Inverse Laplace Transforms — Question 1

Question 1 – Solution

Strategy. Resolve the two poles, then factor the resulting time expression before deciding its sign or shape.

Step 1: Resolve and invert. Writing $F=A/(s+1)+B/(s+3)$ gives $A+B=1$ and $3A+B=5$. Thus $A=2$, $B=-1$, and $$\boxed{f(t)=2e^{-t}-e^{-3t}.}$$ Transforming these two terms gives $2/(s+1)-1/(s+3)$, whose numerator is $2(s+3)-(s+1)=s+5$. This verifies the entire rational expression, not just its pole locations.

Step 2: Retain the actual domain. For $s>-1$, both exponential integrals converge absolutely. Since $f=e^{-t}(2-e^{-2t})\ge e^{-t}$ on $t\ge 0$, its weighted integral diverges for $s\le-1$. Hence the exact real domain is $\boxed{s>-1}$; other algebraic values of $F$ are not ordinary transform values.

Step 3: Determine the sign. The factor $2-e^{-2t}$ lies in $[1,2)$, so $f(t)>0$ everywhere. The negative coefficient belongs to one term in a decomposition; it does not determine the sign of the whole function. Here the faster-decaying negative term never dominates the positive term.

Step 4: Locate the peak. We have $f(0)=1$, $f(t)\to 0$, and $$f'(t)=e^{-t}(-2+3e^{-2t}).$$ The bracket decreases through zero once. Therefore $f$ increases up to, then decreases after, $$\boxed{t_*=\frac 12\ln(3/2),\qquad f(t_*)=\frac 43\sqrt{2/3}.}$$ In particular $f'(0)=1>0$: decay of each mode does not force monotonic decay of their signed combination. The graph marks the small but real overshoot.