Laplace Transform of Identity Mapping/Proof 3

Theorem

Let $\laptrans f$ denote the Laplace transform of a function $f$.

Let $\map {I_\R} t$ denote the identity mapping on $\R$ for $t > 0$.

Then:

$\laptrans {\map {I_\R} t} = \dfrac 1 {s^2}$

for $\map \Re s > 0$.

Proof

From Laplace Transform of Derivative:

$(1): \quad \laptrans {\map {I_\R'} t} = s \laptrans {\map {I_\R} t} - \map {I_\R} 0$

under suitable conditions.

Then:

\(\ds \map {I_\R} t\)

\(=\)

\(\ds t\)

\(\ds \leadsto \ \ \)

\(\ds \map {I_\R'} t\)

\(=\)

\(\ds 1\)

\(\ds \map {I_\R} 0\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds \laptrans 1\)

\(=\)

\(\ds \dfrac 1 s\)

Laplace Transform of 1

\(\ds \)

\(=\)

\(\ds s \laptrans {\map {I_\R} t} - 0\)

from $(1)$, substituting for $\map f t$ and $\map f 0$

\(\ds \leadsto \ \ \)

\(\ds s \laptrans {\map {I_\R} t}\)

\(=\)

\(\ds \dfrac 1 s\)

\(\ds \leadsto \ \ \)

\(\ds \laptrans {\map {I_\R} t}\)

\(=\)

\(\ds \dfrac 1 {s^2}\)

$\blacksquare$

Sources

• 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Solved Problems: Laplace Transform of Derivative: $15 \ \text{(b)}$