Laplace Transform of Identity Mapping/Proof 2

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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

\(\ds \laptrans {\map {I_\R} t}\) \(=\) \(\ds \laptrans t\) 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
\(\ds \) \(=\) \(\ds \frac 1 {s^2}\)

$\blacksquare$


Sources

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