Step Functions — Question 7

Question 7 – Solution

Strategy. Integration by parts across smooth pieces leaves internal boundary terms. A jump is different from a continuous corner.

Step 1: Compute both ordinary integrals. The local shape of $f$ is $1+v$, and the local shape of $h$ is one. Thus $$F(s)=e^{-2s}(1/s+1/s^2),\qquad H(s)=e^{-2s}/s,\qquad s>0.$$ Since $f(0)=0$, the student’s expression is $sF=e^{-2s}(1+1/s)$, larger than $H$ by $\boxed{e^{-2s}}$. Both actual integrals have exact domain $s>0$ because their tails are a positive linear function and a positive constant, respectively.

Step 2: Retain the internal boundary. Only the second piece contributes. For $R>2$, integration by parts gives $$\int_2^R e^{-st}f'(t)\,dt =e^{-sR}f(R)-e^{-2s}f(2+)+s\int_2^R e^{-st}f(t)\,dt.$$ Here $f(2+)=1$. The upper term vanishes for $s>0$, leaving $H=sF-e^{-2s}$. The omitted term records the jump from zero to one, not a value assigned to $h$ at the single point two.

Step 3: Sum the piecewise boundary terms. At an internal time $c_j$, the left piece contributes $e^{-sc_j}f(c_j-)$ and the right piece contributes $-e^{-sc_j}f(c_j+)$. Their sum is $-J_je^{-sc_j}$. Consequently $$\boxed{\mathcal L\{f'_{\mathrm{pw}}\}(s)=sF(s)-f(0+)-\sum_jJ_je^{-sc_j}.}$$ Use finite one-sided limits, convergence of the piecewise function and derivative integrals, and $e^{-sR}f(R)\to 0$. Exponential-order bounds on the pieces and their derivatives suffice for large $s$. If $f$ is right-continuous at zero, $f(0+)=f(0)$. The formula concerns ordinary piecewise derivatives only.

Step 4: Check a continuous corner. For $k$, both one-sided values at two are zero, so its jump is zero. We have $K=e^{-2s}/s^2$ and its ordinary piecewise derivative is $u_2$ except possibly at two. Thus $$\mathcal L\{k'_{\mathrm{pw}}\}=e^{-2s}/s=sK-k(0).$$ A corner changes slopes but introduces no internal value jump. Splitting the integral proves the usual rule here without pretending the derivative exists at the corner.