Ok so I have recently found a transform that produced.
$$x\left( s \right) = \frac{\pi }{2}\frac{{\log s}}{{{s^2} - 1}}$$
However the function was given in an integral parametric form so to call it, (i.e. an integral depending on a parameter) so I want to express the integral in a closed form using the inverse transformation. Since I have proven
$$\mathcal{L}\left( {\log t} \right)\left( s \right) = - \frac{{\gamma + \log s}}{s}$$ where $\gamma$ is Euler's constant. I wrote the following:
$$x\left( s \right) = \frac{\pi }{2}\frac{s}{{{s^2} - 1}}\frac{{\gamma + \log s}}{s} - \frac{\pi }{2}\frac{\gamma }{{{s^2} - 1}}$$
Thus taking the inverse Laplace produces
$$x\left( s \right) = - \frac{\pi }{2}\cosh t * \log t - \frac{\pi }{2}\gamma \sinh t$$
Where $ * $ denotes convolution. I'm guessing I can solve the convolution by splitting the hyperbolic cosine into exponentials, but I'd like to know if anyone can give me a nice straight forward method to solve this. Thanks in advance.