Question

Derive the Laplace transform of f(t) = t

Answer 1

on the risk of sounding like a retard wouldn't that just be the anti derive of t?

Answer 2

LT is defined as int 0 to infinity of f(t)e^-st dt wherever this converges.

Answer 3

\[\huge \int\limits_0^\infty te^{-st} dt\]

I think someone can manage this (integration by parts)

Answer 4

http://www.wolframalpha.com/input/?i=laplacetransfrom+f%28t%29%3Dt

Answer 5

\[\int\limits_0^\infty te^{-st} dt=[\frac{t}{-s}e^{-st}]^{\infty}_{0}-\int\limits_0^\infty \frac{1}{-s}e^{-st}dt=-\int\limits_0^\infty \frac{1}{-s}e^{-st}dt\]

Answer 6

\[=\frac{1}{s^{2}}\]

Answer 7

Yeah, forgot to integrate on the last step.

Answer 8

still one integral left yes?

Answer 9

Hmm, skipping a few steps with the old infinite limits there, right idea though.

Answer 10

http://www.wolframalpha.com/input/?i=integrate+te^%28-st%29dt

Answer 11

Doesn't work like that...

Answer 12

i actually know nothing about this so ignore me

Answer 13

I put it up because I only just strted learning about it myself...

Answer 14

but at first glance it looks like if you compute the improper integral of that thing you will in fact get
\[\frac{1}{s^2}\]

Answer 15

very true

Answer 16

Yes, that's right, there's a few niceties with the infinite limit and u assume s>0 for convergence.

Answer 17

It's better to go from 0 to a, say, then let a tend to infinity after...

Answer 18

yes that is the definition

Answer 19

I know, but I tend to plug infinity as a value in the integral.

Answer 20

I saw...:-)

Answer 21

It works most of the time.

Answer 22

Try doing it with a general f(t) instead of just t....

Answer 23

\[\int_a^{\infty}f(t)dt=\lim_{l\rightarrow \infty}\int_a^lf(t)dt\]man my typing is slow this morning

Answer 24

Integration by parts isn't useful when you have the general f(t).

Answer 25

It is in this case...

Answer 26

The whole idea with Laplace transform is to turn it all into simple algebra, get the answer, then go back to t...

Answer 27

I only just started with it, it looks quite good and it works for quite a lot of des and pdes u encounter in practice...

Answer 28

I learned it a couple of months ago, haven't used it in practice too much yet though.

Answer 29

by "it" you mean turn an ode into a polynomial right?

Answer 30

With f(t) u get L{f(t)} = sL{f(t)}-f(0)

Answer 31

an ode into a polynomial

Yes, easier to solve....

Answer 32

About as much work as without Laplace transform though.

Answer 33

i should get a book. any recommendations?

Answer 34

I came across it recently in Engineering Mathematics by KA Stroud - never heard of it before. Lokks unteresting.

Answer 35

I am using The Laplace Transform:Theory and Applications by Joel L Schiff

Answer 36

About as much work as without Laplace transform though.

This isn't right, it's why u have LT (and Fourier), to work those u can't normally.

Answer 37

For those differential equations you can do two ways then.

Answer 38

Agreed, I wouldn't use it for that though.....

Answer 39

it is more fun.

Answer 40

Another pure mathematician in the making..:-)

Answer 41

:)

Answer 42

Right, L(1) = 1/s so now show that this holds for s= x +iy.