Laplace transform of a piecewise function

[fluidistic]

Homework Statement

I must calculate the Laplace transform of the following function:
$f(x)=1$ for $x \in [0,1] \cap [2,3] \cap [4,5] \cap$ ... , $f(x)=0$ otherwise.

Homework Equations

The Laplace transform is $\mathbb{L} [f(x)]=\int _0 ^{\infty} e^{-sx}f(x)dx$.

The Attempt at a Solution

I've applied the formula of the definition of the Laplace transform to the function. It gave me $\mathbb{L} [f(x)] = \int _0 ^1 e^{-sx}dx+\int _2 ^3 e^{-sx}dx+... =\left ( -\frac{1}{s} \right ) (e^{-s\cdot 1} - e^{-s\cdot 0}+e^{-3s}-e^{-2s}+...)=\frac{1}{s} \sum _{n=0}^{\infty} (-1)^n e^{-ns}$.
What bothers me is that I don't recognize any closed form for this solution, hence the doubt of weather I'm right so far and if I'm missing any simplification.
Any input will be happily received.

 

[vela]

Homework Statement

I must calculate the Laplace transform of the following function:
$f(x)=1$ for $x \in [0,1] \cap [2,3] \cap [4,5] \cap$ ... , $f(x)=0$ otherwise.

Homework Equations

The Laplace transform is $\mathbb{L} [f(x)]=\int _0 ^{\infty} e^{-sx}f(x)dx$.

The Attempt at a Solution

I've applied the formula of the definition of the Laplace transform to the function. It gave me $\mathbb{L} [f(x)] = \int _0 ^1 e^{-sx}dx+\int _2 ^3 e^{-sx}dx+... =\left ( -\frac{1}{s} \right ) (e^{-s\cdot 1} - e^{-s\cdot 0}+e^{-3s}-e^{-2s}+...)=\frac{1}{s} \sum _{n=0}^{\infty} (-1)^n e^{-ns}$.
What bothers me is that I don't recognize any closed form for this solution, hence the doubt of weather I'm right so far and if I'm missing any simplification.
Any input will be happily received.

I think you mean $x \in [0,1] \cup [2,3] \cup \cdots$. According to what you wrote, f(x)=0 for all x, so F(s)=0.

You can write
$$\sum_{n=0}^\infty (-1)^n e^{-ns} = \sum_{n=0}^\infty (-e^{-s})^n.$$ You should recognize the form of that series.

 

[fluidistic]

I think you mean $x \in [0,1] \cup [2,3] \cup \cdots$. According to what you wrote, f(x)=0 for all x, so F(s)=0.

Yes.

You can write
$$\sum_{n=0}^\infty (-1)^n e^{-ns} = \sum_{n=0}^\infty (-e^{-s})^n.$$ You should recognize the form of that series.

I see, so the answer would be $\left ( \frac{1}{s} \right ) \left ( \frac{1}{1+e^{-s}} \right )$ right?
Thanks a million for your help!