Table Of Laplace Transforms — Question 7

Question 7 – Solution

Strategy. Integrating a familiar exponential row over its rate parameter generates a nonrational row.

Step 1: Remove the time singularity. Direct integration in $a$ gives $$\boxed{g(t)=\int_1^3 e^{-at}\,da,\qquad g(0)=2.}$$ This formula also proves continuity at zero without dividing by $t$.

Step 2: Integrate the table row. For real $s>-1$, the nonnegative double integral is finite, and interchange gives $$G(s)=\int_1^3\int_0^\infty e^{-(s+a)t}\,dt\,da =\int_1^3\frac{da}{s+a} =\boxed{\ln\frac{s+3}{s+1}}.$$ Both logarithm arguments are positive in this domain. Finiteness follows also from $s+a\ge s+1>0$ on the finite parameter interval.

Step 3: Establish shape and area. The integrand is positive, so $g(t)>0$. Parameter integration on a bounded interval permits differentiation at every $t\ge 0$: $$g'(t)=-\int_1^3 a e^{-at}\,da<0.$$ Thus the extension decreases strictly from $2$, approaching zero. Since $0$ lies in the transform domain, its area is $G(0)=\boxed{\ln 3}$.

Step 4: Check the boundary from the original integral. At $s=-1$, the transformed integrand is $(1-e^{-2t})/t$ for $t>0$. For $t\ge 1$ it is at least $(1-e^{-2})/t$, whose integral diverges. For $s<-1$ the positive integrand is even larger on the tail. Therefore $\boxed{s>-1}$ is the exact real domain. For example, the logarithm of $(s+3)/(s+1)$ is real again when $s<-3$, but that algebraic expression cannot represent the divergent defining integral. The table entry must travel with its domain and the continuous value at zero.