Inverse laplace transform of $\ln$ Can someone help me with the following inverse Laplace transform, have not had trouble with any others thus far but this one is catching me

$\mathcal{L}^{-1}\{ \ln(s^3 + s) \} = ?$

 A: Let $F(s) =\log(s^3+s)$  Then, certainly we can write
$$\begin{align}
F(s)&=\log(s^3+s)\\\\
&=3\log(s)+\log\left(1+\frac1{s^2}\right)\\\\
&=3H(s)+G(s)
\end{align}$$
where $H(s)=\log(s)$ and $G(s)=\log\left(1+\frac1{s^2}\right)$.

NOTE:
The inverse Laplace Transform, $\displaystyle \mathscr{L}^{-1}\{H\}(t)$, of $H(s)=\log(s)$ does not exist and therefore neither does the inverse Laplace Transform of $F(s)$. 

To see this, we note that $\lim_{s\to \infty}\log(s)=\infty$ whereas  
$$\lim_{s\to \infty}\int_0^\infty h(t)e^{-st}\,dt=0$$.
We can evaluate, however, the integral given by 
$$I(t)=\int_{\sigma -i\infty}^{\sigma +i\infty}\log(s)e^{st}\,dt$$
Choosing the non-positive real axis as the branch cut, we see that 
$$\begin{align}
I(t)&=\frac{1}{2\pi i}\int_0^{-\infty}e^{tx}(\log(|x|)+i\pi)\,dx+\frac{1}{2\pi i}\int_{-\infty}^0 e^{tx}(\log(|x|)-i\pi)\,dx\\\\
&=\int_0^{-\infty}e^{tx}\,dx\\\\
&=-\frac1t
\end{align}$$
But, the integral $\int_0^\infty \frac{e^{-st}}{t}\,dt$ fails to converge due the singularity at $t=0$.

We can evaluate the inverse Laplace Transform, $\mathscr{L}^{-1}\{G\}(t)$, indirectly by noting that 
$$G'(s)=\color{blue}{\frac{2s}{s^2+1}}-\color{red}{\frac2s}$$
Therefore, we find that 
$$\mathscr{L}^{-1}\{G'\}(t)=\color{blue}{2\cos(t)}-\color{red}{2}$$
Inasmuch as , $-\mathscr{L}\{t\cdot g\}(s)=-\int_0^\infty t\,g(t)\,e^{-st}\,dt=G'(s)$, we find immediately that 
$$\bbox[5px,border:2px solid #C0A000]{G(s)=2\left(\frac{1-\cos(t)}{t}\right)}$$

We can also calculate the inverse Laplace Transform of $G$ directly.  For $t>0$, we have 
$$\begin{align}
\mathscr{L}^{-1}\{G\}(t)&=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}G(s)e^{st}\,ds\\\\
&=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}\log\left(1+\frac1{s^2}\right)e^{st}\,ds\\\\
&=\text{Res}\left(\log\left(1+\frac1{s^2}\right)e^{st},s=\infty\right)\\\\
&=\text{Res}\left(\frac1{s^2}\log\left(1+s^2\right)e^{(1/s)t},s=0\right)\\\\
\end{align}$$
Expanding $\frac1{s^2}\log\left(1+s^2\right)e^{(1/s)t}$ is a Laurent Series around $s=0$ yields
$$\frac1{s^2}\log\left(1+s^2\right)e^{(1/s)t}=\left(\sum_{n=0}^\infty \frac{(-1)^{n}}{n+1}s^{2n}\right)\left(\sum_{m=0}^\infty \frac{t^m}{m!\,s^m}\right)$$
It is straightforward to show that the coefficient, $a_{-1}$, on the term $s^{-1}$ is given by
$$\begin{align}
a_{-1}&=2\sum_{n=0}^\infty \frac{(-1)^{n}t^{2n+1}}{(2n+2)!}\\\\
&=2\left(\frac{1-\cos(t)}{t}\right)
\end{align}$$
as was to be shown!


Putting it all together, we see that 
$$\mathscr{L}^{-1}\{F\}=-\frac{1+2\cos(t)}{t}$$
but $\int_0^\infty \left(\frac{1+2\cos(t)}{t}\right)\,e^{-st}\,dt$ does not exist.  Therefore, $\log(s^3+s)$ is not a Laplace Transform. 

A: The inverse transform of $\ln s$ can be computed as here, and the the inverse transform of $\ln(1 + 1/s^2)$ exists in the sense of ordinary functions.
But $\ln(s^3)$ is not the same as $3\ln s$ on a line $\operatorname{Re} s = \sigma > 0$, because $s^3$ will cross the branch cut of the logarithm. It'll be easier to see the issue if we consider $\ln(-s)$ first and define the Laplace transform in terms of the Fourier transform as
$$\mathcal L_{t \to s}[f] = \mathcal F_{t \to \omega}[f(t) e^{-\sigma t}], \\
s = \sigma + i \omega.$$
It follows that
$$\mathcal L_{s \to t}^{-1}[F] =
e^{\sigma t} \mathcal F_{\omega \to t}^{-1}[F].$$
For $\ln s$, we have
$$\mathcal F_{\omega \to t}^{-1}[\ln (\sigma + i \omega)] =
-t_+^{-1} e^{-\sigma t} - \gamma \delta(t),$$
and, since $\ln s$ and $\ln(-s)$ differ by a piecewise constant on the vertical line,
$$\mathcal F_{\omega \to t}^{-1}[\ln (-\sigma - i \omega)] =
\mathcal F_{\omega \to t}^{-1}[
 \ln (\sigma + i \omega) - i \pi \operatorname{sgn} w ] = \\
-t_+^{-1} e^{-\sigma t} - \gamma \delta(t) + t^{-1},$$
which, when multiplied by $e^{\sigma t}$, will not be independent of $\sigma$; also, the support of $t^{-1}$ is not contained in $[0, \infty)$. For these reasons, we cannot construct a functional the Laplace transform of which would be $\ln(-s)$, or $\ln(s^3)$, or $\ln(s^3 + s)$.
An explanation of the dependence on $\sigma$ in terms of the Bromwich integral would be that if $F(s)$ has a branch cut extending to infinity in the right half-plane, then, even if the integral converges, moving the vertical line to the right will change the value of the integral.
