# Energies of Mechanical Fractional-Order Elements: Causal Concept and Kernel Effects
## 5 More General Formulations of the Causal Force-Displacement Relationship
### Fractional Element with Exponential Memory
#### 5.1.1 General Power-Law Force to the Element with Exponential Memory  
Consider a general formulation of the power-law force \(F(t)=t^{m}\), where we can model a unit step (\(m=0\)), a linear ramp (\(m=1\)), and many other functional models \(F(t)\). In such a case, the displacement (we use the version (63) to make the derived expressions easier) is  
\[x(t)=\frac{1}{k_{e}}\int\limits_{0}^{t}e^{-b(t-\tau)}\tau^{m}d\tau. \tag{65}\]  
In the specific case of a unit step when \(m=0\), we have  
\[x(t)=\frac{1}{k_{e}}\frac{1}{b}\Big{(}1-e^{-bt}\Big{)}\Rightarrow\frac{dx(t) }{dt}=\frac{1}{k_{e}}e^{-bt} \tag{66}\]  
with a decreasing rate approaching zero for a long time.  
For a linear ramp \(m=1\), through simple integration by parts we obtain  
\[x(t)=\frac{1}{k_{e}}\bigg{(}\frac{t}{b}+\frac{1}{b^{2}}e^{-bt}-\frac{1}{b^{2} }\Big{)}\Rightarrow\frac{dx(t)}{dt}=\frac{1}{k_{e}}\frac{1}{b}\Big{(}1+e^{- bt}\Big{)}, \tag{67}\]  
that is, for a long time we have a time-independent displacement rate \(\frac{dx(t)}{dt}\rightarrow\frac{1}{k_{e}}\frac{1}{b}\).  
In the general case, we may solve (65) approximately by expansion of the exponential functions as series, namely,\[x(t)=\frac{1}{k_{e}}\int\limits_{0}^{t}e^{-b(t-\tau)}\tau^{m}d\tau= \frac{1}{k_{e}}e^{-bt}\int\limits_{0}^{t}e^{b\tau}\tau^{m}d\tau \tag{68}\] \[\approx\frac{1}{k_{e}}e^{-bt}\int\limits_{0}^{t}\tau^{m}\sum \limits_{k=0}^{\infty}\frac{(b\tau)^{k}}{k!}d\tau\approx\frac{1}{k_{e}}e^{-bt} \int\limits_{0}^{t}\sum\limits_{k=0}^{\infty}\frac{b^{k}\tau^{m+k}}{k!}d\tau= \frac{1}{k_{e}}e^{-bt}\sum\limits_{k=0}^{\infty}\frac{b^{k}t^{m+k+1}}{(m+k+1)k!}.\]  
Then, with \(e^{-bt}\approx\sum\limits_{k=0}^{\infty}\left(-1\right)^{k}\frac{b^{k}t^{k}}{k!}\) we can approximate \(x(t)\) as  
\[x(t)\approx\sum\limits_{k=0}^{\infty}\left(-1\right)^{k}\frac{b^{k}t^{k}}{k!} \sum\limits_{k=0}^{\infty}\frac{b^{k}t^{m+k+1}}{(m+k+1)k!}=\sum\limits_{k=0}^ {\infty}c_{k}t^{k}\cdot\sum\limits_{k=0}^{\infty}g_{k}t^{(m+1)+k}=t^{(m+1)} \sum\limits_{k=0}^{\infty}c_{k}t^{k}\cdot\sum\limits_{k=0}^{\infty}g_{k}t^{k} \tag{69}\]  
with \(c_{k}=(-1)^{k}\frac{b^{k}t^{k}}{k!}\) and \(g_{k}=\frac{b^{k}}{(m+k+1)k!}\).  
Then, from the multiplication of infinite series (we used the same approach in the case with an exponential force ramp; see the result of (58)),  
\[x(t)\approx t^{(m+1)}\sum\limits_{k=0}^{\infty}c_{k}t^{k}\cdot\sum\limits_{k=0 }^{\infty}g_{k}t^{k}=t^{(m+1)}\sum\limits_{k=0}^{\infty}h_{k}t^{k}=\sum\limits _{k=0}^{\infty}h_{k}t^{k+(m+1)}, \tag{70}\]  
where \(h_{k}=\sum\limits_{l=0}^{k}c_{l}g_{k-i}=(c_{0}g_{k}+c_{1}g_{k-1}+\ldots+c_{k}g _{0})\), \(k=0,1,2,\ldots\)[31] (p. 110).  
Consequently,  
\[\frac{dx(t)}{dt}\approx\sum\limits_{k=0}^{\infty}\left(k+m+1\right)h_{k}t^{k +m} \tag{71}\]  
and the energy can be calculated approximately from  
\[W_{exp}\approx\frac{1}{k_{e}}\int\limits_{0}^{t}\sum\limits_{k=0}^{\infty} \left[(k+m+1)h_{k}t^{k+m}\right]t^{m}dt\approx\frac{1}{k_{e}}\sum\limits_{k=0 }^{\infty}\frac{(k+m+1)}{(k+2m+1)}h_{k}t^{k+2m+1}. \tag{72}\]