Solving inverse Laplace Transform with convolution theorem. 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.
 A: Unfortunately, the convolution in above cannot directly solve as it contains some divergent integrals, so you should consider on this approach instead.
With the result of http://eqworld.ipmnet.ru/en/auxiliary/inttrans/LapInv7.pdf,
$\mathcal{L}^{-1}_{s\to t}\left\{\dfrac{\pi}{2}\dfrac{\log s}{s^2-1}\right\}$
$=\mathcal{L}^{-1}_{s\to t}\left\{\dfrac{\pi}{2}\dfrac{\log s}{s^2\left(1-\dfrac{1}{s^2}\right)}\right\}$
$=\mathcal{L}^{-1}_{s\to t}\left\{\dfrac{\pi}{2}\sum\limits_{n=0}^\infty\dfrac{\log s}{s^{2n+2}}\right\}$
$=\dfrac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=1}^{2n+1}\dfrac{t^{2n+1}}{(2n+1)!k}-\dfrac{\pi}{2}\sum\limits_{n=0}^\infty\dfrac{t^{2n+1}(\gamma+\log t)}{(2n+1)!}$
$=\dfrac{\pi t}{2}+\dfrac{\pi}{2}\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{t^{2n+1}}{(2n+1)!2k}+\dfrac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{t^{2n+1}}{(2n+1)!(2k+1)}-\dfrac{\pi(\gamma+\log t)\sinh t}{2}$
$=\dfrac{\pi t}{2}+\sum\limits_{k=1}^\infty\sum\limits_{n=k}^\infty\dfrac{\pi t^{2n+1}}{4^{n+1}n!\left(\dfrac{3}{2}\right)_nk}+\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{\pi t^{2n+1}}{4^{n+1}n!\left(\dfrac{3}{2}\right)_n\left(k+\dfrac{1}{2}\right)}-\dfrac{\pi(\gamma+\log t)\sinh t}{2}$ (according to http://mathworld.wolfram.com/PochhammerSymbol.html)
$=\dfrac{\pi t}{2}+\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{\pi t^{2n+2k+3}}{4^{n+k+2}(n+k+1)!\left(\dfrac{3}{2}\right)_{n+k+1}(k+1)}+\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{\pi t^{2n+2k+1}}{4^{n+k+1}(n+k)!\left(\dfrac{3}{2}\right)_{n+k}\left(k+\dfrac{1}{2}\right)}-\dfrac{\pi(\gamma+\log t)\sinh t}{2}$
$=\dfrac{\pi t}{2}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{\pi t^{2n+2k+3}(1)_k}{2^{2n+2k+3}3(2)_{n+k}\left(\dfrac{5}{2}\right)_{n+k}(2)_k}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{\pi t^{2n+2k+1}\left(\dfrac{1}{2}\right)_k}{2^{2n+2k+1}(1)_{n+k}\left(\dfrac{3}{2}\right)_{n+k}\left(\dfrac{3}{2}\right)_k}-\dfrac{\pi(\gamma+\log t)\sinh t}{2}$
