Inverse Laplace Transforms — Question 8

Question 8 – Solution

Strategy. Invert first, then turn global sign and monotonicity requirements into inequalities for a linear time factor.

Step 1: Separate the numerator at its pole. Since $as+1=a(s+1)+(1-a)$, $$\boxed{f_a(t)=e^{-t}[a+(1-a)t].}$$ Forward transformation gives $a/(s+1)+(1-a)/(s+1)^2=F_a$. The exact domain is $s>-1$ for every $a$: the nonzero linear or constant polynomial tail has eventual fixed sign and diverges at or below $-1$. Thus $s=0$ is allowed, and $$\int_0^\infty f_a(t)\,dt=F_a(0)=1,\qquad f_a(0)=a.$$ Unit signed integral alone does not establish nonnegativity.

Step 2: Classify nonnegative inverses. Because $e^{-t}>0$, the sign is that of $a+(1-a)t$. Nonnegativity at zero requires $a\ge 0$. If $a>1$, the negative slope makes the factor negative for sufficiently large $t$. Conversely, $0\le a\le 1$ makes both its intercept and slope nonnegative. Therefore the exact admissible set is $\boxed{0\le a\le 1}$.

Step 3: Distinguish monotone and peaked designs. For admissible $a$, $$f_a'(t)=e^{-t}[1-2a-(1-a)t].$$ The bracket is nonincreasing in $t$, so it is nonpositive everywhere exactly when its initial value is nonpositive. Hence the nonincreasing designs have $\boxed{1/2\le a\le 1}$. For $0\le a<1/2$, the derivative changes from positive to negative at $$\boxed{t_*=\frac{1-2a}{1-a},\qquad f_a(t_*)=(1-a)\exp(-\frac{1-2a}{1-a}).}$$ This also covers $a=0$, whose maximum occurs at one.

Step 4: Use the extra integral to select a design. Direct exponential moments give $$\int_0^\infty t f_a(t)\,dt=a+2(1-a)=2-a.$$ Equating this to $3/2$ yields $\boxed{a=1/2}$ and $f(t)=(1+t)e^{-t}/2$. This is nonnegative, has unit integral, and is nonincreasing: $f'(t)=-te^{-t}/2\le 0$. Its derivative vanishes only at the initial endpoint. The measurement selects one member, rather than allowing the initial value to be prescribed independently.