The Definition — Question 5

Question 5 – Solution

Strategy. A uniform tail estimate can justify passage to the boundary parameter; absolute and conditional convergence must still be distinguished.

Step 1: Evaluate for positive damping. For $s>0$, the double integral of the absolute integrand is at most $\int_0^\infty e^{-st}\,dt=1/s$, so interchange is justified. Integration by parts gives $\int_0^\infty e^{-st}\cos(ut)\,dt=s/(s^2+u^2)$. Therefore $$F(s)=\int_0^1\frac{s}{s^2+u^2}\,du =\boxed{\arctan(1/s),\qquad s>0.}$$

Step 2: Control the tail uniformly. Set $w(t)=e^{-st}/t$ for $t\ge R$. For every $s\ge 0$, $w$ decreases to zero and $w'\le 0$. Integration by parts on $[R,L]$ gives $$\int_R^Lw(t)\sin t\,dt=[-w(t)\cos t]_R^L+\int_R^Lw'(t)\cos t\,dt.$$ The integral of $|w'|$ is $w(R)-w(L)$, so the right side has magnitude at most $2w(R)\le 2/R$. Applying the same bound with any larger lower endpoint proves the Cauchy criterion as $L\to\infty$, including at $s=0$. The bound persists for the improper tail.

Step 3: Pass to the boundary parameter. On each fixed $[0,R]$, $e^{-st}f(t)\to f(t)$ uniformly as $s\to 0^+$, since $f$ is continuous and bounded there. The tails at both $s$ and zero have magnitude at most $2/R$. Thus $$|F(s)-F(0)|\le\left|\int_0^R(e^{-st}-1)f(t)\,dt\right|+4/R.$$ First choose large $R$, then let $s\to 0^+$. This proves continuity at the boundary and hence $\boxed{F(0)=\pi/2}$.

Step 4: Test absolute convergence. On $[n\pi+\pi/6,n\pi+5\pi/6]$, $|\sin t|\ge 1/2$ and $t\le(n+1)\pi$. Each such interval contributes at least $1/[3(n+1)]$ to $\int|\sin t|/t\,dt$. The harmonic series diverges, so convergence at zero is conditional. For $s>0$, $|\sin t/t|\le 1$ gives absolute convergence by comparison with $e^{-st}$. The boundary value here exists as an ordinary improper integral, not merely as a limit of damped values.