On convolution theorem of Laplace transform: Schiff

psie · Oct 26, 2023

Here follows the theorem and proof:

Theorem (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) \nonumber \\
&=\int_0^\infty \left(\int_0^\infty e^{-s(\tau+u)}f(\tau)g(u)du\right)d\tau \nonumber .
\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 \nonumber \\
&=\int_0^\infty \left(\int_0^t e^{-st}f(\tau)g(t-\tau)d\tau\right)dt \nonumber \\
&=\int_0^\infty e^{-st} \left(\int_0^t f(\tau)g(t-\tau)d\tau\right)dt \nonumber \\ &=\mathcal{L}[(f*g)(t)]. \nonumber
\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. I do not understand the following part "...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 (I assume this is the statement he is making)? If yes, how come?

The definition of absolute convergence given in the book is that ##\int_0^\infty |e^{-st}f(t)|dt## converges for a given real or complex parameter ##s##.

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?

martinbn · Oct 26, 2023

psie said:

1. I do not understand the following part "...and hence, in view of the preceding calculation, ##\int_0^\infty \int_0^\infty |e^{-st}f(\tau)g(t-\tau)|dtd\tau## converges".

You are given that ##|f(\tau)|\le K_1 e^{\alpha\tau}## and ##|g(t-\tau)|\le K_2 e^{\alpha(t-\tau)}##, hence

##|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} ##

Which is integrable for ##Re(s) > \alpha##.

psie · Oct 26, 2023

martinbn said:

Which is integrable for ##Re(s) > \alpha##.

Hmm, I get then that $$\int_0^\infty \int_0^\infty |e^{-st}f(\tau)g(t-\tau)|dtd\tau\leq \int_0^\infty \int_0^\infty K_1K_2e^{(\alpha-Re(s))t}dtd\tau,$$ under the assumption that improper integrals depending on a parameter respect monotonicity. Are you sure the above converges?

By the way, in his book the definition of exponential order ##\alpha## means that it's supposed to hold eventually, so we have ##|f(\tau)|\le K_1 e^{\alpha\tau}## for ##\tau\geq\tau_0## where ##\tau_0\geq 0##.

pasmith · Oct 26, 2023

psie said:

Hmm, I get then that $$\int_0^\infty \int_0^\infty |e^{-st}f(\tau)g(t-\tau)|dtd\tau\leq \int_0^\infty \int_0^\infty K_1K_2e^{(\alpha-Re(s))t}dtd\tau,$$ under the assumption that improper integrals depending on a parameter respect monotonicity. Are you sure the above converges?

Recall that [itex]g(u) = 0 [/itex] for [itex]u < 0[/itex], so the lower limit of the inner integral is really [itex]\tau[/itex]. Also note that convergence requires [itex]\alpha - \operatorname{Re}(s) < 0[/itex], so we can replace it by [itex]-|\alpha - \operatorname{Re}(s)|[/itex]. Then [tex]\begin{split}
\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-\operatorname{Re}(s)|t}\,dt\,d\tau \\
&= K_1K_2\frac1{|\alpha - \operatorname{Re}(s)|}\int_0^\infty e^{-|\alpha-\operatorname{Re}(s)|\tau}\,d\tau.\end{split}[/tex]

psie · Oct 26, 2023

Thank you.

I think I can answer my second question. 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.

On convolution theorem of Laplace transform: Schiff

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