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)

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