Find the inverse laplace transforms of $\frac{\ln(s)}{s^2}$

I have the equation $\mathcal{L}^{-1}=\frac{\ln(s)}{s^2}$. How do I find the Laplace inverse transform? I know that $\mathcal{L}[\ln(t)]=-\frac{1}{s}(\gamma+\ln(s))$, so I need to probably transform the equation into that. How do I start, and how do I arrive at the inverse?

Solution 1:

Let $F(s) = \frac{\ln s} s$ and $f(t) = \mathcal L^{-1}\{F\}(t).$ As you have written $$ \mathcal L \{\ln t\}(s) = -\frac \gamma s - F(s)$$ where $\gamma$ is the Euler-Mascheroni constant. Since $\mathcal L\{1\}(s) = \frac 1 s$ it follows that $$F(s) = \mathcal L \{ \ln t+ \gamma \} $$ so $$f(t) =\ln t+ \gamma. $$

Now, by the convolution property of the Laplace transform \begin{align*} \mathcal L^{-1} \bigg \{ \frac{\ln s}{s^2} \bigg \}(t)&= \mathcal L^{-1} \bigg \{ \frac{F(s)}{s} \bigg \}(t) \\ &= \int_0^t f(\tau) \, d\tau \\ &= \int_0^t \big ( \ln \tau+ \gamma \big )\, d\tau \\ &= t \ln t - t+ \gamma t. \end{align*}