Laplace Transform of Identity Mapping

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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 1

\(\displaystyle \laptrans {\map {I_\R} t}\) \(=\) \(\displaystyle \laptrans t\) Definition of Identity Mapping
\(\displaystyle \) \(=\) \(\displaystyle \int_0^{\to +\infty} t e^{-s t} \rd t\) Definition of Laplace Transform
\(\displaystyle \) \(=\) \(\displaystyle \intlimits {\frac {e^{-s t} } {-s} \paren {t - \frac 1 {-s} } } {t \mathop = 0} {t \mathop \to +\infty}\) Primitive of $x e^{a x}$
\(\displaystyle \) \(=\) \(\displaystyle -\frac 1 s \lim_{t \mathop \to +\infty} \frac t { e^{s t} } - \paren {0 - \frac 1 {s^2} }\) Exponential of Zero and One, Exponent Combination Laws: Negative Power
\(\displaystyle \) \(=\) \(\displaystyle 0 + \frac 1 {s^2}\) Limit at Infinity of Polynomial over Complex Exponential
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {s^2}\)

$\blacksquare$


Proof 2

\(\displaystyle \laptrans {\map {I_\R} t}\) \(=\) \(\displaystyle \laptrans t\) Definition of Identity Mapping
\(\displaystyle \) \(=\) \(\displaystyle \int_0^{\to +\infty} t e^{-st} \rd t\) Definition of Laplace Transform


From Integration by Parts:

$\displaystyle \int f g' \rd t = f g - \int f'g \rd t$

Here:

\(\displaystyle f\) \(=\) \(\displaystyle t\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle f'\) \(=\) \(\displaystyle 1\) Derivative of Identity Function
\(\displaystyle g'\) \(=\) \(\displaystyle e^{-st}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle g\) \(=\) \(\displaystyle -\frac 1 s e^{-s t}\) Primitive of Exponential Function

So:

\(\displaystyle \int t e^{-s t} \rd t\) \(=\) \(\displaystyle -\frac t s e^{-s t} - \frac 1 s \int e^{-s t} \rd t\)
\(\displaystyle \) \(=\) \(\displaystyle -\frac t s e^{-s t} - \frac 1 {s^2} e^{-s t}\) Primitive of Exponential Function


Evaluating at $t = 0$ and $t \to +\infty$:

\(\displaystyle \laptrans t\) \(=\) \(\displaystyle \intlimits {-\frac t s e^{-s t} - \frac 1 {s^2} e^{-s t} } {t \mathop = 0} {t \mathop \to +\infty}\)
\(\displaystyle \) \(=\) \(\displaystyle -\frac 1 s \lim_{t \mathop \to +\infty} \frac t { e^{s t} } - \paren {0 - \frac 1 {s^2} }\) Exponential of Zero and One, Exponent Combination Laws: Negative Power
\(\displaystyle \) \(=\) \(\displaystyle 0 + \frac 1 {s^2}\) Limit at Infinity of Polynomial over Complex Exponential
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {s^2}\)

$\blacksquare$


Proof 3

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:

\(\displaystyle \map {I_\R} t\) \(=\) \(\displaystyle t\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map {I_\R'} t\) \(=\) \(\displaystyle 1\)
\(\displaystyle \map {I_\R} 0\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \laptrans 1\) \(=\) \(\displaystyle \dfrac 1 s\) Laplace Transform of 1
\(\displaystyle \) \(=\) \(\displaystyle s \laptrans {\map {I_\R} t} - 0\) from $(1)$, substituting for $\map f t$ and $\map f 0$
\(\displaystyle \leadsto \ \ \) \(\displaystyle s \laptrans {\map {I_\R} t}\) \(=\) \(\displaystyle \dfrac 1 s\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \laptrans {\map {I_\R} t}\) \(=\) \(\displaystyle \dfrac 1 {s^2}\)

$\blacksquare$


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