inverse Laplace transformation of s in two way I got two results of inverse Laplace Transformation of $s$
1.
$L(tf(t)) = -{d \over ds}(L(f(t)))$
2.
$L(f'(t)) = sL(f(t)) - y(0)$0
I got $f(t)= - { \delta(t) \over t}$ using the first method and $f(t)= {d \over dt} \delta(t)$ using the second.
Why have I got two results? What have I overlooked?
 A: Those two expressions, $\delta'(t)$ and $-\delta(t)/t$, are equivalent.  Here is an explanation.

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function with compact support that is differentiable at 0.  We will show that 
$$
  \int_{-\infty}^{\infty} tf(t)\delta'(t) dt = -\int_{-\infty}^{\infty} f(t) \delta(t) dt
$$
or in other words, $t\delta'(t) = -\delta(t)$.  Let $F$ be an antiderivative of $f$ (not the Laplace transform).  Since 
$$
 (tF\delta')' = F\delta' + tf\delta' + tF\delta'',
$$
we can say 
$$
  \begin{align}
  \int_{-\infty}^{\infty} tf(t)\delta'(t) dt &= \left.tF(t)\delta'(t)\right|_{-\infty}^{\infty} -
  \int_{-\infty}^{\infty}F(t)\delta'(t)dt - \int_{-\infty}^{\infty} tF(t)\delta''(t)dt \\
  &= f(0) - (tF)''(0) \\
  &= -f(0) \\
  &= -\int_{-\infty}^{\infty} f(t)\delta(t) dt.
  \end{align}
$$
The $tF(t)\delta'(t)$ term is zero because $F$ and $\delta'$ are identically zero for large $|t|$.  The $(tF)''(0)$ term is $2f(0)$ since $(tF)'' = 2f + tf'$.  


Let $g(t) = tf(t)$.  Then 
$$
  \int_{-\infty}^{\infty} g(t)\delta'(t) dt = \int_{-\infty}^{\infty} -\frac{g(t)}{t} \delta(t) dt.
$$
So $\delta'(t) = -\delta(t)/t$ for test functions $g$ that are twice differentiable at 0 with $g(0)=0$.

