Laplace Transforms — Question 5

Question 5 – Solution

Strategy. Differentiation in time creates boundary terms. Keep them until the initial data and exponential decay have been used.

Step 1: Retain the boundaries. Integration by parts on $[0,R]$ gives $$\int_0^R e^{-st}f'(t)\,dt=e^{-sR}f(R)-f(0)+s\int_0^R e^{-st}f(t)\,dt.$$ Repeating for $f''$ and then taking $R\to\infty$ yields $$\mathcal L\{f'\}=sF-f(0),\qquad \mathcal L\{f''\}=s^2F-sf(0)-f'(0),$$ provided the upper boundary terms vanish. Here they do for every $s>-1$, since $f,f'$ are polynomials times $e^{-t}$.

Step 2: Compute the functions and transforms. Using elementary exponential integrals, $$F(s)=\frac 1{s+1}+\frac 1{(s+1)^2}=\frac{s+2}{(s+1)^2}.$$ Direct differentiation gives $f'=-te^{-t}$ and $f''=(t-1)e^{-t}$, with $f(0)=1$ and $f'(0)=0$. Hence $$\boxed{\mathcal L\{f'\}=-\frac 1{(s+1)^2},\qquad \mathcal L\{f''\}=-\frac{s}{(s+1)^2},\quad s>-1.}$$ These intervals are exact: at or below $-1$, the resulting polynomial or growing exponential tails do not have convergent improper integrals.

Step 3: Repair and check the claim. The omitted initial contribution is $-sf(0)-f'(0)=-s$. Thus the corrected expression is $s^2F-s=-s/(s+1)^2$. Independently, transforming $(t-1)e^{-t}$ gives $$\frac 1{(s+1)^2}-\frac 1{s+1}=-\frac{s}{(s+1)^2}.$$ This agreement verifies both the boundary terms and the algebra.

Step 4: Check zero-parameter integrals. At zero, the first-derivative transform equals $-1$, consistent with $\int_0^\infty f'=f(\infty)-f(0)=0-1$. The second-derivative transform is zero, consistent with $\int_0^\infty f''=f'(\infty)-f'(0)=0$. The function $f''$ is negative before $t=1$ and positive afterward; its two signed contributions cancel. Zero integral is not the same as a zero function.