What is the Laplace transform of : $t^{\gamma-1} F(\alpha,\beta,\delta,t)$, where $\gamma >0 $ and $F$ is the Gauss' hypergeometric function.
Thanks!
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$\begingroup$ My trivial guess that this laplace transform equals a special function. :-D $\endgroup$– AlanCommented Jul 1, 2014 at 13:28
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$\begingroup$ Could you be more specific ? $\endgroup$– tamCommented Jul 1, 2014 at 14:22
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$\begingroup$ Mathematica seems to be able to evaluate this Laplace transform $\endgroup$– NewbieCommented Jul 1, 2014 at 17:12
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$\begingroup$ That's right, but I need the computation method .. $\endgroup$– tamCommented Jul 1, 2014 at 18:27
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$\begingroup$ Have you tried the Euler integral representation of the hypergeometric function? This will assume $\delta-\beta-\gamma>0$. $\endgroup$– Alex R.Commented Jul 1, 2014 at 18:57
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1 Answer
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There is an explicit formula in the book: A.P. Prudnikov, Yu.A. Brychkov, O.I. Marichev. INTEGRALS AND SERIES, Volume 4. Direct Laplace Transforms. GORDON AND BREACH, 1992.
It is on the page 533 and is in terms of $_{2}F_{2}$ hypergeometric function. For special values of parameters for sure it can be simplified using the same book volume 3.
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$\begingroup$ I don't have this book and it seems impossible to find it for free on the internet. Could you please put a capture of page 533. $\endgroup$– tamCommented Jul 4, 2014 at 12:53
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$\begingroup$ OK. Connect me by [email protected] $\endgroup$– SergeiCommented Jul 4, 2014 at 13:00
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1$\begingroup$ Any Idea if I need to compute $\int_{d}^{\infty}e^{-st}t^{\gamma-1}F(\alpha,\beta,\delta,t)dt$ where $d>0$? $\endgroup$– tamCommented Jul 6, 2014 at 13:25