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.
Best Answer
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)
$=\dfrac{\pi t}{2}+\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{\pi t^{2n+2k+3}}{4^{n+k+2}(n+k+1)!\left(\dfrac{3}{2}\right)_{n+k+1}(k+1)}+\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{\pi t^{2n+2k+1}}{4^{n+k+1}(n+k)!\left(\dfrac{3}{2}\right)_{n+k}\left(k+\dfrac{1}{2}\right)}-\dfrac{\pi(\gamma+\log t)\sinh t}{2}$
$=\dfrac{\pi t}{2}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{\pi t^{2n+2k+3}(1)_k}{2^{2n+2k+3}3(2)_{n+k}\left(\dfrac{5}{2}\right)_{n+k}(2)_k}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{\pi t^{2n+2k+1}\left(\dfrac{1}{2}\right)_k}{2^{2n+2k+1}(1)_{n+k}\left(\dfrac{3}{2}\right)_{n+k}\left(\dfrac{3}{2}\right)_k}-\dfrac{\pi(\gamma+\log t)\sinh t}{2}$