Question

Integral of transform.

Answer 1

Find the laplace of \[\frac{ 2sinht }{ t }\]

Answer 2

@ganeshie8 can you?

Answer 3

@wio can you?

Answer 4

Find the laplace transform?

Answer 5

Are you stuck?

Answer 6

Yes. So far I am doing this.
\[2\int\limits_{s}^{\infty}\frac{ 1 }{ u^2-1 } = 2\tanh^{-1}t\]

Answer 7

Is there something wrong with that?

Answer 8

It is a definite integral, so \(t\) should go away. Where is your \(e^{-st}\)?

Answer 9

Ooops sorry. t there must be u. And I haven't evaluated yet the limits.

Answer 10

Isn't it laplace of sinht is \[\frac{ 1 }{ u^2-1 }\] ?

Answer 11

First of all, what is sinh?

Answer 12

\[
\mathcal L \left[\frac{2\sinh t}{t}\right] = \int_0^{\infty}\frac{2\sinh t}{t}(e^{-st})\;dt
\]

Answer 13

\[
\sinh(t) = \frac{e^{t}-e^{-t}}{2}
\]

Answer 14

It splits into two integrals which aren't too bad.

Answer 15

\[
\int_0^{\infty}t^{-1}e^{(1-s)t}\;dt - \int_0^{\infty}t^{-1}e^{-(s+1)t}\;dt
\]

Answer 16

Can you do it now?

Answer 17

Wait wait wait. Why is there a exponential e, there?

Answer 18

definition of laplace

Answer 19

I have no idea how you got that.

Answer 20

Just integrate.

Answer 21

But that will give me lower points coz my professor doesn't want us to use the typical method.

Answer 22

^ what i said is like property of all sinht function when converting to laplace.

Answer 23

Which properties can you use then?

Answer 24

There is a property wherein when we are to convert laplce of sin at is same as 1/(u^2-a^2)

Answer 25

I mean sinh at

Answer 26

after conversion, then i will do the integration so far, i am with
\[2\tanh^{-1}u\] from s to infinity
but i don't know what is arc tanh infinity equal to, and I believe it is equal to 1

Answer 27

Yow.

Answer 28

\[\Large 2 \int\limits_{s}^{\infty}\frac{ 1 }{ \tanh }ds\] you mean?

Answer 29

No. Arc tangent of t ds is what I mean

Answer 30

http://www.wolframalpha.com/input/?i=lapalce+transform+2+sinh%28t%29%2Ft

Answer 31

By Wolframalpha or Mathematica
\[

\mathcal{L}_t\left[\frac{e^t-e^{-t}}{t}\right](s)=\ln
\left(\frac{s+1}{s-1}\right)
\]

Answer 32

Notice that
\[
\frac{2 \sinh (t)}{t}=\frac{e^t-e^{-t}}{t}
\]

Answer 33

You can show that
\[
2 \tanh ^{-1}\left(\frac{1}{s}\right)=\ln \left(\frac{s+1}{s-1}\right)
\]

Answer 34

The answer is one of the expression above/