Solving $x'(t)=x(t-1)$ using Laplace transform! I'm trying to solve 

$$x'(t)=x(t-1)$$

where $x(t)=1$ for $x\in[-1,0]$. I first need to show that $s\bar{x}(s)-1= \frac{1}{s}(1-e^{-s})+e^{-s}\bar{x}(s)$ where $\bar{x}(s)=\int_{0}^{\infty}x(t)e^{-st}dt$ then show that:
$$\bar{x}(s)=\frac{1}{s}+\frac{1}{s^2}\sum_{k=0}^\infty\frac{e^{-ks}}{s^k}$$
I've tried integrating $\bar{x}(s)$ by parts but I can't seem to get the right expression for $s\bar{x}(s)-1$. Please can someone give me a hint? Thanks.
 A: Using integration by parts,
\begin{align*}
  \bar{x}(s) &= \int_0^\infty x(t)e^{-st}\,dt \\
             &= -\frac{1}{s}x(t)e^{-st}\bigg\lvert_0^\infty
                  + \frac{1}{s}\int_0^\infty x'(t)e^{-st}\,dt \\
             &= \frac{1}{s}x(0) 
                  + \frac{1}{s}e^{-s}\int_0^\infty x(t-1)e^{-s(t-1)}\,dt \\
             &= \frac{1}{s} 
                  + \frac{1}{s}e^{-s}\int_{-1}^\infty x(t)e^{-st}\,dt \\
             &= \frac{1}{s}
                  + \frac{1}{s}e^{-s}\int_{-1}^0 x(t)e^{-st}\,dt
                  + \frac{1}{s}e^{-s}\int_0^\infty x(t)e^{-st}\,dt \\
             &= \frac{1}{s}
                  + \frac{1}{s}e^{-s}\int_{-1}^0 e^{-st}\,dt
                  + \frac{1}{s}e^{-s}\int_0^\infty x(t)e^{-st}\,dt \\
             &= \frac{1}{s} + \frac{1}{s}e^{-s}\left(-\frac{1}{s}e^{-st}\right)\bigg\lvert_{-1}^0 + \frac{1}{s}e^{-s}\bar{x}(s) \\
             &= \frac{1}{s} + \frac{1}{s}e^{-s}\left(-\frac{1}{s} + \frac{1}{s}e^{s}\right) + \frac{1}{s}e^{-s}\bar{x}(s).
\end{align*}
Multiply through by $s$ and collect terms, giving
\begin{align*}
  s\bar{x}(s)-1 &= e^{-s}\left(-\frac{1}{s} + \frac{1}{s}e^{s}\right) + e^{-s}\bar{x}(s) \\
      &= \frac{1}{s}(1-e^{-s}) + e^{-s}\bar{x}(s).
\end{align*}
A: 
Obtaining a Series Representation of the Inverse Laplace Transform

Given the expression $s\bar x(s)-1=\frac1s(1-e^{-s})+e^{-s}\bar x(s)$, we find directly that 
$$\begin{align}
\bar x(s)&=\frac{\frac1s(1-e^{-s})+1}{s-e^{-s}}\\\\&=\frac1s\left( \frac{\frac1s(1-e^{-s})+1}{1-\frac{e^{-s}}{s}}\right)\tag 1
\end{align}$$
Note that from $(1)$ we have
$$\bar x(s)-\frac1s =\frac1{s^2}\frac{1}{1-\frac{e^{-s}}{s}}\tag 2$$
Next, we can expand the denominator term $1-\frac{e^{-s}}{s}$ and write
$$\frac1{1-\frac{e^{-s}}{s}}=\sum_{k=0}^\infty \frac{e^{-ks}}{s^k}\tag3$$
Using $(3)$ in $(2)$ reveals
$$\bar x(s)-\frac1s=\frac1{s^2}\sum_{k=0}^\infty \frac{e^{-ks}}{s^k}$$
whence we find that 
$$\bar x(s)=\frac1s+\frac1{s^2}\sum_{k=0}^\infty \frac{e^{-ks}}{s^k}$$
as was to be shown!


NOTE: Inverse Laplace Transform

The inverse Laplace Transform of $\frac1{s^n}$ is $\frac{t^{n-1}}{(n-1)!}u(t)$ and the inverse Laplace Transform of $F(s)e^{-s\tau}$ is $f(t-\tau)u(t-\tau)$.  
Hence, the inverse Laplace Transform of $\frac{e^{-ks}}{s^{k+2}}$ is $\frac{(t-k)^{k+1}}{(k+1)!}u(t-k)$.
Finally, we have
$$x(t)=1+\sum_{k=0}^\infty \frac{(t-k)^{k+1}}{(k+1)!}u(t-k)$$
whereupon shifting the summation index yields
$$\begin{align}
x(t)&=1+\sum_{k=1}^\infty \frac{(t+1-k)^k}{k!}u(t+1-k)\\\\
&=\sum_{k=0}^{\lfloor t+1\rfloor} \frac{(t+1-k)^k}{k!}
\end{align}$$
A: From $$s\bar{x}(s)-1= \frac{1}{s}(1-e^{-s})+e^{-s}\bar{x}(s)$$
you get 
$$s\bar{x}(s)-e^{-s}\bar{x}(s)=1+\frac{1}{s}(1-e^{-s})$$
collect $\bar{x}(s)$
$$(s-e^{-s})\bar{x}(s)=1+\frac{1}{s}(1-e^{-s})$$
$$\color{red}{\bar{x}(s)=\frac{1+s-e^{-s}}{s(s-e^{-s})}}$$
Now prove the second part
$$\sum _{k=0}^{\infty } \frac{e^{-k s}}{s^k}=\sum _{k=0}^{\infty } \left(\frac{e^{- s}}{s}\right)^k=\frac{1}{1-\frac{e^{- s}}{s}}=\frac{s}{s-e^{-s}}$$
Therefore 
$$\bar{x}(s)=\frac{1}{s}+\frac{1}{s^2}\sum _{k=0}^{\infty } \frac{e^{-k s}}{s^k}=\frac{1}{s}+\frac{1}{s^2}\frac{s}{s-e^{-s}}=\frac{1}{s}+\frac{1}{s(s-e^{-s})}=\color{red}{\frac{1+s-e^{-s}}{s(s-e^{-s})}}$$
