Inverse laplace transform of $\frac{1}{s^3(s-1)}$

Question

I have a problem with this transform. I know that using this theorem (see the theroem) I can get the answer, and i have the resolution. But I don't understand why the put a $-1$ in the integral.

IMG with the theorem

After know this they do:

Transform to integral I don't know why they put the -1 in that integral.

Answer 1

See that you have to apply the Inverse Laplace of $1/(s^2(s-1))$ and then plug that into the integral. So we have that with partial fractions:

$$\frac{1}{s^2(s-1)}=\frac{1}{s-1}-\frac{1}{s^2}-\frac 1s $$

So, $$\mathcal{L}^{-1}\left[\frac{1}{s^2(s-1)}\right ](t) =e^{t}-t-1, \space t>0$$

Answer 2

With the theorem we can say $${\cal L}^{-1}\left(\dfrac{F(s)}{s^3}\right)=\int_0^t\int_0^t\int_0^t{\cal L}^{-1}F(s)\ dt\ dt\ dt$$ then \begin{align} {\cal L}^{-1}\left(\dfrac{\frac{1}{s-1}}{s^3}\right) &= \int_0^t\int_0^t\int_0^t{\cal L}^{-1}\frac{1}{s-1}\ dt\ dt\ dt \\ &= \int_0^t\int_0^t\int_0^t e^t \ dt\ dt\ dt \\ &= \int_0^t\int_0^t e^t\Big|_0^t \ dt\ dt \\ &= \int_0^t\int_0^t e^t-1 \ dt\ dt \\ &= \int_0^t e^t-t\Big|_0^t\ dt \\ &= \int_0^t e^t-t-1\ dt \\ &= e^t-\dfrac12t^2-t\Big|_0^t \\ &= \color{blue}{e^t-\dfrac12t^2-t-1} \end{align}