Inverse Laplace Transform with Division and Integration

[_N3WTON_]

Homework Statement

Division by s Equals integration by t:
For this problem use the following property (see relevant equations) to find the inverse transform of the given function: $F(s) = \frac{1}{s(s-1)}$

Homework Equations

$L^{-1}(\frac{F(s)}{s}) = \int_{0}^{t} f(\tau)\,d \tau$

The Attempt at a Solution

Using the above formula I have: $F(s) = \frac{1}{s(s-1)}$
So : $\frac{F(s)}{s} = \frac{\frac{1}{s(s-1)}}{s} = \frac{1}{s^{2}(s-1)}$
Next I used a partial fraction decomposition:
$\frac{1}{s^{2}(s-1)} = \frac{A}{s} + \frac{B}{s^{2}} + \frac{C}{(s-1)}$
$As(s-1) + B(s-1) + Cs^{2} = 1$
Solving for A, B, and C I found $A=-1 \hspace{1 mm} B=-1 \hspace{1 mm} C = 1$
So the decomposition is:
$\frac{1}{s^{2}(s-1)} = -\frac{1}{s} - \frac{1}{s^{2}} + \frac{1}{(s-1)}$
The inverse Laplace transform of the above equation is:
$-t - e^{t} - 1$
However, the answer given in my textbook is:
$e^{t} - 1$
Coincidentally, when I perform the partial fraction decomposition on $F(s)$, I arrive at the answer given in the back of the text, which is really confusing me, do I not need to divide $F(s)$ by $s$?

 

[LCKurtz]

Homework Statement

Division by s Equals integration by t:
For this problem use the following property (see relevant equations) to find the inverse transform of the given function: $F(s) = \frac{1}{s(s-1)}$

Homework Equations

$L^{-1}(F(s)) = \int_{0}^{t} f(\tau)\,d \tau$

The Attempt at a Solution

Using the above formula I have: $F(s) = \frac{1}{s(s-1)}$

You have stated the formula wrong. The formula you want is$$
\mathcal L^{-1}\frac 1 s F(s) = \int_0^t f(\tau)~d\tau$$where $f$ is the inverse of $F$. In your problem $F(s) =\frac 1 {s-1}$. Informally, this says if you suppress a factor of $\frac 1 s$ you must integrate the result to get the original inverse.

 

[_N3WTON_]

You have stated the formula wrong. The formula you want is$$
\mathcal L^{-1}\frac 1 s F(s) = \int_0^t f(\tau)~d\tau$$where $f$ is the inverse of $F$. In your problem $F(s) =\frac 1 {s-1}$. Informally, this says if you suppress a factor of $\frac 1 s$ you must integrate the result to get the original inverse.

ah, thank you very much, the problem makes much more sense now. I am blaming my book though, the formula they provided isn't very clear, they formatted it in the same way that I did above...although to be fair I guess I just completely misinterpreted the problem :)