Question

Case closed lol

Answer 1

keep getting lost in the integration by parts

Answer 2

I know this is just a basic integ by parts

Answer 3

I just forgot everything

Answer 4

You just game me the transform, didn't you?

Answer 5

gave**

Answer 6

yes yes i really want to know the derivation of the transform

Answer 7

my internet is poor, I apologize for any delays due to this.

Answer 8

Its okay. I am also trying to figure this thing out.

Answer 9

\(\color{#000000 }{ \displaystyle \lim_{{\rm N}\to\infty}\int\limits_0^{\rm N}e^{st}\sinh(at)dt }\)

I would apply some basics, before integrating:
(1) \(\color{#000000 }{ \displaystyle \sinh(x)=\frac{e^x-e^{-x}}{2}=i\sin(x) }\)

Why? Because....
\(\color{#000000 }{ \displaystyle e^{ix}=\cos(x)+i\sin(x) }\)
(Derivation: Taylor series for e^x, and sub x=iz). Next,
\(\color{#000000 }{ \displaystyle e^{-ix}=\cos(x)-i\sin(x) }\)
(Cosine is even)

Thus,
\(\color{#000000 }{ \displaystyle e^{ix}-e^{-ix}=2i\sin(x) }\)
(this is the reasoning for (1).)

\(\color{#000000 }{ \displaystyle \lim_{{\rm N}\to\infty}\int\limits_0^{\rm N}ie^{st}\sin(t)dt }\)

Answer 10

Then, you can do integration b parts twice, and then algerba.

Answer 11

You will see, it is one of those on Khan academy, where you don't take the integral apart, but rather integration by parts allows to solve for the integral algebriacly.

Answer 12

I can show an example of such integral if you like.

Answer 13

you want to go backwards?

Answer 14

yes yes thank you

Answer 15

Ok, I am back after reloading:)

Answer 16

By the way, anyone knows a latex for laplace?

Answer 17

I have no idea 'bout the latex thing sorry

Answer 18

Oh wait the laplace transfrom for e^at is 1/s-a, right?

Answer 19

Yes.
(s>a)

Answer 20

Before I go on with inverse, just wanted to note my integral is incorrect.
It should be sin(at) (not just t).

Answer 21

ah yes

Answer 22

Now im totally lost..

Answer 23

How can i get the right transform with this?

Answer 24

What you drew, is correct, actually ...

Answer 25

I mean for the laplace transform of sinh(at)

Answer 26

Comment removed.. but get the inverse laplace transofrm with that (what you wrote?)
that is the same thing, if you combine fractions.

Answer 27

Was this just a small typo Sol?\[\large\rm \color{#000000 }{ \displaystyle \mathscr L\{\sinh(at)\}\quad=\quad\lim_{{\rm N}\to\infty}\int\limits\limits_0^{\rm N}e^{\color{red}{-}st}\sinh(at)dt }\]negative in the exponent if I remember correctly, ya?
It's been a while since I've done Laplace :) lol