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

Question

What is the Laplace transform of : $t^{\gamma-1} F(\alpha,\beta,\delta,\frac{t}{d})$, where $\gamma >0 $ and $F$ is the Gauss' hypergeometric function.

Note that I have the Laplace transform of : $t^{\gamma-1} F(\alpha,\beta,\delta,-t)$.

Thanks!

Answer 1

Since $F\left( {\alpha ,\beta ,\delta ;t} \right) = \sum {\frac{{\left( \alpha \right)_n \left( \beta \right)_n }}{{n!\left( \delta \right)_n }}t^n } $. Assuming the uniform convergence of the series, then term by term integration yields that \begin{align} \int_0^\infty {e^{ - st} } t^{\gamma - 1} F\left( {\alpha ,\beta ,\delta ;t} \right)dt \\ &= \int_0^\infty {e^{ - st} } t^{\gamma - 1} \left( {\sum {\frac{{\left( \alpha \right)_n \left( \beta \right)_n }}{{n!\left( \delta \right)_n }}t^n } } \right)dt \\ &= \sum {\frac{{\left( \alpha \right)_n \left( \beta \right)_n }}{{n!\left( \delta \right)_n }}\left( {\int_0^\infty {e^{ - st} } t^{n + \gamma - 1} dt} \right)} \\ &= \sum {\frac{{\left( \alpha \right)_n \left( \beta \right)_n }}{{n!\left( \delta \right)_n }}\frac{{\Gamma \left( {n + \gamma } \right)}}{{s^{n + \gamma } }}} \end{align} where $$ \int_0^\infty {e^{ - st} } t^{n + \gamma - 1} dt = \frac{{\Gamma \left( {n + \gamma } \right)}}{{s^{n + \gamma } }} ,$$ valid if $n+\gamma-1>-1$. To write the answer in more general form, we may use the generalized Hypergeometric function as follows:$$ _p F_q \left( {a_1 ,a_2 , \ldots ,a_p ;b_1 ,b_2 , \ldots ,b_q ;t} \right) = \sum\limits_{n = 0}^\infty {\frac{{\left( {a_1 } \right)_n \left( {a_2 } \right)_n \cdots \left( {a_p } \right)_n }}{{\left( {b_1 } \right)_n \left( {b_2 } \right)_n \cdots \left( {b_q } \right)_n }}\frac{{t^n }}{{n!}}} ,$$ so in our case we have $ p = 3,q = 1,a_1 = \alpha ,a_2 = \beta ,a_3 = \gamma ,b_1 = \delta ,t = \frac{1}{s} $ and since $ \left( \alpha \right)_n = \frac{{\Gamma \left( {\alpha + n} \right)}}{{\Gamma \left( \alpha \right)}} $, then we may write the above series as: \begin{align} \sum {\frac{{\left( \alpha \right)_n \left( \beta \right)_n }}{{n!\left( \delta \right)_n }}\frac{{\Gamma \left( {n + \gamma } \right)}}{{s^{n + \gamma } }}} &= \frac{{\Gamma \left( \delta \right)}}{{\Gamma \left( \alpha \right)\Gamma \left( \beta \right)}} \sum {\frac{{\Gamma \left( {\alpha + n} \right)\Gamma \left( {\beta + n} \right)}}{{n!\Gamma \left( {\delta + n} \right)}}\frac{{\Gamma \left( {n + \gamma } \right)}}{{\Gamma \left( \gamma \right)}}\frac{{\Gamma \left( \gamma \right)}}{{s^{n + \gamma } }}} \\ & = \frac{{\Gamma \left( \delta \right)\Gamma \left( \gamma \right)}}{{\Gamma \left( \alpha \right)\Gamma \left( \beta \right)}}s^\gamma \sum {\frac{{\Gamma \left( {\alpha + n} \right)\Gamma \left( {\beta + n} \right)}}{{n!\Gamma \left( {\delta + n} \right)}}\frac{1}{{s^n }}} \\ &= \frac{{\Gamma \left( \delta \right)\Gamma \left( \gamma \right)}}{{\Gamma \left( \alpha \right)\Gamma \left( \beta \right)}}s^\gamma \cdot {}_3 F_1 \left( {\alpha ,\beta ,\gamma ;\delta ;\frac{1}{s}} \right) \end{align}

Answer 2

Although M. W. Alomari's result is correct, the reasoning is critically flawed.

