Multiplication by Power of t | Laplace Transform

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

$\mathcal{L} \left\{ t^n f(t) \right\} = (-1)^n \dfrac{d^n}{ds^n} F(s) = (-1)^n F^{(n)}(s)$

where   $n = 1, \, 2, \, 3, \, ...$
 

Proof of Multiplication by Power of $t$
$F(s) = \mathcal{L} \left\{ f(t) \right\}$

$\displaystyle F(s) = \int_0^\infty e^{-st} f(t) \, dt$
 

Differentiate both sides in   $s$,
$\displaystyle \dfrac{dF}{ds} = \dfrac{d}{ds} \int_0^\infty e^{-st} f(t) \, dt$
 

From Leibniz rule of differentiation under integral sign,
$\displaystyle \dfrac{dF}{ds} = \int_0^\infty \dfrac{\partial}{\partial s} e^{-st} f(t) \, dt$

$\displaystyle \dfrac{dF}{ds} = \int_0^\infty (-te^{-st}) f(t) \, dt$

$\displaystyle \dfrac{dF}{ds} = -\int_0^\infty e^{-st} [ \, t \, f(t) \, ] \, dt$

$\dfrac{dF}{ds} = -\mathcal{L} \left\{ t \, f(t) \right\}$
 

Thus,
$\mathcal{L} \left\{ t \, f(t) \right\} = -\dfrac{dF}{ds} = -F'(s)$       → Equation (1)

which proves the theorem for   $n = 1$.
 

Assuming the theorem is true for   $n = k$   then,
$\mathcal{L} \left\{ t^k f(t) \right\} = (-1)^k \dfrac{d^kF}{ds^k} = (-1)^k F^{(k)}(s)$

$\displaystyle \int_0^\infty e^{-st} [ \, t^k f(t) \, ] \, dt = (-1)^k \dfrac{d^kF}{ds^k} = (-1)^k F^{(k)}(s)$       → Equation (2)
 

Differentiate both sides in   $s$,
$\dfrac{d}{ds} \displaystyle \int_0^\infty e^{-st} [ \, t^k f(t) \, ] \, dt = (-1)^k \dfrac{d}{ds} \left[ \dfrac{d^kF}{ds^k} \right] = (-1)^k \dfrac{d}{ds} \left[ F^{(k)}(s) \right]$

$\displaystyle \int_0^\infty \dfrac{\partial}{\partial s} e^{-st} [ \, t^k f(t) \, ] \, dt = (-1)^k \dfrac{d^{(k + 1)}F}{ds^{(k + 1)}} = (-1)^k F^{(k + 1)}(s)$

$\displaystyle \int_0^\infty (-te^{-st}) [ \, t^k f(t) \, ] \, dt = (-1)^k \dfrac{d^{(k + 1)}F}{ds^{(k + 1)}} = (-1)^k F^{(k + 1)}(s)$

$-\displaystyle \int_0^\infty e^{-st} [ \, t^{(k + 1)} f(t) \, ] \, dt = (-1)^k \dfrac{d^{(k + 1)}F}{ds^{(k + 1)}} = (-1)^k F^{(k + 1)}(s)$

$\displaystyle \int_0^\infty e^{-st} [ \, t^{(k + 1)} f(t) \, ] \, dt = (-1)^{(k + 1)} \dfrac{d^{(k + 1)}F}{ds^{(k + 1)}} = (-1)^k F^{(k + 1)}(s)$       → Equation (3)
 

This shows that the theorem is true for   $n = k$   from Equation (2) and for   $n = k + 1$,   from Equation (3). From Equation (1), the theorem is true for   $n = 1$.   Hence, it is true for   $n = 1 + 1 = 2$   and   $n = 2 + 1 = 3$,   and so on, and thus, for all positive integer values of   $n$.
 

Therefore,
$\mathcal{L} \left\{ t^n f(t) \right\} = (-1)^n \dfrac{d^n}{ds^n} F(s) = (-1)^n F^{(n)}(s)$

where   $n = 1, \, 2, \, 3, \, ...$       okay
 

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