Question

does the last part of this require knowledge of Laplace transformations?

Answer 1

you're sure? you answered very quickly :P

Answer 2

can anyone help me

Answer 3

not here. post your own question

Answer 4

I did

Answer 5

well stop clogging this post up then please

Answer 6

@TuringTest ?

Answer 7

Uh Sorry then :)

Answer 8

Not my strong point but I'm thinking on it...
sure looks like a Laplace situation to me

Answer 9

i havent done much laplace before, i basically know the definition and that's about it.

Answer 10

i dont want to launch into laplace now if this can be explained without it

Answer 11

@KingGeorge what are your thoughts?

Answer 12

I've never done Laplace. I've got no clue.

Answer 13

actually this doesn't look like laplace after all, at least not to me...

Answer 14

I think I'd have to do the whole thing to understand what they are asking, but I don't see where R and theta come from

Answer 15

the first section is just a few improper integrals that can be handled with integration by parts obviously

Answer 16

basically the formula for the first part is n!

and evaluating at theta = 0 gets you n! as well

Answer 17

but

\[A( \theta) = \frac{1}{1- \theta}\]

Answer 18

yeah I see the n! thing...

Answer 19

\[A^n( \theta ) = \frac{n!}{(1- \theta)^{n+1}}\]

which is what made me think maybe the explanation is something Laplace-like

Answer 20

one thing I guess they want is for you to notice that \(0<\theta<1\) if the integral is to converge as \(R\to\infty\), but I figure you have already noticed that

Answer 21

yeah i have that, im just stuck on explaining the link between the two parts

i have a book describing the transformation, it looks to me like

\[L(x^{n}e^{-x}) = A^n( \theta) \]

although this does not help me explain it a great deal..

Answer 22

@Hero