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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 $\endgroup$ – Alan Jul 1 '14 at 13:28
  • $\begingroup$ Could you be more specific ? $\endgroup$ – tam Jul 1 '14 at 14:22
  • $\begingroup$ Mathematica seems to be able to evaluate this Laplace transform $\endgroup$ – Newbie Jul 1 '14 at 17:12
  • $\begingroup$ That's right, but I need the computation method .. $\endgroup$ – tam Jul 1 '14 at 18:27
  • $\begingroup$ Have you tried the Euler integral representation of the hypergeometric function? This will assume $\delta-\beta-\gamma>0$. $\endgroup$ – Alex R. Jul 1 '14 at 18:57
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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. $\endgroup$ – tam Jul 4 '14 at 12:53
  • $\begingroup$ OK. Connect me by mathsms@yandex.ru $\endgroup$ – Sergei Jul 4 '14 at 13:00
  • $\begingroup$ Any Idea if I need to compute $\int_{d}^{\infty}e^{-st}t^{\gamma-1}F(\alpha,\beta,\delta,t)dt$ where $d>0$? $\endgroup$ – tam Jul 6 '14 at 13:25

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