# Laplace transform of : $t^{\gamma-1} F(\alpha,\beta,\delta,t)$, where $F$ is the Gauss' hypergeometric function

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!

• My trivial guess that this laplace transform equals a special function. :-D – Alan Jul 1 '14 at 13:28
• Could you be more specific ? – tam Jul 1 '14 at 14:22
• Mathematica seems to be able to evaluate this Laplace transform – Newbie Jul 1 '14 at 17:12
• That's right, but I need the computation method .. – tam Jul 1 '14 at 18:27
• Have you tried the Euler integral representation of the hypergeometric function? This will assume $\delta-\beta-\gamma>0$. – Alex R. Jul 1 '14 at 18:57

## 1 Answer

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.

• 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. – tam Jul 4 '14 at 12:53
• OK. Connect me by mathsms@yandex.ru – Sergei Jul 4 '14 at 13:00
• Any Idea if I need to compute $\int_{d}^{\infty}e^{-st}t^{\gamma-1}F(\alpha,\beta,\delta,t)dt$ where $d>0$? – tam Jul 6 '14 at 13:25