laplace transform
The common statement of the Laplace transform for the convolution is:
$$ L(f*g)(s) = F(s)G(s), $$
where
$$ (f*g)(t) = \int_0^t f(\tau)g(t-\tau)\, d\tau. $$
Can this property be used for a convolution where the upper limit is infinity? Can the Laplace transform of:
$$ (f*g)(t) = \int_0^\infty f(\tau)g(t-\tau)\, d\tau $$
be expressed in a similar fashion as the product of the individual transforms of $f$ and $g$?
You can change variables $t^2=x$:
$$ H(s) = \int_0^\infty f(t^2) e^{-st}\; dt = \int_0^\infty \frac{f(x)}{2\sqrt{x}} e^{-s\sqrt{x}}\; dx $$ Now we can express $e^{-s \sqrt{x}}/\sqrt{x}$ as a Laplace transform, so: $$ \eqalign{ H(s) &= \int_0^\infty \int_0^\infty \frac{f(x)}{2 \sqrt{\pi t}} e^{-s^2/(4 t)} e^{-xt}\; dt\; dx \cr &= \int_0^\infty\frac{F(t)}{2 \sqrt{\pi t}} e^{-s^2/(4t)}\; dt}$$
\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
$$0\leq v\leq {{L}},\qquad 0\leq u\leq {{L}},\qquad v\geq u.$$
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*}
Best Answer
Suppose you take the upper limit to be $\infty$. Then for $\tau>t$ you’d be considering $g(t)$ for $t<0$. But the Laplace transform of $g(t)$ is only taken over positive times, and so cannot carry any information about $g(t)$ during this time. Hence there’s no way that the “full-time convolution” can be expressed in terms of the Laplace transforms of $f(t)$ and $g(t)$.
Notes:
1) This issue pertains to the one-sided transform. The two-sided transform doesn’t have this quirk since it extends over all times.
2) One quirk of the “full-time convolution” is that it is manifestly asymmetric between $f$ and $g$: the substitution $\tau’=t-\tau$ does not swap $f$ and $g$ owing to the integration endpoints. Hence one would have $f\ast g\neq g\ast f$, which is hardly appealing.
3) One “exception” to the above is if you insist that $g(t)$ and $f(t)$ both vanish for $t<0$. Under that convention the convolution can be written in the usual way as an integral over the real line. But the integrand is still only supported for $\tau\in[0,t]$ so there’s really no point.