I'm trying to find out whether there is a way to get this into the s domain? $$\mathcal{L} \left\{ 1 / \log \ x\right\}$$ (1)
now I know using Laplace Tranforms I can get $$\mathcal{L} \left\{ \log x\right\}=-\frac{1}{s}\left(\log s + \gamma\right)$$ (2)
but the reciprocal of log is causing a problem since $$\log_bc=\frac{1}{\log_cb}$$ simply changing the base.
Is there any trick one can use? I tried getting it all in terms of log but I could not get that to work.
answer is : $\exp(-1/s)$, use the Feynman integration, you will get the integral of $\dfrac {t! }{ s ^ t}$, use the Melin transform and substitute (after calculating the integral) $t = -1$