What is the Laplace Inverse Transform of $\ln(s)/(s(s+a))$? I've come across a term $\frac{\ln(s)}{s(s+a)}$ in a transformed differential equation.
I am trying to take the inverse transform of it, but I have no idea how to approach it. 
If anyone knows how to solve this, that would be greatly appreciated.

The derivation from the properties of the convolution would proceed as follows. Let $1/t$ be the distribution defined as
$$\left (\frac 1 t, \phi \right) =
\int_0^1 \frac {\phi(t) - \phi(0)} t dt +
 \int_1^\infty \frac {\phi(t)} t dt.$$
Also, let
$$f(t) = \frac {1 - e^{-at}} a H(t).$$
Then
$$\mathcal L^{-1}[\ln p] = -\frac 1 t - \gamma \delta(t), \\
\mathcal L^{-1}\left[ \frac 1 {p(p+a)} \right] = f(t), \\
\mathcal L^{-1}\left[ \frac {\ln p} {p(p+a)} \right] =
-\frac 1 t * f(t) - \gamma \delta(t) * f(t) = \\
-\left( \frac 1 \tau, f(t - \tau) \right) -
 \gamma(\delta(\tau), f(t - \tau)) = \\
-\int_0^t \frac {f(t - \tau) - f(t)} \tau d\tau -
 \int_1^t \frac {f(t)} \tau d\tau -
 \gamma f(t).$$
And carrying out the integration indeed yields
$$-\frac {e^{-a t} E_1(-a t) + e^{-a t} \ln(-a) + \ln t +\gamma} a,$$
valid for all real $a$, with a removable singularity at zero.
