Question

Laplace Transform Proof:

Answer 1

Suppose that f and f' are continuous for \(t \ge 0\) and of exponential order as \(t \rightarrow \infty\). Use integration by parts to show that if \(F(s)= \mathcal{L}(f(t))\) , then \(\lim_{s \rightarrow \infty} F(s)=0\).

Answer 2

hey @CShrix!

you probably need to add a definition to get this going for many people....eg

\[F(s) = \mathcal{L_s}\{f(t)\}= \int\limits_{t=0}^{\infty} \;dt \; f(t) \; e^{-st} \]

\[= \left[ -f(t) \dfrac{1}{s}e^{-st}\right]_{t=0}^{\infty} - \int\limits_{0}^{\infty} \;dt \; f'(t) \; e^{-st} \]

\[= \left[ -f(t) \dfrac{1}{s}e^{-st}\right]_{t=0}^{\infty} - \int\limits_{0}^{\infty} \;dt \; f'(t) \; e^{-st} \]

\[= \left[ f(t) \dfrac{1}{s}\right] - \int\limits_{0}^{\infty} \;dt \; f'(t) \; e^{-st} \]

Answer 3

we can apply the "initial value theorem" which states this:
\[\Large f\left( 0 \right) = \mathop {\lim }\limits_{t \to 0 + } f\left( t \right) = \mathop {\lim }\limits_{\operatorname{Re} \left( s \right) \to + \infty } sF\left( s \right)\]

Answer 4

hint:
we can write this:
\[\Large F\left( s \right) = \frac{{sF\left( s \right)}}{s} = \left\{ {sF\left( s \right)} \right\} \cdot \left( {\frac{1}{s}} \right)\]

Answer 5

I think I understand it so far..