Laplace Transforms — Question 6

Question 6 – Solution

Strategy. Cancellation removes the origin singularity. Preserve that cancellation by integrating over an exponential parameter.

Step 1: Remove the apparent singularity. Expansion at zero, or l’Hopital’s rule, gives $\boxed{h(0)=3}$. Direct integration in $u$ gives $$h(t)=\int_1^4e^{-ut}\,du,$$ including at $t=0$ by continuity. In particular, $h$ is positive and bounded above by three.

Step 2: Derive the logarithmic transform. For $s>-1$, $s+u>0$ throughout $[1,4]$. The absolute double integral is finite, as seen by integrating first in $t$: $$H(s)=\int_1^4\int_0^\infty e^{-(s+u)t}\,dt\,du =\int_1^4\frac{du}{s+u} =\boxed{\ln\frac{s+4}{s+1},\qquad s>-1.}$$ The finite value justifies interchange; both logarithm arguments are positive on this domain.

Step 3: Check the domain and invalid separation. At $s=-1$, the integrand is $(1-e^{-3t})/t$, whose tail is at least $1/(2t)$ for sufficiently large $t$. It diverges. For smaller $s$, a larger positive exponential weight makes divergence persist. Thus $s>-1$ is the full real convergence interval. Each separate term $e^{-t}/t$ or $e^{-4t}/t$ has a nonintegrable origin singularity for every $s$. Their divergent transforms cannot be subtracted; it is the difference of the functions that is transformable.

Step 4: Interpret derivative and initial scale. Differentiating the logarithm gives $$\boxed{H'(s)=\frac 1{s+4}-\frac 1{s+1} =-\frac 3{(s+1)(s+4)}<0.}$$ This agrees with $H'(s)=-\int_0^\infty te^{-st}h(t)\,dt$, since $h>0$. Also $H(s)=\ln[1+3/(s+1)]$, so $\ln(1+x)/x\to 1$ implies $\boxed{sH(s)\to 3=h(0)}$. The graph displays the continuous signal; its apparent division-by-zero point has been correctly filled in.