Question

@ganeshie8 Laplace

Answer 1

How do I do this...suppose that f and f' are continuous functions for \(t \ge 0\) and of exponential order as \(t \rightarrow \infty \). Use integration by parts to show that if \(F(s) = L{f(t)}\), then \[\lim_{s \rightarrow \infty} F(s) = 0 \].

Answer 2

That L is laplace I don't know how to make the fancy L

Answer 3

`\mathcal{L}` gives \(\mathcal{L}\)

Answer 4

\[\mathcal{L}(f(t)) = \int\limits_0^{\infty} f(t)e^{-st}\, dt\]

Answer 5

Oh so I am using that?

Answer 6

try IBP once and see if it gives some clue...

Answer 7

ok

Answer 8

Ah, u = f' du = f' dt

Answer 9

u=f*

Answer 10

\[\mathcal{L}(f(t)) = \int\limits_0^{\infty} f(t)e^{-st}\, dt \\~\\=f(t) \dfrac{e^{-st}}{-s}\Bigg|_0^{\infty} +\dfrac{1}{s}\int\limits_0^{\infty} f'(t)e^{-st}\, dt \\~\\ \]

Answer 11

Yes I got that

Answer 12

What do I do with the integral though lol

Answer 13

Since f(t) is of exponential order, the first term evaluates to 0, yes ?

Answer 14

lim t - > infinity

Answer 15

Yes

Answer 16

\[\mathcal{L}(f(t)) = \int\limits_0^{\infty} f(t)e^{-st}\, dt \\~\\=f(t) \dfrac{e^{-st}}{-s}\Bigg|_0^{\infty} +\dfrac{1}{s}\int\limits_0^{\infty} f'(t)e^{-st}\, dt \\~\\
=0+\dfrac{1}{s}\int\limits_0^{\infty} f'(t)e^{-st}\, dt \\~\\
\]

Answer 17

Since f'(t) is also of exponential order, the definite integral, \(\int\limits_0^{\infty} f'(t)\,e^{-st}\, dt\), converges to some finite number, yes ?

Answer 18

no wait, the definite integral must be a function of \(s\)

Answer 19

\[F(s) = \mathcal{L} {f}(t)\]

Answer 20

So what just take this as lim s -> infinity?

Answer 21

\[\lim_{s \rightarrow \infty} \left( \frac{ -fe^{-st} }{ s } |_0^s+\int\limits_{0}^{s}\frac{ e^{-st} }{ s }f'dt\right)\]

Answer 22

Yeah

Answer 23

\[\lim\limits_{s\to\infty}F(s) = \lim\limits_{s\to\infty}\dfrac{1}{s}\int\limits_0^{\infty} f'(t)e^{-st}\, dt\]

Answer 24

f'(t) = 0

Answer 25

how do you know ?

Answer 26

Because of exponential order?

Answer 27

oh nvm that's as t-> ifninity

Answer 28

f(x) is of exponential order just means that the function e^x can overtake f(x)

Answer 29

Right, also I see it says here the result is actually true under less restrictive conditions

Answer 30

There's a theorem, f is piecewise continuous on the interval \(0 \le t \le A\) for any positive A.
\(|f(t)| \le Ke^{at}\) when \(t \ge M\), K, a , and M are real constants, K and M are positive then the Laplace \[\mathcal{L} f(t) = F(s)\]

Answer 31

Not sure how to apply that here though or even if you could

Answer 32

that theorem is just saying that the laplace transform exists for exponential type functions

Answer 33

in other words its saying that the improper integral \(\int\limits_0^{\infty}f(t)e^{-st}\,dt\) converges if \(f(t)\) is of exponnential type

Answer 34

Oooh ok

Answer 35

\[\lim\limits_{s\to\infty}F(s) = \lim\limits_{s\to\infty}\dfrac{1}{s}\int\limits\limits_0^{\infty} f'(t)e^{-st}\, dt\] what can we do with f' here then

Answer 36

No clue, if you do IBP again we get

\[\lim\limits_{s\to\infty}F(s) = \lim\limits_{s\to\infty}\dfrac{1}{s^2}\int\limits\limits_0^{\infty} f''(t)e^{-st}\, dt\]

Answer 37

Yeah I know haha

Answer 38

after doing IBP n times :

\[\lim\limits_{s\to\infty}F(s) = \lim\limits_{s\to\infty}\dfrac{1}{s^{n}}\int\limits\limits_0^{\infty} f^{(n)}(t)e^{-st}\, dt\]

Answer 39

maybe we can "recover" f

Answer 40

Not sure how to conclude that equals 0

Answer 41

Yeah I don't know, I'm sort of making things up and that's of course not working lol

Answer 42

When math becomes more letters than numbers... *shivers*

Answer 43

@ShadowLegendX No worries. You'll realise the importance of letters once you turn 10. :)

Answer 44

Iʻm just a kitteh, idk what any of this stuff is >.<