Difficult Inverse Laplace Transform I've had this question in my exam, which most of my batch mates couldn't solve it.The question by the way is the Laplace Transform inverse of  
$$\frac{\ln s}{(s+1)^2}$$
A Hint was also given, which includes the Laplace Transform of ln t.
 A: $f(s,a) = L(t^a) = \frac{\Gamma(a+1)}{s^{a+1}} $  
Differentiating with respect to a, we get
$L(t^a\cdot lnt) = \frac{\Gamma'(a+1) - \Gamma(a+1)\cdot lns}{s^{a+1}} $
set a = 1.
A: Let $f(t)=\ln t$, then $F(s)=L\{f\}=-\frac{\gamma+\ln s}{s}$. So $\ln s=-sL\{f\}-\gamma$. Let $G(s)=\frac{s}{(s+1)^2}$ and then $g(t)=L^{-1}\{G\}=(t-1)e^{-t}$. Thus
$$ \frac{\ln s}{(s+1)^2}=-L\{f\}\frac{s}{(s+1)^2}-\frac{\gamma}{(s+1)^2}=-F(s)G(s)-\frac{\gamma}{(s+1)^2}. $$
Using
$$ F(s)G(s)=L\{\int_0^tf(\tau)g(t-\tau)d\tau\} $$
one has
\begin{eqnarray}
L^{-1}\{\frac{\ln s}{(s+1)^2}\}&=&-\int_0^tf(t)g(t-\tau)d\tau-\gamma L^{-1}\{\frac{1}{(s+1)^2}\} \\
&=&-\int_0^t\ln\tau(t-\tau-1)e^{-(t-\tau)}d\tau-\gamma te^{-t}\\
&=&e^{-t} (e^t-1-t\text{Chi}(t)-t\text{Shi}(t)).
\end{eqnarray}
Here $\text{Chi}(t), \text{Shi}(t)$ are CoshIntegaral and SinhIntegral.
A: $$x(t) \rightleftharpoons X(s)$$
$$tx(t) \rightleftharpoons -\frac{dX(s)}{ds}$$
$$x_1(t) \rightleftharpoons ln(s)$$
$$tx_1(t) \rightleftharpoons -\frac{1}{s}$$
$$tx_1(t) = -u(t)$$
$$x_1(t) = -\frac{u(t)}{t}$$
$$x_2(t) \rightleftharpoons \frac{1}{(s+2)^2}$$
$$e^{-2t} u(t) \rightleftharpoons \frac{1}{s+2}$$
$$te^{-2t} u(t) \rightleftharpoons \frac{1}{(s+2)^2}$$
$$x_2(t) = te^{-2t}u(t)$$
$$x_1(t)*x_2(t) \rightleftharpoons X_1(s) X_2(s)$$
$$-\frac{u(t)}{t} * te^{-2t}u(t) \rightleftharpoons \frac{ln(s)}{(s+2)^2}$$
