Table Of Laplace Transforms — Question 5

Question 5 – Solution

Strategy. Removing an old tail requires its original age, not a fresh copy starting at zero.

Step 1: Identify the tail that must be removed. For $t=2+u$, the old signal is $e^{-2}(u+2)e^{-u}$. Thus $$w=te^{-t}-e^{-2}H_2(t)[(t-2)+2]e^{-(t-2)}.$$ The proposed subtraction instead removes $H_2(t)(t-2)e^{-(t-2)}$, which has both the wrong age polynomial and the wrong exponential scale.

Step 2: Transform and check directly. For $s\ne-1$, the correct expression is $$\boxed{W(s)=\frac{1-e^{-2(s+1)}[1+2(s+1)]}{(s+1)^2}.}$$ Initially table operations justify it for $s>-1$. Independently put $p=s+1$: $W=\int_0^2 t e^{-pt}\,dt$. Integration by parts gives $[1-e^{-2p}(1+2p)]/p^2$, agreeing with the table calculation and extending its formula to $p\ne 0$ of either sign.

Step 3: Remove the singularity and identify the domain. A bounded finite interval makes this transform finite for every real $s$. At $s=-1$ the defining integral is $\int_0^2 t\,dt=\boxed{2}$. The numerator’s Taylor expansion is $2p^2+O(p^3)$, so the apparent pole is removable. The two infinite-tail terms need only be transformed separately on their common domain; their difference has a larger actual domain.

Step 4: Record the geometric checks. The area is $W(0)=\boxed{1-3e^{-2}}$. Since $(te^{-t})'=(1-t)e^{-t}$, the unique global maximum is $1/e$ at $t=1$. At the cutoff, $w(2^-)=2e^{-2}$ while $w(2)=w(2^+)=0$. The open dot marks the excluded left-hand value and the filled dot the actual value at $2$; no slanted segment bridges the jump.