## 2 Definition and Properties of Distributed Order Calculus
### Establishment of the Distributed Order Rheological Model and Operator Analysis
\begin{array}{l}f(t)=L^{-1}\left[\frac{1}{v+\int_{a}^{b}\omega(\gamma)\tau s \gamma d\gamma}\right]=f_{1}(t)-f_{2}(t)\\ =\frac{1}{2\pi i\int_{0}^{\infty}}\frac{e^{-xt}}{v+\int_{a}^{b}\omega(\gamma) \tau e^{-i\pi\gamma d\gamma}}dx-\frac{1}{2\pi i\int_{0}^{\infty}}\frac{e^{-xt}} {v+\int_{a}^{b}\omega(\gamma)\tau c^{i\pi\gamma d\gamma}}dx\\ =\frac{1}{2\pi i\int_{0}^{\infty}}\frac{e^{-xt}}{v+\int_{a}^{b}\omega(\gamma) \tau^{i\gamma}\cos(\gamma\gamma)d\gamma-i\int_{a}^{b}\omega(\gamma)\tau^{i }\sin(\gamma\gamma)d\gamma}dx\\ -\frac{1}{2\pi i\int_{0}^{\infty}}\frac{e^{-xt}}{v+\int_{a}^{b}\omega(\gamma) \tau^{i}\cos(\gamma\gamma)d\gamma+i\int_{a}^{b}\omega(\gamma)\tau^{i}\sin( \gamma\gamma)d\gamma}dx\\ =\frac{1}{2\pi i\int_{0}^{\infty}}\frac{e^{-xt}2i\int_{a}^{b}\omega(\gamma) \tau^{i\pi}\sin(\gamma\gamma)d\gamma}{(v+\int_{a}^{b}\omega(\gamma)\tau)\gamma \cos(\gamma\gamma)d\gamma)+\left(\int_{a}^{b}\omega(\gamma)\tau^{i}\sin( \gamma\gamma)d\gamma\right)^{2}}dx\\ =\frac{1}{\pi}\int_{0}^{\infty}\frac{e^{-xt}\int_{a}^{b}\omega(\gamma)\tau^{i }\sin(\gamma\gamma)d\gamma}{[v+\int_{a}^{b}\omega(\gamma)\tau^{i}\cos(\gamma \gamma)d\gamma]^{2}+\left[\int_{a}^{b}\omega(\gamma)\tau^{i}\sin(\gamma\gamma) d\gamma\right]^{2}}dx\\ \end{array}  \tag{35}\]  
\(f(t)\) was an utterly monotone function, so the upper boundary could be obtained for arbitrary \(t\in[0,\infty)\), \((-1)^{n}\big{[}f(t)\big{]}^{(n)}\geq 0\), according to the time-domain expression. \[ \begin{array}{l}f(t)=\frac{1}{\pi}\int_{0}^{\infty}\frac{e^{-xt}\int_{a}^{b }\omega(\gamma)\tau^{i}\sin(\gamma\gamma)d\gamma}{[v+\int_{a}^{b}\omega(\gamma )\tau^{i}\cos(\gamma\gamma)d\gamma]^{2}+[\int_{a}^{b}\omega(\gamma)\tau^{i} \sin(\gamma\gamma)d\gamma]^{2}}dx\\ \leq\frac{1}{\pi}\int_{0}^{\infty}\frac{e^{-xt}\int_{a}^{b}\omega(\gamma) \tau^{i}\sin(\gamma\gamma)d\gamma}{[\int_{a}^{b}\omega(\gamma)\tau^{i}\sin( \gamma\gamma)d\gamma]^{2}}dx=\frac{1}{\pi}\int_{0}^{\infty}\frac{e^{-xt}}{ \int_{a}^{b}\omega(\gamma)\tau^{i}\sin(\gamma\gamma)d\gamma}dx\\ =\frac{1}{\pi}\int_{0}^{\infty}\frac{e^{-xt}}{\omega(\tilde{\gamma})\tau^{i} \sin(\gamma\tilde{\tau})(b-a)}dx\\ =\frac{1}{\pi}\frac{t^{\tilde{\gamma}-1}}{\omega(\tilde{\gamma})\sin(\gamma \tilde{\tau})(b-a)}\int_{0}^{\infty}e^{-xt}(tx)^{-\tilde{\gamma}}d(xt)\\ =\frac{1}{\pi}\frac{t^{\tilde{\gamma}-1}}{\omega(\tilde{\gamma})\sin(\gamma \tilde{\tau})(b-a)}\Gamma(1-\hat{\gamma})\leq\frac{1}{\pi}\frac{t^{\tilde{ \gamma}-1}}{\omega(\tilde{\gamma})\sin(\gamma\tilde{\tau})(b-a)}\Gamma(1- \hat{\gamma})\qquad\hat{\gamma}\in(a,b)\end{array}  \tag{36}\]  
Similarly, the lower boundary could be obtained as follows:  
\[ \begin{array}{l}f(t)=\frac{1}{\pi}\int_{0}^{\infty}\frac{e^{-xt}\int_{a}^{b} \omega(\gamma)x^{\gamma}\sin(\gamma\gamma)d\gamma}{[v+\int_{a}^{b}\omega( \gamma)x^{\gamma}\cos(\pi\gamma)d\gamma]^{2}+[\int_{a}^{b}\omega(\gamma)x^{ \gamma}\sin(\pi\gamma)d\gamma]^{2}}dx\\ \geq\frac{1}{\pi}\int_{0}^{\infty}\frac{e^{-xt}\int_{a}^{b}\omega(\gamma)x^{ \gamma}\sin(\gamma\tilde{\tau})d\gamma}{[v+\int_{a}^{b}\omega(\gamma)x^{\gamma }d\gamma]^{2}+[\int_{a}^{b}\omega(\gamma)x^{\gamma}d\gamma]^{2}}dx\\ =\frac{1}{\pi}\int_{0}^{\infty}\frac{e^{-xt}\tilde{\gamma}\sin(\gamma\tilde{ \tau})(b-a)}{2[v+\int_{a}^{b}\omega(\gamma)x^{\gamma}d\gamma]^{2}}dx\\ =\frac{1}{\pi}\frac{\omega(\tilde{\gamma})\sin(\gamma\tilde{\tau})(b-a)}{2[v+ \omega(\tilde{\gamma})\tau^{i}]^{2}}\int_{0}^{\infty}e^{-xt}x\tilde{\gamma}dx \\ =\frac{1}{\pi}\frac{\omega(\tilde{\gamma})\sin(\gamma\tilde{\tau})(b-a)}{2[v+ \omega(\tilde{\gamma})\tau^{i}]^{2}}t^{-1-\tilde{\gamma}}\Gamma(1+\tilde{ \gamma})\\ \geq\frac{1}{\pi}\frac{\omega(\tilde{\gamma})\sin(\gamma\tilde{\tau})(b-a)}{2[v +\omega(\tilde{\gamma})\tau^{i}]^{2}}t^{-1-a}\Gamma(1+\tilde{\gamma})\\ \end{array}  \tag{37}\]  
A calculation example: With \(\
u=0.1\), \(a=0.3\), \(b=0.8\), \(\omega(\gamma)=1\) and \(\hat{\gamma}=0.6\), \(\widetilde{\gamma}=0.5\), \(\overline{x}=0.5\), the analytic values, upper boundary, and lower boundary values of \(f(t)\) could be calculated, as shown in Figure 5.