Laplace Transform of Intergrals

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

$\displaystyle \mathcal{L} \left[ \int_0^t f(u) \, du \right] = \dfrac{F(s)}{s}$

 

Proof
Let   $\displaystyle g(t) = \int_0^t f(u) \, du$

then,   $g'(t) = f(t)$   and   $g(0) = 0$
 

Taking the Laplace transform of both sides,
$\mathcal{L} \left\{ g'(t) \right\} = \mathcal{L} \left\{ f(t) \right\}$
 

From Laplace transform of derivative,   $\mathcal{L} \left\{ g'(t) \right\} = s \, \mathcal{L} \left\{ g(t) \right\} - g(0)$   and from the Theorem above, $\mathcal{L} \left\{ f(t) \right\} = F(s)$
 

Thus,
$s \, \mathcal{L} \left\{ g(t) \right\} - g(0) = F(s)$

$s \, \mathcal{L} \left\{ g(t) \right\} - 0 = F(s)$

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

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

$\displaystyle \mathcal{L} \left[ \int_0^t f(u) \, du \right] = \dfrac{F(s)}{s}$
 

Tags
Laplace
proof
Laplace transform
transform of integral