Question

Use Problem 41 and the change of variable $u=st$to obtain the generalization

$\mathcal{L}\left\{t^{\alpha }\right\}=\frac{\Gamma (\alpha +1)}{s^{\alpha +1}},\alpha >-1,$

of the result in Theorem 7.1.1(b).

Short answer

$\mathcal{L}\left\{t^{\alpha }\right\}=\frac{\Gamma (\alpha +1)}{s^{\alpha +1}},\alpha >-1$

Step-by-step solution

Definition of Laplace Transform

Let f be a function defined for $t⩾0$ . Then the integral

$\mathcal{L}\left\{f\right(t\left)\right\}={\int }_0^x{e}^{-xt}f\left(t\right)dt$

is said to be the Laplace transform of f, provided that the integral converges.

Use Laplace Transform

The formula for Laplace transform

$\mathcal{L}\left\{f\right(t\left)\right\}={\int }_0^{\infty }{e}^{-st}f\left(t\right)\text{d}t$

We have that$f\left(t\right)=t^{\alpha }$, so use the above formula.

$\mathcal{L}\left\{t^{\alpha }\right\}={\int }_0^{\infty }{e}^{st}{t}^{\alpha }\text{d}t$

Let$u=st\Rightarrow \text{d}u=s\text{d}t$, then

$$\begin{gathered} \mathcal{L}\left\{t^{\alpha }\right\}={\int }_0^{\infty }{e}^{st}{t}^{\alpha }\text{d}t \\ =\frac{1}{s}{\int }_0^{\infty }{e}^u{\left(\frac{u}{s}\right)}^{\alpha }\text{d}u \\ =\frac{1}{s^{\alpha +1}{\int }_0^{\infty }{e}^u{u}^{\alpha }\text{d}u}\left(\Gamma (\alpha +1)={\int }_0^{\infty }{e}^t{t}^{\alpha }\text{d}t,\alpha >-1\right) \\ =\frac{\Gamma (\alpha +1)}{s^{\alpha +1}},\alpha >-1 \end{gathered}$$

Therefore,$\mathcal{L}\left\{t^{\alpha }\right\}=\frac{\Gamma (\alpha +1)}{s^{\alpha +1}},\alpha >-1$