Laplace Transforms — Question 8

Question 8 – Solution

Strategy. Positivity of the transform is only one consequence of a nonnegative signal; parameter derivatives supply stronger necessary tests.

Step 1: Derive the necessary signs. If $0\le f\le M$, then the kernels obtained by differentiating once or twice are dominated locally in $s>0$ by constant multiples of $te^{-s_0t/2}$ and $t^2e^{-s_0t/2}$. Differentiation under the integral is therefore justified, and $$F(s)\ge 0,\qquad F'(s)=-\int_0^\infty te^{-st}f(t)\,dt\le 0,$$ $$F''(s)=\int_0^\infty t^2e^{-st}f(t)\,dt\ge 0.$$ If $f$ is not identically zero, continuity gives a positive interval contribution and hence $F(s)>0$ for every $s>0$.

Step 2: Test the positive candidate. Although $R(s)>0$, $$R'(s)=\frac{1-s}{(s+1)^3}>0\qquad(0<s<1).$$ This contradicts the required nonpositive derivative. Thus no continuous bounded nonnegative signal can have this transform.

Step 3: Find a valid signed signal. Rewrite $R=1/(s+1)-1/(s+1)^2$. The basic exponential integrals give $$\boxed{g(t)=e^{-t}(1-t),\qquad \mathcal L\{g\}(s)=R(s),\quad s>-1.}$$ The function is positive for $0\le t<1$, zero at one, and negative for $t>1$. It is continuous and bounded. At or below $s=-1$, its eventually negative polynomial or growing tail has no finite improper integral, so the domain is exact.

Step 4: Explain the weighted cancellation. At $s=0$, the integral exists and equals $R(0)=0$, also verified by $\int e^{-t}=1$ and $\int te^{-t}=1$. For positive $s$, the extra weight suppresses the late negative portion more strongly, producing a positive transform. Consequently $F>0$ alone is insufficient to infer $f\ge 0$. The two plots show a sign-changing signal and its everywhere-positive transform on $s>0$.