Laplace Transforms — Question 1

Question 1 – Solution

Strategy. Linearity applies to sums and scalar multiples, not to pointwise products. A large-parameter check can expose the distinction.

Step 1: Rewrite the function before transforming. Since $\sin^2(3t)=(1-\cos 6t)/2$, linearity gives $$F(s)=\frac 1{2s}-\frac{s}{2(s^2+36)} =\boxed{\frac{18}{s(s^2+36)},\qquad s>0.}$$

Step 2: Check convergence and the proposed rule. For $s>0$, boundedness of $f$ gives absolute convergence. For $s\le 0$, the weight is at least one and $\sin^2(3t)$ has positive integral on each period, so the nonnegative improper integral diverges. Thus the real convergence set is exactly $(0,\infty)$. The proposed expression is $$\frac 9{(s^2+9)^2},$$ which is not the correct transform. For example, at $s=3$ the correct value is $2/15$, while the proposed value is $1/36$. Squaring a time function is not transformed by squaring its transform.

Step 3: Obtain a bound from the integral. The inequality $0\le\sin^2(3t)\le 9t^2$ yields $$0<F(s)\le 9\int_0^\infty t^2e^{-st}\,dt =\boxed{18/s^3}.$$ The last integral is $2/s^3$, by two integrations by parts. Its boundary terms vanish for $s>0$.

Step 4: Recover the leading scale independently. With $u=st$, $$s^3F(s)=\int_0^\infty e^{-u}s^2\sin^2(3u/s)\,du.$$ The integrand tends to $9u^2e^{-u}$ and is bounded by that integrable function. Dominated convergence therefore gives $\boxed{s^3F(s)\to 18}$. This reflects the local behavior $\sin^2(3t)\sim 9t^2$. In contrast, multiplying the proposed expression by $s^3$ gives a limit of zero. The proposed rule even predicts the wrong leading power of $1/s$.