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 August 2nd, 2013, 09:46 AM #1 Member   Joined: Nov 2010 Posts: 77 Thanks: 0 inverse laplace transform could you find the $\mathcal{L}_s^{-1}\left[\frac{\cos \left(\frac{1}{s}+1\right)}{s}\right](x)$ thanks
August 2nd, 2013, 10:03 PM   #2
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Re: inverse laplace transform

Hi !
The inverse Laplace is expressed in terms of Kelvin's functions.
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 August 3rd, 2013, 12:40 AM #3 Member   Joined: Nov 2010 Posts: 77 Thanks: 0 Re: inverse laplace transform thanks good work
 August 3rd, 2013, 12:56 AM #4 Member   Joined: Nov 2010 Posts: 77 Thanks: 0 Re: inverse laplace transform hi again could you calculate $\mathcal{L}_s^{-1}\left[\frac{s \log \left(\Gamma \left(\frac{s-1}{s}\right)\right)-\gamma }{s^2}\right](x)$ Thanks.
August 3rd, 2013, 07:04 AM   #5
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Re: inverse laplace transform

Below, expressed on the form of infinite series.
I don't know if a close form exists.
Certainely, also it could be expressed on the form of an integral (a rather complicated one, with exponential and Bessel function)
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 August 3rd, 2013, 02:30 PM #6 Member   Joined: Nov 2010 Posts: 77 Thanks: 0 Re: inverse laplace transform well done

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