Laplace Transforms — Question 2

Question 2 – Solution

Strategy. Account for the exponential shift first, then differentiate the correct parameter expression and retain its convergence domain.

Step 1: Justify parameter differentiation. The factor $e^{-2t}$ changes the effective parameter to $p=s+2$. For any $p_0>0$, a sufficiently small neighborhood has $p\ge p_0/2$. Differentiating the kernel twice produces $t^2e^{-pt}\cos 3t$, dominated by $t^2e^{-p_0t/2}$, an integrable function. The first derivative is justified similarly. Thus, for $s>-2$, the desired transform is $C''(s+2)$.

Step 2: Differentiate and simplify. Direct calculation gives $$C'(p)=\frac{9-p^2}{(p^2+9)^2},\qquad C''(p)=\frac{2p(p^2-27)}{(p^2+9)^3}.$$ Therefore $$\boxed{F(s)=\frac{2(s+2)[(s+2)^2-27]}{[(s+2)^2+9]^3},\qquad s>-2.}$$ The exponential times a polynomial gives absolute convergence on this interval.

Step 3: Interpret its zero and sign. For $p>0$, the denominator and $2p$ are positive. Hence the only zero in the convergence interval is $\boxed{s=3\sqrt 3-2}$. The transform is negative below that value and positive above it. At the zero, weighted positive and negative portions of $t^2e^{-2t}\cos 3t$ cancel; the function itself is certainly not identically zero.

Step 4: Test the boundary rather than the rational expression. At $s=-2$, the integral becomes $\int_0^R t^2\cos 3t\,dt$. Integration by parts gives $$\frac{R^2\sin 3R}{3}+\frac{2R\cos 3R}{9}-\frac{2\sin 3R}{27}.$$ At $R_n=2n\pi/3$, this equals $2R_n/9\to\infty$. For $p=s+2<0$, consider $[R_n,R_n+\pi/9]$, where $\cos 3t\ge 1/2$. The integral on that interval is at least $(\pi/18)R_n^2e^{-pR_n}$ and does not tend to zero, violating the Cauchy criterion. Thus no real $s\le-2$ works; the algebraic boundary value is not a transform value.