#223に示した理由により、本稿の議論は全て取り消します。

前稿の続きで、ISO 26262のPMHFの導出の場合、確率積分を実行する際に次の(226.1)が出てくるため、あらかじめ結果を導出しておき、後程積分公式として使用します。 $$ \img[-1.35em]{/images/withinseminar.png} \tag{226.1} $$ $t=i\tau+u, i=0,1,...,n-1, n:=\frac{T_\text{lifetime}}{\tau}$とおいて変数変換すれば、 $$ (226.1)=\frac{1}{T_\text{lifetime}}\sum_{i=0}^{n-1}\int_0^\tau(i\tau+u)e^{-\lambda_\text{IF}(i\tau+u)-\lambda_\text{SM}u}du\\ =\tau\sum_{i=0}^{n-1}i e^{-\lambda_\text{IF}i\tau} \frac{1}{T_\text{lifetime}}\int_0^\tau e^{-(\lambda_\text{IF}+\lambda_\text{SM})u}du +\sum_{i=0}^{n-1}e^{-\lambda_\text{IF}i\tau}\frac{1}{T_\text{lifetime}}\int_0^\tau ue^{-(\lambda_\text{IF}+\lambda_\text{SM})u}du\\ \tag{226.2} $$ ここで、(226.2)右辺第1項の級数の和を求めるため和を$x$とおけば、 $$ x:=\sum_{i=0}^{n-1}i e^{-\lambda_\text{IF}i\tau}=e^{-\lambda_\text{IF}\tau}+2e^{-\lambda_\text{IF}2\tau}+...+(n-1)e^{-\lambda_\text{IF}(n-1)\tau}\tag{226.3} $$ となり、 $$ e^{-\lambda_\text{IF}\tau}x=\sum_{i=0}^{n-1}i e^{-\lambda_\text{IF}(i+1)\tau}=e^{-\lambda_\text{IF}2\tau}+...+(n-2)e^{-\lambda_\text{IF}(n-1)\tau}+(n-1)e^{-\lambda_\text{IF}n\tau}\tag{226.4} $$ よって、(226.3)-(226.4)より、 $$ x(1- e^{-\lambda_\text{IF}\tau})=e^{-\lambda_\text{IF}\tau}+e^{-\lambda_\text{IF}2\tau}+...+e^{-\lambda_\text{IF}(n-1)\tau}-(n-1)e^{-\lambda_\text{IF}n\tau}\\ =\underbrace{e^{-\lambda_\text{IF}\tau}+e^{-\lambda_\text{IF}2\tau}+...+e^{-\lambda_\text{IF}(n-1)\tau}+e^{-\lambda_\text{IF}n\tau}}_{\text{n terms}}-ne^{-\lambda_\text{IF}n\tau}\\ =e^{-\lambda_\text{IF}\tau}\frac{1-e^{-\lambda_\text{IF}T_\text{lifetime}}}{1-e^{-\lambda_\text{IF}\tau}}-n e^{-\lambda_\text{IF}T_\text{lifetime}}\tag{226.5} $$ よって、Maclaurin展開の1次近似を用いれば、 $$ \require{cancel} x\approx\frac{\bcancel{\lambda_\text{IF}}T_\text{lifetime}}{\lambda_\text{IF}^\bcancel{2}\tau^2}(1-\lambda_\text{IF}\tau)-\frac{n(1-\lambda_\text{IF}T_\text{lifetime})}{\lambda_\text{IF}\tau}\\ =\frac{T_\text{lifetime}(\bcancel{1}-\bcancel{\lambda_\text{IF}}\tau)-T_\text{lifetime}(\bcancel{1}-\bcancel{\lambda_\text{IF}}T_\text{lifetime})}{\bcancel{\lambda_\text{IF}}\tau^2}=\frac{T_\text{lifetime}(T_\text{lifetime}-\tau)}{\tau^2}\tag{226.6} $$ 次に、(226.2)右辺第2項の級数の和は、 $$ \sum_{i=0}^{n-1}e^{-\lambda_\text{IF}i\tau}=e^{-\lambda_\text{IF}\tau}+...+e^{-\lambda_\text{IF}(n-1)\tau}=\frac{1-e^{-\lambda_\text{IF}T_\text{lifetime}}}{1-e^{-\lambda_\text{IF}\tau}} \approx\frac{\bcancel{\lambda_\text{IF}}T_\text{lifetime}}{\bcancel{\lambda_\text{IF}}\tau} \tag{226.7} $$ 次に、(226.2)右辺第1項の定積分の値は、 $$ \int_0^\tau e^{-(\lambda_\text{IF}+\lambda_\text{SM})u}du =\left[\frac{e^{-(\lambda_\text{IF}+\lambda_\text{SM})u}}{-(\lambda_\text{IF}+\lambda_\text{SM})}\right]_0^\tau =\frac{e^{-(\lambda_\text{IF}+\lambda_\text{SM})\tau}-1}{-(\lambda_\text{IF}+\lambda_\text{SM})}\\ \approx\frac{1}{\bcancel{\lambda_\text{IF}+\lambda_\text{SM}}}\left(\bcancel{(\lambda_\text{IF}+\lambda_\text{SM})}\tau-\frac{1}{2}(\lambda_\text{IF}+\lambda_\text{SM})^\bcancel{2}\tau^2\right) =\tau\left(1-\frac{1}{2}(\lambda_\text{IF}+\lambda_\text{SM})\tau\right) \tag{226.8} $$ 以上から、$\color{red}{(226.5)}$と$\color{green}{(226.6)}$を(226.2)に適用し、$\color{blue}{(226.7})$と部分積分の結果$\color{purple}{(225.1)}$を用いれば、 $$ (226.2)=\bcancel{\tau}\color{red}{\left(\frac{\bcancel{T_\text{lifetime}}(T_\text{lifetime}-\tau)}{\bcancel{\tau^2}}\right)}\frac{1}{\bcancel{T_\text{lifetime}}}\color{blue}{\bcancel{\tau}\left(1-\frac{1}{2}(\lambda_\text{IF}+\lambda_\text{SM})\tau\right)}\\ +\color{green}{\frac{\bcancel{T_\text{lifetime}}}{\bcancel{\tau}}}\frac{1}{\bcancel{T_\text{lifetime}}} \color{purple}{\left(\frac{\tau^\bcancel{2}}{2}- \frac{(\lambda_\text{IF}+\lambda_\text{SM})\tau^{\bcancel{3}2}}{3}\right)}\\ =(T_\text{lifetime}-\tau)\left(1-\frac{1}{2}(\lambda_\text{IF}+\lambda_\text{SM})\tau\right)+\left(\frac{\tau}{2}-\frac{1}{3}(\lambda_\text{IF}+\lambda_\text{SM})\tau^2\right)\\ =\left(1-\frac{1}{2}(\lambda_\text{IF}+\lambda_\text{SM})\tau\right)T_\text{lifetime}-\frac{\tau}{2}+\frac{1}{6}(\lambda_\text{IF}+\lambda_\text{SM})\tau^2 \tag{226.9} $$

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