Inverse Laplace Transforms — Question 5

Question 5 – Solution

Strategy. A repeated quadratic can be handled with parameter differentiation and linear combinations, without assuming a product rule for inverses.

Step 1: Produce the repeated denominator. Differentiating the cosine pair gives $$\mathcal L\{t\cos 2t\}=-\frac d{ds}\frac{s}{s^2+4} =\frac{s^2-4}{(s^2+4)^2}.$$ Polynomial factors times the damping kernel are integrable for $s>0$, justifying the differentiation.

Step 2: Cancel the unwanted numerator. The transform of $A\sin 2t+B t\cos 2t$ has numerator $2A(s^2+4)+B(s^2-4)$. Requiring it to equal one gives $2A+B=0$, $8A-4B=1$, hence $$\boxed{f(t)=\frac{\sin 2t-2t\cos 2t}{16}.}$$ Substitution gives $(s^2+4)/8-(s^2-4)/8=1$, verifying the numerator exactly. Inverting a product by multiplying inverses would not give this result.

Step 3: Exhibit growth, not just an envelope. The bound $|f(t)|\le(1+2t)/16$ is only an upper bound. Actual unboundedness follows from $$\boxed{f(n\pi)=-n\pi/8\longrightarrow-\infty.}$$ The factor $t$ produced by the repeated poles causes increasing oscillation size. The graph displays several oscillations at their actual scale.

Step 4: Check onset and convergence. Taylor expansion gives $\sin 2t-2t\cos 2t=(8/3)t^3+O(t^5)$, so $f(t)=t^3/6+O(t^5)$. Since $\mathcal L\{t^3/6\}=s^{-4}$, the initial scale agrees with $F(s)\sim s^{-4}$. Absolute convergence holds for $s>0$ by the linear envelope. At $s=0$, direct integration gives $$\int_0^R f(t)\,dt=\frac{1-\cos 2R-R\sin 2R}{16},$$ which has no limit. For $s<0$, intervals near successive positive peaks of $-\cos 2t$ have eventually positive integrand of order $t e^{|s|t}$; their integrals fail the Cauchy criterion. Thus $\boxed{s>0}$ is exact.