Convolution Integrals — Question 1

Question 1 – Solution

Strategy. Convolution integrates over a moving interval; pointwise multiplication does not.

Step 1: Integrate and reverse the factors. For $t\ge 0$, $$c(t)=e^{-t}\int_0^t e^{-u}\,du =\boxed{e^{-t}-e^{-2t}}.$$ The substitution $v=t-u$ changes $\int_0^t f(t-u)g(u)\,du$ to $\int_0^t g(t-v)f(v)\,dv$, proving commutativity directly.

Step 2: Test the student’s claim. Pointwise multiplication gives $p(t)=e^{-3t}$, so $p(0)=1$ whereas $c(0)=0$. Their transforms are $$P(s)=\frac 1{s+3},\qquad C(s)=\frac 1{s+1}-\frac 1{s+2}=\frac 1{(s+1)(s+2)}.$$ The product $F(s)G(s)$ equals $C(s)$, not $P(s)$. For example, at $s=0$ they have values $1/2$ and $1/3$ respectively.

Step 3: Derive the product formula. In the triangular region $0\le u\le t<\infty$, put $v=t-u$. Its image is the quadrant $u,v\ge 0$, and its Jacobian has absolute value $1$. For real $s>-1$, the resulting integrand has a finite absolute integral: $$\begin{aligned} \mathcal L\{f*g\}(s) &=\int_0^\infty\int_0^t e^{-st}e^{-(t-u)}e^{-2u}\,du\,dt\\ &=\int_0^\infty\int_0^\infty e^{-(s+1)v}e^{-(s+2)u}\,du\,dv =F(s)G(s). \end{aligned}$$ Thus the interchange is justified by absolute integrability; it is not a formal rule for pointwise products.

Step 4: Locate the peak and inspect the tail. Since $c'=e^{-t}(2e^{-t}-1)$, the derivative changes from positive to negative only at $t=\ln 2$. Consequently $$\boxed{\max_{t\ge 0}c(t)=\tfrac 14,\qquad \int_0^\infty c(t)\,dt=1-\tfrac 12=\tfrac 12.}$$ The positive tail is asymptotic to $e^{-t}$, giving exact real domain $\boxed{s>-1}$; at $s=-1$ the transformed integrand tends to $1$. At zero the integration interval has length zero. Finite endpoint values cannot produce a nonzero integral over that interval.