Difficult Inverse Laplace Transform

Question

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.

Answer 1

$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.

Answer 2

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.