inverse laplace known but dunno how to prove it

L-[ In ((s^2 + 1) / s^2) ] = 2(1 - cost) / t

how do u prove this?

korhaniski,

L-[ In ((s^2 + 1) / s^2) ] = 2(1 - cost) / t

how do u prove this?

Click to expand...

You do it by finding the Laplace transform of the expression. Look at the Frequency integration transform in this link https://en.wikipedia.org/wiki/Laplace_transform , under the Properties and theorems section. So you first find the Laplace of 2(1 - cos(t)), which is 2/s -(2s)/(s^2+1). Then find the ∫ from s to ∞ of 2/σ -(2σ)/(σ^2+1) with respect to dσ, and you will get ln ((s^2 + 1) / s^2) .

Ratch

Example i had shown in another thread back in May where f(t) is the inverse transform:

[LATEX]
\[\int_{0}^{\infty }f(t)\ e^{-s t}\,\mathit{dt} = \int_{0}^{\infty} (\, 2\ -\,2 e^{-6 t}\, cos(12 t)\, )\, e^{-s t}\,\, \mathit{dt} = \frac{12 s+360}{s^3+12 s^2+180 s}[/LATEX]