First, the power series representation for the Gauss hypergeometric function F (also denoted $_2F_1$) is not defined (or is infinity) when neither $\alpha$ nor $\beta$ is a negative integer and $|t|>|d|$. As the Laplace transform consists in integrating along the positive half-line, an undefined function (except for the [0,|d|] segment) is being integrated in M. W. Alomari's answer; this does not make sense.

Second, generalized hypergeometric functions $_pF_q$ just do not converge as power series for $p > q+1$ (see [1]).

There is one exception for polynomials, when at least one of the first two parameters is a negative integer. Then $_3F_1$ as a series makes sense as a polynomial, as the series terminates.

Otherwise you must use Mellin-Barnes integrals along a Bromwich path, or at least a Euler-type integral representations, see [2]. You will arrive at the same result, with different intermediate steps (basically the same argument comes into play, except that you have integrals instead of series).

Then if you want to use generalized hypergeometic power series, $_3F_1$ may be expressed as a combination of confluent hypergeometric functions of the first kind and a hypergeometric function $_2F_2$ in the simple case of $\gamma=1$ and $d=1$. Otherwise you need three $_2F_2$ functions. These two kinds of hypergeometric functions have power series representations, unlike $_3F_1$.

As an incidental note, one has to suppose that $\delta$ is not a negative or null integer for the question to make sense. With this extra assumption, the result given by Mathematica is:

$$ \frac{(1-\delta) \left[\, _{2}F_{2}(1,2-\delta ;2-\alpha ,2-\beta ;-s)\right]}{(\alpha -1) (\beta -1)}+\frac{\pi \Gamma (\delta ) \left(\frac{e^{-i \pi \beta } \csc(\pi \beta ) s^{\beta } \Gamma (\alpha -\beta ) \, _{1}F_{1}(\beta -\delta +1;-\alpha +\beta +1;-s)}{\Gamma (\delta -\beta )}+\frac{e^{-i \pi \alpha } \csc(\pi \alpha ) s^{\alpha } \Gamma (\beta -\alpha ) \, _{1}F_{1}(\alpha -\delta +1;\alpha -\beta +1;-s)}{\Gamma (\delta -\alpha )}\right)}{s \Gamma (\alpha ) \Gamma (\beta )} $$ if:
$$\\ \gamma = 1 \land d = 1 \land (\alpha,\beta) \notin \mathbb{Z}^2 \land (\delta-\alpha,\delta-\beta) \notin \mathbb{Z^-}^2 \land \Re(\alpha +\beta -\delta) <1 \land \space\space \Re(s) > 0. $$ and in the general case: $$ \Gamma (\delta ) \left(\frac{(-d)^{\alpha } \Gamma (\beta -\alpha ) \Gamma (\gamma -\alpha ) s^{\alpha -\gamma } \, _2F_2(\alpha ,\alpha -\delta +1;\alpha -\beta +1,\alpha -\gamma +1;-d s)}{\Gamma (\beta ) \Gamma (\delta -\alpha )}+\frac{\frac{(-d)^{\beta } \Gamma (\alpha -\beta ) \Gamma (\gamma -\beta ) s^{\beta -\gamma } \, _2F_2(\beta ,\beta -\delta +1;-\alpha +\beta +1,\beta -\gamma +1;-d s)}{\Gamma (\delta -\beta )}+\frac{\Gamma (\gamma ) (-d)^{\gamma } \Gamma (\alpha -\gamma ) \Gamma (\beta -\gamma ) \, _2F_2(\gamma ,\gamma -\delta +1;-\alpha +\gamma +1,-\beta +\gamma +1;-d s)}{\Gamma (\beta ) \Gamma (\delta -\gamma )}}{\Gamma (\alpha )}\right) $$ if:
$$ (\Re(d)<0\lor d\notin \mathbb{R})\land \Re(\gamma )>0 \land (\alpha,\beta, \gamma) \notin \mathbb{Z}^3 \land (\delta-\alpha,\delta-\beta, \delta-\gamma) \notin \mathbb{Z^-}^3 \land \space\space \Re(s) > 0 $$ There is no such thing as a "closed-form" solution without some generalized hypergeometric functions around, unless you want to use the even more sophisticated Merjer-G or Fox-H functions, but I guess this is not what you want.

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$[1]$ NIST Handbook of Mathematical Functions, Cambridge University Press, Olver and alii, $2010$, p. $404$, $16.2$(iv)

$[2]$ NIST p. $408$, §$16.5$, or Slater's classic book Generalized Hypergeometric Series, Cambridge University Press, $1966$, pages $22$ to $25$