Inverse Laplace transform of s/s-1 Finding the inverse laplace transform:
$$L^{-1}\left\{\frac{s}{s-1}\right\}$$
I wrote:
$$L^{-1}\left\{\frac{s}{s-1}\right\}=L^{-1}\left\{\frac{1}{s-1}\right\} + L^{-1}\{1\}=L^{-1}\{1\} + e^{t}$$ 
And i don't know how to complete, what is the inverse Laplace of 1? And is another method  you can provide me with?  Please, I need help :) 
 A: There is another way of looking at this that removes the mystery around the delta function, i.e., why do we need to introduce it in the first place?  This reason may not be all that obvious to those working from tables.
The clue is that $s/(s-1)$ doesn't vanish as $|s| \to \infty$.  That alone should put the problem solver on some alert regarding the existence of nicely-behaved analytic functions in the inverse transform.
However, let's do some analysis to see where we can derive a delta function without knowing that we need to do a partial fraction decomposition to properly obtain the inverse LT.  (This is going to involve complex integration, so if I am going beyond the OP's original intent, please understand that this is for the greater M.SE public.)  Consider the integral
$$\oint_C dz \frac{z}{z-1} e^{z t} $$
where $t \gt 0$ and $c$ consists of the line $[c-i \sqrt{R^2-c^2},c+ i \sqrt{R^2-c^2}]$ and the circular arc $R e^{i \theta}$, $\theta \in \left [ \frac{\pi}{2} - \arcsin{\frac{c}{R}},\frac{3\pi}{2} + \arcsin{\frac{c}{R}} \right ]$, as illustrated in the Figure below.

The contour integral is thus equal to
$$\int_{c-i \sqrt{R^2-c^2}}^{c+i \sqrt{R^2-c^2}} ds \, \frac{s}{s-1} e^{s t} + i R \int_{\frac{\pi}{2} - \arcsin{\frac{c}{R}}}^{\frac{3\pi}{2} + \arcsin{\frac{c}{R}}} d\theta \, e^{i \theta} \frac{R e^{i \theta}}{R e^{i \theta}-1} e^{R t e^{i \theta}}$$
The delta function will emerge from this second integral.  
The contour integral is also equal to $i 2 \pi$ times the residue of the integrand at the pole $z=1$, or $e^t$.  Thus,
$$\begin{align}\int_{c-i \sqrt{R^2-c^2}}^{c+i \sqrt{R^2-c^2}} ds \, \frac{s}{s-1} e^{s t} &= i 2 \pi e^t - i R \int_{\frac{\pi}{2} - \arcsin{\frac{c}{R}}}^{\frac{3\pi}{2} + \arcsin{\frac{c}{R}}} d\theta \, e^{i \theta} \frac{R e^{i \theta}}{R e^{i \theta}-1} e^{R t e^{i \theta}}\\ &= i 2 \pi e^t - i R \int_{\frac{\pi}{2} - \arcsin{\frac{c}{R}}}^{\frac{3\pi}{2} + \arcsin{\frac{c}{R}}} d\theta \, e^{i \theta} \frac{e^{R t e^{i \theta}}}{1-\frac1{R e^{i \theta}}}\end{align}$$
Now take the limit as $R \to \infty$ and shift the integral limits by $\pi/2$.  We may then write down the ILT:
$$\begin{align}\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, \frac{s}{s-1} e^{s t} &= e^t + \lim_{R \to \infty}\frac{R}{i 2 \pi} \int_0^{\pi} d\theta \, e^{i \theta} \, e^{i R t e^{i \theta}}\\ &= e^t +\lim_{R \to \infty}\frac{R}{i 2 \pi} (-i) \int_0^{\pi} d(e^{i \theta}) \, e^{i R t e^{i \theta}}\\ &=  e^t +\lim_{R \to \infty}\frac{R}{i 2 \pi} (-i) \frac1{i R t} \left [e^{i R t e^{i \theta}} \right ]_0^{\pi} \\ &= e^t+\lim_{R \to \infty}\frac{R}{i 2 \pi} (-i) \frac1{i R t} (-i 2 \sin{R t}) \\ &= e^t +\lim_{R \to \infty} \frac{\sin{R t}}{\pi t} \end{align} $$
The latter term on the RHS is equal to $\delta(t)$.
Although this seemed like a chore, it does illustrate how the delta function emerges from the LT.  
A: Here's an proof why is $\ L \left\{\delta (t)\right\} =1 $
Defined as $\begin{cases} \delta(t) = \infty &when \; t=0 \\\delta(t)=0 &when \;t\neq 0 \end{cases}$
Now, you must be thinking this is indeed a bizarre looking function, infact it's in exact, is not even a function. :)
Now how we will take the laplace transform of this function even? 
Let us define an neighbour hood of $0$ defined in $t \in [-\epsilon,\epsilon]$ 
as $\epsilon \to 0$ . 
Now by the definition of Laplace transform.
$\int^\infty_0 e^{-st} \delta (t) dt$ is our required Laplace transform.  
Now there's a thing you will need some logic here, think about the dirac delta function here's it's $0$ everywhere except the point $t=0$ hence our integral is only defined at. 
$\lim_{\epsilon \to 0}\int^{0+\epsilon}_{0}e^{-st}\delta(t) dt$ 
Now in such an small range the function $e^{-st}$ is essentially constant and you can safely factor it at evaluated at $0$.  
$\lim_{\epsilon \to 0}e^{0}\int^{0+\epsilon}_{0}\delta(t) dt = \lim_{\epsilon \to 0}\int^{0+\epsilon}_{0}\delta(t) dt$  
And now another trouble is how to find this bizarre integral of $\delta(t)$ well i will give you the result for now which is $1$ amazingly. But if you need proof. Comment and i will add. 
