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As I understand it, the theorem and proof for the Laplace transform of positive integer powers is as follows:

Theorem

Let $t^n : \mathbb{R} \to \mathbb{R}$ be $t$ to the $n$th power for some $n \in \mathbb{N}_{\ge 0}$

Let $\mathcal{L}$ be the Laplace transform.

Then

$$\mathcal{L} \{ t^n \} = \dfrac{n!}{s^{n + 1}}$$

for $\Re(s) > 0$

Proof

The proof proceeds by induction on $n$ for $t^n$.

Base case:

When $n = 0$, we have

$$t^0 = 1$$

\begin{align*} \therefore \mathcal{L} \{ t^0 \} &= \mathcal{L} \{ 1 \} \\\ &= \dfrac{1}{s} \ \ \ \text{(Laplace transform of a constant.)} \\\ &= \dfrac{0!}{s^{0 + 1}} \ \ \ \text{(By the definition of $0!$.)} \end{align*}

Induction Hypothesis:

Let $n \in \mathbb{N}$ with $n \ge 1$.

Then

$$\mathcal{L} \{ t^n \} = \dfrac{n!}{s^{n + 1}}$$

Induction Step:

$$\mathcal{L} \{ t^{n + 1} \} = \int_0^\infty t^{n + 1} e^{-st} \ dt \ \ \ \text{(By the definition of Laplace transform.)}$$

We now use integration by parts, which says that

$$\int fg' = fg - \int f'g \ dt$$

Let $f = t^{n + 1}$ and $g' = e^{-st}$

Therefore, $f' = (n + 1)t^n$ and $g = - \dfrac{1}{s} e^{-st}$

So we have

$$\int t^{n + 1} e^{-st} \ dt = - \dfrac{t^{n + 1}}{s}e^{-st} + \dfrac{n + 1}{s} \int t^n e^{-st} \ dt$$

Evaluating $t = 0$ and $t \to \infty$, we get

\begin{align*} \mathcal{L} \{ t^{n + 1} \} &= \left. - \dfrac{1}{s} t^{n + 1}e^{-st} \right|_0^\infty + \dfrac{n + 1}{s} \mathcal{L} \{ t^n \} \\ &= \left. -\dfrac{s^{-1}t^{n + 1}}{e^{st}} \right|_0^\infty + \dfrac{n + 1}{s} \mathcal{L} \{ t^n \} \\ &= 0 - 0 + \dfrac{n + 1}{s} \mathcal{L} \{ t^n \} \\ &= \dfrac{n + 1}{s} \times \dfrac{n!}{s^{n + 1}} \\ &= \dfrac{(n + 1)!}{s^{n + 1 + 1}} \end{align*}

I'm now looking for the proof of the Laplace transform of real number powers, rather than just positive integer powers. I've been unable to find this in my research.

I would greatly appreciate it if people could please take the time to demonstrate this.

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1 Answer 1

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\begin{align*} \mathcal{L} \{ t^\alpha \} &= \int_0^\infty e^{-st} t^\alpha\ dt\\ &= \int_0^\infty e^{-w} \left(\dfrac{w}{s}\right)^\alpha\ \dfrac{dw}{s}\\ &= \dfrac{1}{s^{\alpha+1}}\int_0^\infty e^{-w} w^\alpha\ dw\\ &= \dfrac{\Gamma(\alpha+1)}{s^{\alpha+1}} \end{align*} where $s>0$

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  • $\begingroup$ Can you please explain what's going on here? $\endgroup$ Sep 3, 2018 at 3:28
  • $\begingroup$ Did you want this or another thing? $\endgroup$
    – Nosrati
    Sep 3, 2018 at 3:30
  • $\begingroup$ I meant a worded explanation. $\endgroup$ Sep 3, 2018 at 3:35
  • $\begingroup$ this is Laplace definition, used to $t^\alpha$. $\endgroup$
    – Nosrati
    Sep 3, 2018 at 3:37
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    $\begingroup$ @ThePointer This computation is making use of the definition of the gamma function: $$ \Gamma(z) = \int_0^\infty e^{-t} t^{z-1}\ \mathsf dt $$ $\endgroup$
    – Math1000
    Sep 3, 2018 at 7:46

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