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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.

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up vote 2 down vote accepted

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,

$\mathcal{L}^{-1}_{s\to t}\left\{\dfrac{\pi}{2}\dfrac{\log s}{s^2-1}\right\}$

$=\mathcal{L}^{-1}_{s\to t}\left\{\dfrac{\pi}{2}\dfrac{\log s}{s^2\left(1-\dfrac{1}{s^2}\right)}\right\}$

$=\mathcal{L}^{-1}_{s\to t}\left\{\dfrac{\pi}{2}\sum\limits_{n=0}^\infty\dfrac{\log s}{s^{2n+2}}\right\}$

$=\dfrac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=1}^{2n+1}\dfrac{t^{2n+1}}{(2n+1)!k}-\dfrac{\pi}{2}\sum\limits_{n=0}^\infty\dfrac{t^{2n+1}(\gamma+\log t)}{(2n+1)!}$

$=\dfrac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n}\dfrac{t^{2n+1}}{(2n+1)!(k+1)}-\dfrac{\pi(\gamma+\log t)\sinh t}{2}$

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