show that $$\mathcal L \lbrace \cosh kt \rbrace = \frac{s}{s^2 - k^2}; \; \text{for} \; s |k|$$

what does s |k| mean? is it important in the solution or is it just there to say that the integral converges?

Question

show that $$\mathcal L \lbrace \cosh kt \rbrace = \frac{s}{s^2 - k^2}; \; \text{for} \; s> |k|$$

what does s> |k| mean? is it important in the solution or is it just there to say that the integral converges?

anonymous · Answer

here it is

anonymous · Answer

if s will be less than k then the exponential will get positive  (and the function will not be of Exponential order) and for such functions Laplace does not Exists.

lgbasallote · Answer

...i was just asking if s> |k| mattered in the solution or not =))

lgbasallote · Answer

why would the laplace not exist if $s \le |k|$

anonymous · Answer

because in this case the exponential will be positive . and the condition for Laplace existence is it should be of Exponential order .when s http://www.efunda.com/math/laplace_transform/index.cfm you can verify that for positive exponent http://www.wolframalpha.com/input/?i=integrate+%28e%5E%2B%28s%2Bt%29%29++t%3D1..infinity

lgbasallote · Answer

my brain hurts =_=

anyway i got $$\large \frac{-e^{-st} \left[k\sinh kt + s\cosh kt\right]}{s^2 - k^2} |_0^{\infty}$$
that sounds wrong...is it?

lgbasallote · Answer

yes that

lgbasallote · Answer

and i didnt need to do that cosh infinity...e^-infty is 0

lgbasallote · Answer

anyway..i was abke to get s/s^2 - k^2 so i guess it's right

but i still dont get these "limits" if that's how it's called