Suppose I have the following piecewise function: $$Q(t) = \begin{cases} W(t) & t<T \\ 1 & t=T \\ 0 & t>T \end{cases}$$ Where, $W(t)$ is a non-decreasing function, in fact $W(t)$ is the cumulative probability distribution function of a Weibull distribution.
My question: I need to take Laplace transformation of $Q(t)$ and calculate its first and second derivatives at point 0. Let's assume $q(S) \triangleq LS\{Q(t)\}$ represents Laplace transform of $Q(t)$.
My attepmt: $$\dot{Q}(t) = \begin{cases} W(t) & t<T \\ 1 & T-\epsilon <t< T + \epsilon \\ 0 & t>T + \epsilon \end{cases} $$ Where, $\epsilon \to 0^+$. Then I have the following equation: \begin{equation} \dot{Q}(t) = W(t) + U_{T-\epsilon}(t) (1 - W(t))+ U_{T+\epsilon}(t)(0-1) \end{equation} Following is the Laplace transformation: $$ \dot{q}(S) = w(S) +e^{-S(T-\epsilon)}(\frac{1}{S}-LS\{W(t+T+\epsilon)\}- e^{-S(T+\epsilon)} \frac{1}{S} $$ After simplification we have:
$$\dot{q}(S)=w(S) - e^{-ST}\bar{w}(S)$$ Where, $\bar{w}(S) \triangleq LS\{W(t+T+\epsilon)\}$. Hence, we have the following equations: $$\dot{q}(0) = 0$$ and $$\dot{q}'(S)=w'(S) + T e^{-ST}\bar{w}(S)-e^{-ST}\bar{w}'(S)$$ Because $W(t)$ is a cumulative Weibull distribution, $w(0)=1$, $\bar{w}(0)=1$, $w'(0)=-\mu$, and $\bar{w}'(0)=-\mu -T$. Here, we assume $\mu$ is the expected value of Weibull distribution. Thus: $$\dot{q}'(0) = -\mu + T + \mu + T$$.
What I expected: $\dot{q}(0) = 1$ and $\dot{q}'(0)<0$.
I wish someone can assist me with this issue.
One more thing: I have searched, and still could not find any close-form for $\int_{0}^{T}e^{-St}W(t)d(t)$, if you have any approximation, that could be a huge help too.