x,a)=0 \tag{94}\]The contributions of the two arcs tend to 0 as \(r\to 0\) and \(R\to\infty\). Hence the contributions of the two line segments have equal modulus. This allows us to obtain the following bounds:  
\[\left|G^{(n)}(\xi)\right| = \left|(-1)^{n}\,\text{Re}\,\int_{0}^{\infty}\!\!\mathrm{d}s\,\gamma (s,\xi,n\beta)\right|\] \[\leq \left|\int_{0}^{\infty}\!\mathrm{d}s\,\gamma(s,\xi,n\beta)\right|\] \[= \left|\int_{0}^{\infty}\!\mathrm{d}s\,\gamma(s\mathrm{e}^{i\phi },\xi,n\beta)\right|\] \[G(x):=\text{Re}\,\int_{0}^{\infty}\!\mathrm{d}s\,\gamma(s,x,0) \tag{90}\]0 \[G(x):=\text{Re}\,\int_{0}^{\infty}\!\mathrm{d}s\,\gamma(s,x,0) \tag{90}\]1 \[G(x):=\text{Re}\,\int_{0}^{\infty}\!\mathrm{d}s\,\gamma(s,x,0) \tag{90}\]2  
At this point we choose \(\phi\) as in Equation (33), which ensures \(\cos(\phi\beta)\geq 0\). The bound  
\[G(x):=\text{Re}\,\int_{0}^{\infty}\!\mathrm{d}s\,\gamma(s,x,0) \tag{90}\]3  
is independent of \(\xi\). The integral is the same as in Equation (11).