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initial version

A simple implementation using the local heaviside function (see also generalized functions), is presented in this question, and is such that:

var('t s')
u=piecewise([[(0,3),0],[(3,infinity),exp(-2*(t-3))]], var=t)
F(s) = u.laplace(t,s);
InverseLaplace(F, s, t)

produces

\newcommand{\Bold}[1]{\mathbf{#1}}t \ {\mapsto}\ e^{\left(-2 \, t + 6\right)} H\left(t - 3\right).

click to hide/show revision 2
No.2 Revision

A simple implementation using the local heaviside function (see also generalized functions), ) is presented in this question, and is such that:

var('t s')
u=piecewise([[(0,3),0],[(3,infinity),exp(-2*(t-3))]], var=t)
F(s) = u.laplace(t,s);
InverseLaplace(F, s, t)

produces

\newcommand{\Bold}[1]{\mathbf{#1}}t \ {\mapsto}\ e^{\left(-2 \, t + 6\right)} H\left(t - 3\right).

click to hide/show revision 3
No.3 Revision

A simple implementation using the local heaviside function (see also generalized functions) is presented proposed in this question, and is such that:

var('t s')
u=piecewise([[(0,3),0],[(3,infinity),exp(-2*(t-3))]], var=t)
F(s) = u.laplace(t,s);
InverseLaplace(F, s, t)

produces

\newcommand{\Bold}[1]{\mathbf{#1}}t \ {\mapsto}\ e^{\left(-2 \, t + 6\right)} H\left(t - 3\right).

click to hide/show revision 4
No.4 Revision

A simple implementation using the local heaviside function (see also generalized functions) is proposed written in this question, and is such that:

var('t s')
u=piecewise([[(0,3),0],[(3,infinity),exp(-2*(t-3))]], var=t)
F(s) = u.laplace(t,s);
InverseLaplace(F, s, t)

produces

\newcommand{\Bold}[1]{\mathbf{#1}}t \ {\mapsto}\ e^{\left(-2 \, t + 6\right)} H\left(t - 3\right).

click to hide/show revision 5
No.5 Revision

A simple implementation using the local heaviside function (see also generalized functions) is written in this question, and is such that:

var('t s')
u=piecewise([[(0,3),0],[(3,infinity),exp(-2*(t-3))]], var=t)
F(s) = u.laplace(t,s);
InverseLaplace(F, s, t)

produces

$$ \newcommand{\Bold}[1]{\mathbf{#1}}t produces $\newcommand{\Bold}[1]{\mathbf{#1}}t \ {\mapsto}\ e^{\left(-2 \, t + 6\right)} H\left(t - 3\right). $$3\right)$.