Laplace Transform of Identity Mapping/Proof 1

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 $\map \Re s > 0$.

$\ds \laptrans {\map {I_\R} t}$

$=$

$\ds \laptrans t$

$\ds $

$=$

$\ds \int_0^{\to +\infty} t e^{-s t} \rd t$

$\ds $

$=$

$\ds \intlimits {\frac {e^{-s t} } {-s} \paren {t - \frac 1 {-s} } } {t \mathop = 0} {t \mathop \to +\infty}$

$\ds $

$=$

$\ds -\frac 1 s \lim_{t \mathop \to +\infty} \frac t { e^{s t} } - \paren {0 - \frac 1 {s^2} }$

$\ds $

$=$

$\ds 0 + \frac 1 {s^2}$

$\ds $

$=$

$\ds \frac 1 {s^2}$

$\blacksquare$