# Time-Dependent Probabilistic Approach of Failure Mode and Effect Analysis
## Appendix A  
\[\begin{split} E[L]&=\int_{0}^{\infty}\int_{0}^{ \infty}L(s,t)g_{\text{S}\uparrow=t}(s)h_{\text{T}}(t)dsdt\\ &=\int_{0}^{\infty}\int_{0}^{t}L(s,t)g_{\text{S}\uparrow=t}(s)h_ {\text{T}}(t)dsdt+\int_{0}^{\infty}\int_{t}^{\infty}L(s,t)g_{\text{S}\uparrow=t }(s)h_{\text{T}}(t)dsdt\\ &=\int_{0}^{\infty}\int_{0}^{t}\alpha\frac{\mu}{c+\mu t}\lambda e ^{-\lambda t}dsdt+\int_{0}^{\infty}\int_{t}^{\infty}[\alpha+\beta+\theta(s-t)] \frac{c\mu}{c+\mu}\ e^{-\mu(s-t)}\lambda e^{-\lambda t}dsdt\\ &=\int_{0}^{\infty}\alpha\bigg{[}\int_{0}^{t}\frac{\mu}{c+\mu t} ds+\int_{t}^{\infty}\frac{c\mu}{c+\mu t}\ e^{-\mu(s-t)}ds\bigg{]}\lambda e^{-\lambda t}dt+ \int_{0}^{\infty}\int_{t}^{\infty}[\beta+\theta(s-t)]\frac{c\mu}{c+\mu t}\ e^{- \mu(s-t)}\lambda e^{-\lambda t}dsdt.\end{split} \tag{1}\]  
\[\begin{split}&\int_{0}^{\infty}\alpha\bigg{[}\int_{0}^{t} \frac{\mu}{c+\mu t}ds+\int_{t}^{\infty}\frac{c\mu}{c+\mu t}\ e^{-\mu(s-t)}ds \bigg{]}\lambda e^{-\lambda t}dt\\ &=\int_{0}^{\infty}\alpha\bigg{[}\int_{0}^{t}\frac{\mu}{c+\mu t} ds+\int_{0}^{\infty}\frac{c\mu}{c+\mu t}\ e^{-\mu(s-t)}ds\bigg{]}\lambda e^{-\lambda t}dt\\ &=\int_{0}^{\infty}\alpha\bigg{[}\int_{0}^{t}\frac{\mu}{c+\mu t} ds+\int_{0}^{\infty}\frac{c\mu}{c+\mu t}\ e^{-\mu(s-t)}ds\bigg{]}\lambda e^{-\lambda t}dt \\ &=\int_{0}^{\infty}\alpha\bigg{[}\frac{\mu t}{c+\mu t}+\frac{c}{c+ \mu t}\lambda e^{-\lambda t}dt\\ &=\alpha\frac{c+\mu t}{c+\mu t}\\ &=\alpha.\end{split} \tag{2}\]  
\[\begin{split}&\int_{0}^{\infty}\int_{t}^{\infty}[\beta+\theta (s-t)]\frac{c\mu}{c+\mu t}\ e^{-\mu(s-t)}\lambda e^{-\lambda t}dsdt\\ &=\int_{0}^{\infty}\frac{c\lambda}{c+\mu t}\lambda e^{-\lambda t }dt\int_{0}^{\infty}[\beta+\theta(s-t)]\mu e^{-\mu(s-t)}ds\\ &=\int_{0}^{\infty}\frac{c\lambda}{c+\mu t}e^{-\lambda t}dt \int_{0}^{\infty}[\beta+\theta\beta]\mu e^{-\mu\frac{c}{c+\mu t}}ds\end{split} \tag{3}\]  
\[\begin{split}&=\int_{0}^{\infty}\frac{c\lambda}{c+\mu t}e^{- \lambda t}dt\bigg{[}(\beta+\theta\beta)\left(-e^{-\mu\beta}\right)\right]_{0} ^{\infty}-\theta\left(\frac{1}{\mu}e^{-\mu\beta}\right)\right]_{0}^{\infty} \bigg{]};\text{ integration by parts}\\ &=\left(\beta+\frac{\theta}{\mu}\right)\int_{0}^{\infty}\frac{c \lambda}{c+\mu t}e^{-\lambda t}dt\\ &=\left(\beta+\frac{\theta}{\mu}\right)\int_{0}^{\infty}\frac{c \lambda^{2}}{c+\mu t}e^{\frac{\lambda t}{c}-\lambda t}\lambda t\tau;\tau= \frac{c\lambda}{\mu}+\lambda t=\frac{1}{\mu}(c+\mu t)\\ &=\left(\beta+\frac{\theta}{\mu}\right)\frac{c\lambda}{\mu}e^{ \frac{\lambda t}{c}}\int_{\frac{c\mu}{c}}^{\infty}\frac{c-\tau}{\mu}d\tau\\ &=\left(\beta+\frac{\theta}{\mu}\right)\frac{c\lambda}{\mu}e^{ \frac{\lambda t}{c}}E_{1}\bigg{(}\frac{c\lambda}{\mu}\bigg{)}.\end{split} \tag{4}\]  
For \(x>0\), \(E_{1}(x)\) is the exponential integral function defined by  
\[E_{1}(x)=\int_{1}^{\infty}\frac{e^{-\lambda t}}{t}dt=\int_{x}^{\infty}\frac{e^{ -t}}{t}dt=-\gamma-\ln x-\sum_{k=1}^{\infty}\frac{(-x)^{k}}{k!}, \tag{5}\]  
where \(\gamma=\lim\limits_{n\to\infty}\left(\sum_{k=1}^{n}\frac{1}{k}-\ln n\right)=0.5 77\ldots\) is an irrational number called the Euler constant. Therefore,  
\[E[L]=\alpha+\left(\beta+\frac{\theta}{\mu}\right)\frac{c\lambda}{\mu}e^{\frac{c \lambda}{\mu}}E_{1}\bigg{(}\frac{c\lambda}{\mu}\bigg{)}. \tag{6}\]