Proof of the Transform of Derivatives

For this post, I wanted to upload an interesting proof of the Laplace transform of Derivatives that is used for transforming a given differential equation in the unknown function $f(t)$ into an algebraic equation in $F(s)$. The Theorem is stated as follows: Suppose that the function $f(t)$ is continuous and piecewise smooth for $t \geq 0$ and is of exponential order as $t \to \infty$, so that there exist nonnegative constants $M$, $c$, and $T$ such that $|f(t)| \leq Me^{ct}$ for $t \geq T$. Then $\mathcal{L}\{f'(t)\}$ exists for $s>c$, and $\mathcal{L}\{f'(t)\}=s\mathcal{L}\{f(t)\}-f(0)=sF(s) - f(0)$ Now for the proof of the Theorem in the general case in which $f'$ is merely piecewise continuous.  To do this it must be proven that the limit $\displaystyle \lim_{b\to\infty} \int_{0}^{b}e^{-st}f'(t) dt$ exists and we also need to find its value. With $b$ fixed, let $t_{1},t_{2},...,t_{k-1}$ be the points interior to the interval $[0,b]$ at which $f'$ is discontinuous. Let $t_{0} = 0$ and $t_{k} = b$. Then the interval can be integrated by parts on each interval $(t_{n-1},t_{n})$ where $f'$ is continuous. This yields $\int_{0}^{b} e^{-st}f'(t)dt = \displaystyle \sum_{n=1}^{k} \int_{t_{n-1}}^{t_{n}}e^{-st}f'(t)dt$ and is equivalent to $\displaystyle \sum^{k}_{n=1}[e^{-st}f(t)]^{t_{n}}_{t_{n-1}} + \displaystyle \sum^{k}_{n=1}s \int^{t_{n}}_{t_{n-1}}e^{-st} f(t) dt$, which I will denote as equation $!$. Now the first summation in equation $!$ is equal to $\displaystyle \sum_{n=1}^{k}[e^{-st}f(t)]^{t_{n}}_{t_{n-1}} = [-f(t_{0})+e^{-st_{1}}]+[-e^{-st_{1}}f(t_{1})+e^{-st_{2}}f(t_{2})]+...+[-e^{st_{k-2}}f(t_{k-2})+e^{-st_{k-1}}f(t_{k-1})] +[-e^{st_{k-1}}f(t_{k-1})+e^{-st_{k}}f(t_{k})]$ and telescopes down to $-f(t_{0}) + e^{-st_{k}}f(t_{k}) = -f(0) +e^{-sb}f(b)$. The second summation adds up to $s$ times the integral from $t_{0} = 0$ to $t_{k} = b$. Therefore equation $!$ reduces to $\int_{0}^{b}e^{-st}f'(t) dt = -f(0) + e^{-sb}f(b) +s\int^{b}_{0}e^{-st}f(t) dt$ From the Theorem above I obtain the expression $|e^{-sb}f(b)| \leq e^{-sb}$ where $Me^{cb}=Me^{-b(s-c)} \Rightarrow 0$ if $s>c$. Therefore, finally taking the limits as $b \Rightarrow \infty$ in the preceding equation, we get the desired result $\mathcal{L}\{f'(t)\} = s \mathcal\{f(t)\}-f(0)$