Inverse Laplace Transforms — Question 2

Question 2 – Solution

Strategy. A pole of order three allows all three reciprocal powers, and each power has its own factorial normalization.

Step 1: Expand around the pole. Set $p=s+1$. Then $s^2+3s+4=p^2+p+2$, so $$F(s)=\frac 1{s+1}+\frac 1{(s+1)^2}+\frac 2{(s+1)^3}.$$ Equivalently, multiplying by $(s+1)^3$ gives the polynomial identity $s^2+3s+4=(s+1)^2+(s+1)+2$.

Step 2: Apply the correctly normalized pairs. The inverse of $(s+1)^{-3}$ is $t^2e^{-t}/2!$. Thus $$\boxed{f(t)=e^{-t}(1+t+t^2).}$$ Its three forward transforms are $1/(s+1)$, $1/(s+1)^2$ and $2/(s+1)^3$, confirming the result. Forgetting $2!$ would double the quadratic term incorrectly.

Step 3: Explain the incomplete ansatz. An expression $C/(s+1)^3$ has constant numerator after multiplication by the denominator. It cannot equal the nonconstant polynomial $s^2+3s+4$ for every $s$. The proposed inverse $Ct^2e^{-t}/2$ lacks both $e^{-t}$ and $te^{-t}$; no choice of $C$ can repair that omission.

Step 4: Check initial data and convergence. Differentiation gives $$f'=e^{-t}(t-t^2),\qquad f''=e^{-t}(1-3t+t^2),$$ so $\boxed{f(0)=1,\ f'(0)=0,\ f''(0)=1}$. Independently, expansion at large positive $s$ gives $$F(s)=\frac 1s+\frac 0{s^2}+\frac 1{s^3}+O(s^{-4}).$$ These are the initial-value coefficients obtained by repeated integration by parts; this polynomial-exponential function satisfies all needed derivative bounds. Its defining integral converges absolutely for $s>-1$. At $s=-1$ the integrand is the positive polynomial $1+t+t^2$, and for smaller $s$ it grows still faster. The exact domain is $\boxed{s>-1}$.