The Definition — Question 3

Question 3

For real $s$, use $F(s)=\int_0^\infty e^{-st}f(t)\,dt$ whenever this ordinary improper integral converges.

Let $f(t)=\sin 3t$ for $t\ge 0$. Its truncated transform is $F_R(s)=\int_0^R e^{-st}\sin 3t\,dt$.

Tasks

1. Evaluate $F_R(s)$ for real $s$, retaining its upper-endpoint contribution.

2. Find the exact real convergence interval for $F(s)$ and determine whether convergence there is absolute.

3. At $s=0$, exhibit two sequences of upper endpoints giving different values of $F_R(0)$.

4. Compare $\lim_{s\to 0^+}F(s)$ with the existence of $F(0)$. Why does a finite parameter limit not justify interchanging the two limits $R\to\infty$ and $s\to 0^+$?

Question 3 – Solution

Strategy. Damping by $e^{-st}$ can create convergence that disappears at the boundary parameter.

Step 1: Keep the endpoint term. Two integrations by parts, or differentiation of the resulting antiderivative, give $$\boxed{F_R(s)=\frac{3-e^{-sR}[s\sin 3R+3\cos 3R]}{s^2+9}.}$$ There is no real zero of the denominator, but that alone says nothing about the limit as $R\to\infty$.

Step 2: Classify convergence. If $s>0$, the endpoint term vanishes and $$\boxed{F(s)=\frac 3{s^2+9},\qquad s>0.}$$ Absolute convergence follows from $|e^{-st}\sin 3t|\le e^{-st}$. If $s<0$, take $R_n=2n\pi/3$. Then $F_{R_n}(s)=3(1-e^{|s|R_n})/(s^2+9)\to-\infty$, so the improper integral cannot converge.

Step 3: Test the boundary directly. At $s=0$, $F_R(0)=(1-\cos 3R)/3$. For $$R_n=2n\pi/3,\quad F_{R_n}(0)=0;\qquad Q_n=(2n+1)\pi/3,\quad F_{Q_n}(0)=2/3.$$ The two sequences tend to infinity but have different integral values. Thus $F(0)$ does not exist, and the full real convergence interval is $(0,\infty)$.

Step 4: Separate the two limiting procedures. The damped transforms satisfy $\lim_{s\to 0^+}F(s)=1/3$. In contrast, setting $s=0$ first produces an oscillating truncated integral with no limit. For fixed $R$, the limit in $s$ exists, but convergence in $R$ is not uniform as damping disappears. The value $1/3$ is a limit of damped integrals, not the value of the undamped improper integral. The graph compares accumulated integrals, not the original sine signal.