Division by t | Laplace Transform

Division by $t$
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   then,
 

$\displaystyle \mathcal{L} \left\{ \dfrac{f(t)}{t} \right\} = \int_s^\infty F(u) \, du$

 

provided   $\displaystyle \lim_{t \rightarrow 0} \left[ \dfrac{f(t)}{t} \right]$   exists.
 

Proof of Division by $t$
Let
$g(t) = \dfrac{f(t)}{t}$,   then

$f(t) = t \, g(t)$

$\mathcal{L} \left\{ f(t) \right\} = \mathcal{L} \left\{ t \, g(t) \right\}$
 

From Multiplication by Power of $t$
$\mathcal{L} \left\{ f(t) \right\} = (-1)^1 \dfrac{d}{ds}\mathcal{L} \left\{ g(t) \right\}$

$F(s) = - \dfrac{d}{ds}\mathcal{L} \left\{ g(t) \right\}$

$F(s) \, ds = -d\big[ \mathcal{L} \left\{ g(t) \right\} \big]$

$d\big[ \mathcal{L} \left\{ g(t) \right\} \big] = -F(s) \, ds$
 

Integrate both sides
$\displaystyle \int d\big[ \mathcal{L} \left\{ g(t) \right\} \big] = -\int F(s) \, ds$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = -\int_\infty^s F(u) \, du$

$\displaystyle \mathcal{L} \left\{ \dfrac{f(t)}{t} \right\} = \int_s^\infty F(u) \, du$
 

Tags: