Laplace Transforms — Question 4

Question 4 – Solution

Strategy. A cumulative integral may retain a nonzero constant even when its input decays. Its transform domain must be checked anew.

Step 1: Derive the cumulative rule. Here $y(0)=0$, $y'=g$, and $y$ is bounded because $g$ is absolutely integrable. For $s>0$, integration by parts gives $$G(s)=[e^{-st}y(t)]_0^\infty+s\int_0^\infty e^{-st}y(t)\,dt=sY(s).$$ The initial term is zero and the upper term vanishes because $y$ is bounded. Thus $Y=G/s$ on this half-line.

Step 2: Determine the response and its domain. An antiderivative gives $$y(t)=\frac{1+e^{-t}(\sin t-\cos t)}2,\qquad \boxed{Y(s)=\frac{s+1}{s[(s+1)^2+1]},\quad s>0.}$$ The formula satisfies $y(0)=0$ and $y'=g$. Since $y(t)\to 1/2$, it is eventually bounded below by a positive constant. Its transform therefore diverges for every $s\le 0$. The cumulative response has a smaller convergence interval than the input.

Step 3: Remove the persistent constant. Now $z=e^{-t}(\sin t-\cos t)/2$. Transforming these two terms directly gives $$\boxed{Z(s)=-\frac{s}{2[(s+1)^2+1]},\qquad s>-1.}$$ It converges absolutely there; at and below $-1$ the nondecaying or growing trigonometric tail prevents ordinary convergence. In particular, $Z(0)=0$: its signed areas cancel. The identity $Z=Y-1/(2s)$ is initially valid only for $s>0$. The wider domain follows from the actual centered function, not by subtracting divergent integrals at $s=0$.

Step 4: Locate the overshoot. Since $y'=e^{-t}\cos t$, the first positive turning time is $t=\pi/2$, with derivative changing from positive to negative. Its height is $$\boxed{y(\pi/2)=\tfrac 12(1+e^{-\pi/2})>\tfrac 12.}$$ The response overshoots and oscillates about its limiting constant, so it is neither monotone nor decaying to zero. Its centered part does decay exponentially.