Laplace transform of 1/cos(t)

[banutraul]

Homework Statement

You have to calculated the Laplace transformation for 1/ cos(t)

Homework Equations

That's all

The Attempt at a Solution

i tryed whit some trigonometric formulas but i don't get anywhere : 1/cos(t) = cos(t) / (1- sin ^2 (t)) or 1/cos(t) = cos(t) + sin(t) x tg(t) or 1/cos(t)= (tg(t))' cos(t) ...

 

[Mark44]

Homework Statement

You have to calculated the Laplace transformation for 1/ cos(t)

Homework Equations

That's all

The Attempt at a Solution

i tryed whit some trigonometric formulas but i don't get anywhere : 1/cos(t) = cos(t) / (1- sin ^2 (t)) or 1/cos(t) = cos(t) + sin(t) x tg(t) or 1/cos(t)= (tg(t))' cos(t) ...

Did you try using the definition of the Laplace Transform?
Note that $\frac 1 {\cos(t)} = \sec(t)$, so using the definition would entail evaluating this integral:
$$ \int_0^\infty \sec(t)e^{-st}dt$$
I haven't attempted doing this integration, so don't know how easy or difficult it would be. Possibly it could be done using integration by parts.

 

[Ray Vickson]

Homework Statement

You have to calculated the Laplace transformation for 1/ cos(t)

Homework Equations

That's all

The Attempt at a Solution

i tryed whit some trigonometric formulas but i don't get anywhere : 1/cos(t) = cos(t) / (1- sin ^2 (t)) or 1/cos(t) = cos(t) + sin(t) x tg(t) or 1/cos(t)= (tg(t))' cos(t) ...

I don't know whether the Laplace transform of your $f(t) = 1/ \cos(t)$ exists in any sense, either as an ordinary function or as a "generalized function". The problem is that $f(t)$ has singularities at $t = (2n+1) \pi/2, n = 0,2,3, \ldots$ because $\cos(t)$ passes through $0$ at those values of $t$. Maybe something like an infinite sum of principal-value integrals will work, but it will not be straightforward at all!

Where did you get this problem? It looks ill-conceived to me.

 

[vela]

Is this even defined? Secant is undefined at odd multiples of $\pi/2$.

 

[banutraul]

I don't know whether the Laplace transform of your $f(t) = 1/ \cos(t)$ exists in any sense, either as an ordinary function or as a "generalized function". The problem is that $f(t)$ has singularities at $t = (2n+1) \pi/2, n = 0,2,3, \ldots$ because $\cos(t)$ passes through $0$ at those values of $t$. Maybe something like an infinite sum of principal-value integrals will work, but it will not be straightforward at all!

Where did you get this problem? It looks ill-conceived to me.

Now i know that the Laplace transform dose'nt exist , there was a diferential ecuation but i solved it without this transformation , thank you

 

[banutraul]

Did you try using the definition of the Laplace Transform?
Note that $\frac 1 {\cos(t)} = \sec(t)$, so using the definition would entail evaluating this integral:
$$ \int_0^\infty \sec(t)e^{-st}dt$$
I haven't attempted doing this integration, so don't know how easy or difficult it would be. Possibly it could be done using integration by parts.

This transformation doesn't exist , i find this on other website , thank you