Convolution theorem of Laplace transform; Schiff

Question

I'm reading Schiff's The Laplace Transform and I have some questions about the convolution theorem he proves on page 92 to 93.

Theorem and proof

Theorem 2.39 (Convolution Theorem). If $f$ and $g$ are piecewise continuous on $[0,\infty)$ and of exponential order $\alpha$, then $$\mathcal{L}\left[(f*g)(t)\right]=\mathcal{L}\big(f(t)\big)\cdot\mathcal{L}\big(g(t)\big)\quad \Big(Re(s)>\alpha\Big).$$

Proof. Let us start with the product \begin{align} \mathcal{L}\big(f(t)\big)\cdot\mathcal{L}\big(g(t)\big)&=\left(\int_0^\infty e^{-s\tau}f(\tau)d\tau\right)\left(\int_0^\infty e^{-su}g(u)du\right) \\ &=\int_0^\infty \left(\int_0^\infty e^{-s(\tau+u)}f(\tau)g(u)du\right)d\tau .\end{align} Substituting $t=\tau+u$, and noting that $\tau$ is fixed in the interior integral, so that $du=dt$, we have $$\mathcal{L}\big(f(t)\big)\cdot\mathcal{L}\big(g(t)\big)=\int_0^\infty \left(\int_\tau^\infty e^{-st}f(\tau)g(t-\tau)dt\right)d\tau .\tag 1$$ If we define $g(t)=0$ for $t<0$, then $g(t-\tau)=0$ for $t<\tau$ and we can write $(1)$ as $$\mathcal{L}\big(f(t)\big)\cdot\mathcal{L}\big(g(t)\big)=\int_0^\infty \int_0^\infty e^{-st}f(\tau)g(t-\tau)dtd\tau .$$ Due to the hypotheses on $f$ and $g$, the Laplace integrals of $f$ and $g$ converge absolutely and hence, in view of the preceding calculation, $$\int_0^\infty \int_0^\infty |e^{-st}f(\tau)g(t-\tau)|dtd\tau$$ converges. This fact allows us to reverse the order of integration,* so that \begin{align} \mathcal{L}\big(f(t)\big)\cdot\mathcal{L}\big(g(t)\big)&=\int_0^\infty \int_0^\infty e^{-st}f(\tau)g(t-\tau)d\tau dt \\ &=\int_0^\infty \left(\int_0^t e^{-st}f(\tau)g(t-\tau)d\tau\right)dt \\ &=\int_0^\infty e^{-st} \left(\int_0^t f(\tau)g(t-\tau)d\tau\right)dt \\ &=\mathcal{L}[(f*g)(t)].\end{align} *Let $$a_{mn}=\int_n^{n+1}\int_m^{m+1} |h(t,\tau)|dtd\tau,\quad b_{mn}=\int_n^{n+1}\int_m^{m+1} h(t,\tau) dtd\tau,$$ so that $|b_{mn}|\leq a_{mn}$. If $$\int_0^\infty\int_0^\infty |h(t,\tau)|dtd\tau <\infty,$$ then $\sum_{n=0}^\infty\sum_{m=0}^\infty a_{mn}<\infty$, implying $\sum_{n=0}^\infty\sum_{m=0}^\infty |b_{mn}|<\infty$. Hence, by a standard result on double series, the order of summation can be interchanged $$\sum_{n=0}^\infty\sum_{m=0}^\infty b_{mn}=\sum_{m=0}^\infty\sum_{n=0}^\infty b_{mn},$$ i.e., $$\int_0^\infty\int_0^\infty h(t,\tau) dtd\tau =\int_0^\infty\int_0^\infty h(t,\tau) d\tau dt.$$

Questions

1. ...and hence, in view of the preceding calculation, $$\int_0^\infty \int_0^\infty |e^{-st}f(\tau)g(t-\tau)|dtd\tau$$ converges.

We know that $\mathcal{L}\big(f(t)\big)$ and $\mathcal{L}\big(g(t)\big)$ converge absolutely. So does their product converge absolutely? If yes, I'd be grateful for a brief justification of this.

2. Regarding the footnote, if $I=\int_0^\infty\int_0^\infty f(x,y)dxdy$, can we then always write $I$ as a double series, i.e. $I=\sum_{n=0}^\infty\sum_{m=0}^\infty c_{mn}$ where $c_{mn}=\int_n^{n+1}\int_m^{m+1} f(x,y)dxdy$ (I assume this is what Schiff is doing)? If not, what justifies that we can in this case and how?

Answer 1

1. It is given that $|f(\tau)|\le K_1 e^{\alpha\tau}$ and $|g(t-\tau)|\le K_2 e^{\alpha(t-\tau)}$. So \begin{align}|e^{-st}f(\tau)g(t-\tau)|&\le K_1K_2 e^{-Re(s)t} e^{\alpha\tau} e^{\alpha(t-\tau)}\\ &=K_1K_2 e^{(\alpha-Re(s))t} .\end{align} Since $\mathcal{L}\big(f(t)\big)$ and $\mathcal{L}\big(g(t)\big)$ exist for $Re(s)>\alpha$, we have $\alpha-Re(s)<0$. Also note that, since $g(t-\tau)=0$ for $t<\tau$, we have \begin{align} \int_0^\infty \int_0^\infty |e^{-st}f(\tau)g(t-\tau)|\,dt\,d\tau &\leq \int_0^\infty \int_\tau^\infty K_1K_2e^{(\alpha-Re(s))t}\,dt\,d\tau \\ &= \frac{K_1K_2}{(\alpha - Re(s))}\int_0^\infty e^{(\alpha-Re(s))\tau}\,d\tau \\ &=\frac{K_1K_2}{(\alpha-Re(s))^2}.\end{align}

2. Let $\int_0^\infty f(x) dx=\lim_{t\to\infty}\int_0^t f(x)dx$ be an improper Riemann integral. Then, if we let $t=n\in\mathbb N$, we get $\int_0^n f(x)dx = \sum_{i=0}^{n-1} \int_i^{i+1} f(x)dx$, and in the limit $n\to\infty$, we get $$\int_0^\infty f(x)dx=\sum_{i=0}^\infty \int_i^{i+1} f(x)dx.$$ So if the integral converges, then the sum converges. However, consider $f(x)=\sin(2\pi x)$. Then the integral over the interval $[i,i+1]$ is $0$, so the sum converges, but the improper integral does not. Hence the converse does not hold.