Theorem
1.Let $f$ be a real function that has continuous $n-1$ derivatives $f^{(n-1)}$ (and hence $f,f',\ldots,f^{(n-2)}$ are also continuous for $t\geq 0 $; assume that $f,f',\ldots,f^{(n-1)}$ are also of exponential order $e^{at}$)
2.Suppose $f^n$ is piecewise continuous in every closed interval $0\leq t \leq b$
Conclusion:
$ℒ\left\{f^{(n)}\right\}$ exists for all $s>a$
$$ℒ\left\{f^{n}(t)\right\}=s^{n}ℒ\left\{f(t)\right\}-s^{n-1}f'(0)-s^{n-2} f''(0)-\cdots-f^{(n-1)}(0)$$
I need to use prove by induction for this one
My work:
- For $n=1$ case
Use the definition of Laplace transform
$$ℒ\left\{f'(t)\right\}=\lim_{R \rightarrow \infty}\int_0^R e^{-st} f'(t) \, \mathrm{d}t$$
We assume that there are so many discontinuities such that they are finite.
$$0\leq t_1 < t_2 <\cdots<t_n\leq R$$
The integral we have above then can be write in such a way that
$$\int_0^R e^{-st}f'(t)\,\mathrm{d}t=\int_0^{t_1}e^{-st} f'(t) \, \mathrm{d}t+\int_{t_1}^{t_2}e^{-st} f'(t)\,\mathrm{d}t+\cdots+\int_{t_n}^R e^{-st} f'(t) \, \mathrm{d}t$$
Performing integration by parts,
\begin{align} & \int_0^R e^{-st}f'(t)\,\mathrm{d}t= \Big[ e^{-st} f(t) \Big]_0^{t_1^-} + s\int_0^{t_1} e^{-st} f(t) \, \mathrm{d}t + \Big[ e^{-st} f(t) \Big]_{t_1^+}^{t_2^-} \\[10pt] & {} + s\int_{t_1}^{t_2}e^{-st} f(t) \, \mathrm{d}t +\cdots+ \Big[ e^{-st} f(t) \Big]_{R^-}^{t_n^+} + s\int_R^{t_n}e^{-st} f(t)\,\mathrm{d}t \end{align}
Since they are continuous, they will cancel out each other nicely. Leaving us with
$$\int_0^R e^{-st}f'(t) \, \mathrm{d}t = sℒ\left\{f(t)\right\}-f(0)$$
For $n=k$ case
$$ℒ\left\{f^{k}(t)\right\}=s^k ℒ\left\{f(t)\right\}-s^{k-1}f'(0)-s^{k-2}f''(0)-\cdots-f^{(k-1)}(0)$$
We assume this case to be true.
Now for For $n=k+1$ case
$$ℒ\left\{f^{k+1}(t)\right\}=s^{k+1} ℒ\{f(t)\} -s^k f'(0)-s^{k-1} f''(0) -\cdots-f^{(k)}(0)$$
I am kind of stuck here. Any help will be appreciated.