# Laplace Transform of Identity Mapping

## 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

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

$\blacksquare$

## Proof 2

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

From Integration by Parts:

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

Here:

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

So:

 $\ds \int t e^{-s t} \rd t$ $=$ $\ds -\frac t s e^{-s t} - \frac 1 s \int e^{-s t} \rd t$ $\ds$ $=$ $\ds -\frac t s e^{-s t} - \frac 1 {s^2} e^{-s t}$ Primitive of Exponential Function

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

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

$\blacksquare$

## Proof 3

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

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

$\blacksquare$