# Division by t | Laplace Transform

**Division by $t$**

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

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

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

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

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

## How have you taken the limits

How have you taken the limits from s to infinity instead of 0 to infinity