Laplace Transforms — Question 3

Question 3 – Solution

Strategy. Time compression changes both the exponential weight and the integration measure; amplitude compensation affects area separately.

Step 1: Derive the scaling rule. For $a>0$, substituting $u=at$ gives $$\int_0^\infty e^{-st}g(at)\,dt =\frac 1a\int_0^\infty e^{-(s/a)u}g(u)\,du =\boxed{\frac 1aG(s/a)}.$$ The effective parameter is $s/a$, so $s/a>-1$ here. Both the reciprocal parameter scaling and the factor $1/a$ are essential.

Step 2: Apply the rule to both signals. For $a=3$, $$\boxed{F(s)=\frac 6{(s+3)^2+36},\qquad H(s)=\frac{18}{(s+3)^2+36},\qquad s>-3.}$$ These are the transforms of $e^{-3t}\sin 6t$ and three times that signal. At the boundary the undamped sine integral fails to converge; below it the exponentially growing sine fails the Cauchy criterion on positive half-wave subintervals. Thus $s>-3$ is exact. $G(3s)$ would stretch the parameter in the opposite direction and omit the measure factor.

Step 3: Compare signed areas. Since zero lies in the convergence intervals, substitution of $s=0$ gives $$\int_0^\infty g=\frac 25,\qquad \int_0^\infty f=\frac 2{15},\qquad \int_0^\infty h=\frac 25.$$ Compression alone reduces signed area by three; multiplying the amplitude by three restores it. These are signed integrals, not integrals of absolute value.

Step 4: Track first moments. Exponential decay justifies differentiation near $s=0$, so $G'(s)=-\int_0^\infty te^{-st}g(t)\,dt$. As $G'(0)=-4/25$, the original first moment is $4/25$. Differentiating $F(s)=G(s/3)/3$ and $H(s)=G(s/3)$ yields $$\boxed{\int_0^\infty t f(t)\,dt=\frac 4{225},\qquad \int_0^\infty t h(t)\,dt=\frac 4{75}.}$$ For $f$, one factor $1/3$ comes from $dt$ and another from $t=u/3$. The amplitude factor in $h$ cancels only one of them.