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The convolution theorem for Laplace transform states that $$\mathcal{L}\{f*g\}=\mathcal{L}\{f\}\cdot\mathcal{L}\{g\}.$$ The standard proof uses Fubini-like argument of switching the order of integration: $$\int_0^\infty d\tau \int_{\tau}^\infty e^{-st}f(t-\tau)g(\tau)\,dt=\int_0^\infty dt\int_0^te^{-st}f(t-\tau)g(\tau)\,d\tau$$ Fubini's theorem says that one can switch the order of integration. But what we have in the iterated integrals are not integrals, but limits of integrals (i.e., improper integrals). Are we justified to treat them like "proper" integrals and switch their order?

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    $\begingroup$ Fubini is generally proven using Lebesgue integration which just needs the double integral to be absolutely convergent, with no limiting process at infinity being involved. $\endgroup$
    – Ian
    Nov 5, 2019 at 10:45

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\begin{align*}\mathcal{L}\{f\}({s})\cdot\mathcal{L}\{g\}({s})&=\left(\int_0^\infty e^{-{s}t}{f}(t)\,dt\right)\left(\int_0^\infty e^{-{s}u}{g}(u)\,du\right)\\ &=\lim_{{{L}}\to\infty}\left(\int_0^{{L}}e^{-{s}t}{f}(t)\,dt\right)\left(\int_0^{{L}}e^{-{s}u}{g}(u)\,du\right) \end{align*} By Fubini's theorem, \begin{align*}\left(\int_0^{{L}}e^{-{s}t}{f}(t)\,dt\right)\left(\int_0^{{L}}e^{-{s}u}{g}(u)\,du\right)&=\int_0^{{L}}\int_0^{{L}}e^{-{s}(t+u)}{f}(t){g}(u)\,dt\,du\\ &=\iint_{R_{{L}}}e^{-{s}(t+u)}{f}(t){g}(u)\,dt\,du, \end{align*} where $R_{{L}}$ is the square region $$0\leq t\leq {{L}},\qquad 0\leq u\leq {{L}}.$$ Let $T_{{L}}$ be the triangular region $$0\leq t\leq {{L}},\qquad 0\leq u\leq {{L}},\qquad t+u\leq {{L}}.$$ Provided that ${f}$ and ${g}$ are bounded by exponential functions, $$\lim_{{{L}}\to\infty}\iint_{R_{{L}}}e^{-{s}(t+u)}{f}(t){g}(u)\,dt\,du=\lim_{{{L}}\to\infty}\iint_{T_{{L}}}e^{-{s}(t+u)}{f}(t){g}(u)\,dt\,du.$$ Now, the function $$\varphi(v,u)=(v-u,u)$$ maps $D_{{L}}$ bijectively onto $T_{{L}}$, where $D_{{L}}$ is the triangular region $$0\leq v\leq {{L}},\qquad 0\leq u\leq {{L}},\qquad v\geq u.$$ enter image description here The component functions of $\varphi$ are $$t(v,u)=v-u\qquad\mbox{and}\qquad u(v,u)=u,$$ so the Jacobian of $\varphi$ is $$J\varphi=\det\begin{bmatrix}t_v&t_u\\u_v&u_u\end{bmatrix}=\det\begin{bmatrix}1&-1\\0&1\end{bmatrix}=1.$$ Hence, $$\iint_{T_{{L}}}e^{-{s}(t+u)}{f}(t){g}(u)\,dt\,du=\iint_{D_{{L}}}e^{-{s}v}{f}(v-u){g}(u)\,dv\,du.$$ By Fubini's theorem, \begin{align*}&\iint_{D_{{L}}}e^{-{s}v}{f}(v-u){g}(u)\,dv\,du=\int_0^{{L}}\int_0^ve^{-{s}v}{f}(v-u){g}(u)\,du\,dv\\ \\ &=\int_0^{{L}}e^{-{s}v}\int_0^v{f}(v-u){g}(u)\,du\,dv=\int_0^{{L}}e^{-{s}v}({f}\ast{g})(v)\,dv. \end{align*} Hence, \begin{align*}\lim_{{{L}}\to\infty}\iint_{D_{{L}}}e^{-{s}v}{f}(v-u){g}(u)\,dv\,du&=\lim_{{{L}}\to\infty}\int_0^{{L}}e^{-{s}v}({f}\ast{g})(v)\,dv\\ \\ &=\int_0^\infty e^{-{s}v}({f}\ast{g})(v)\,dv\\ \\ &=\mathcal{L}\{{f}\ast{g}\}({s}). \end{align*}

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  • $\begingroup$ I know its super late but could you tell me why $$\lim_{{{L}}\to\infty}\iint_{R_{{L}}}e^{-{s}(t+u)}{f}(t){g}(u)\,dt\,du=\lim_{{{L}}\to\infty}\iint_{T_{{L}}}e^{-{s}(t+u)}{f}(t){g}(u)\,dt\,du$$@ashpool. How the integration over square region are same with triangle region? $\endgroup$
    – falamiw
    Jul 11, 2021 at 6:08
  • $\begingroup$ The integrations over $R_L$ and $T_L$ may be different, but their limits are the same; see Lax & Terrel's Multivariable Calculus with Applications, section 6.5. $\endgroup$
    – ashpool
    Jul 12, 2021 at 2:21

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