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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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1 Answer 1

up vote 1 down vote accepted

With the result of http://eqworld.ipmnet.ru/en/auxiliary/inttrans/LapInv7.pdf,

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

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

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

$=\dfrac{\pi}{2}\left(\cosh t*(-\log t-\gamma)\right)$

$=\dfrac{\pi}{2}\int_0^t\cosh(t-u)(-\log u-\gamma)~du$

$=-\dfrac{\gamma\pi}{2}\int_0^t\cosh(t-u)~du-\dfrac{\pi}{2}\int_0^t\cosh(t-u)\log u~du$

$=\dfrac{\gamma\pi}{2}[\sinh(t-u)]_0^t-\dfrac{\pi}{2}\int_0^t(\cosh t\cosh u-\sinh t\sinh u)\log u~du$

$=-\dfrac{\gamma\pi\sinh t}{2}+\dfrac{\pi\sinh t}{2}\int_0^t\sinh u\log u~du-\dfrac{\pi\cosh t}{2}\int_0^t\cosh u\log u~du$

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