Inverse Laplace Transforms — Question 9

Question 9 – Solution

Strategy. Check whether an ordinary inverse can exist before matching a rational remainder to a table. Keep the continuity hypothesis in any uniqueness claim.

Step 1: Divide before attempting inversion. Because $(s+1)^2=s^2+2s+1$, $$\boxed{G(s)=1+\frac 1{(s+1)^2}.}$$ The nonzero constant part cannot be silently discarded. In particular $G(s)\to 1$, rather than zero, at large positive $s$.

Step 2: Apply a necessary existence test. For an allowed $g$ and $s>c$, absolute convergence and the global bound yield $$|\mathcal L\{g\}(s)|\le\int_0^\infty e^{-st}|g(t)|\,dt \le M\int_0^\infty e^{-(s-c)t}\,dt=\frac M{s-c}\longrightarrow 0.$$ A continuous function of exponential order admits this global bound by enlarging $M$ on an initial compact interval. Since $G$ has limit one, $\boxed{G\text{ has no inverse in the stated class.}}$ This conclusion concerns ordinary continuous functions under the explicit hypotheses.

Step 3: Invert only the actual remainder. The strictly proper part has inverse $\boxed{h(t)=te^{-t}}$. Direct integration by parts gives $$\int_0^\infty te^{-(s+1)t}\,dt=\frac 1{(s+1)^2},\qquad s>-1.$$ The integral diverges at or below $-1$. Its transform is $G-1$, not $G$. The missing constant does not arise from changing $h(0)$: changing one finite point value changes no ordinary integral.

Step 4: State uniqueness with its hypothesis. Here $h(1)=e^{-1}\ne 7$. Changing that one value leaves every finite-interval integral, and hence every convergent improper transform, unchanged. Thus $k$ and $h$ have identical transforms on $s>-1$. However, $k$ is discontinuous at one, so it is not in the continuous class specified in the question. There is no contradiction: uniqueness within that class compares two continuous functions. The example separates equality of transforms from unrestricted pointwise equality. No generalized transform is needed to reach any of these conclusions.