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Let $\alpha>1$ and $z>0$. I am looking for a solution to the following Laplace transofrm: \begin{equation} \alpha\cdot\mathcal{L}\left\{{_{2}F_{1}(1-\alpha,1;2;-t)}\right\}(z)=\mathcal{L}\left\{\frac{(1+t)^\alpha-1}{t}\right\}(z) =\int_{0}^{\infty}\frac{(1+t)^\alpha-1}{t}\,e^{-zt}\,\mathrm{d}t \end{equation} I found a solution to the Laplace transform of ${_{2}F_{1}}(-t)$ here but it does not seems to work.

I also found this (Gradshteyn & Ryzhik, ed. 8, #7.522.1): \begin{equation} \int_{0}^{\infty}e^{-\lambda x}x^{\gamma-1}{_{2}F_1}(\alpha,\beta;\delta;-x)\,\mathrm{d}x=\frac{\Gamma(\delta)\lambda^{-\gamma}}{\Gamma(\alpha)\Gamma(\beta)}E(\alpha\,;\beta:\gamma\,;\delta:\lambda), \end{equation} for $\Re \lambda>0$ and $\Re \gamma>0$ where $E(p\,;\alpha_{r}:q\,;\varrho_{s}:x)$ is the MacRobert E-Function. Apparently, the E-function canbe expressed as MeijerG; however, I am having trouble understanding how to do so. In the Table of Integral Transforms (vol. 1, pg. 384) we have \begin{equation} E(p\,;\alpha_{r}:q\,;\beta_{s}:x)= G_{\,q+1,p}^{\,p,1}\left(x\,\middle\vert \begin{array}{c} 1,\beta_1,\dots,\beta_q\\ \alpha_1,\dots,\alpha_p \end{array} \right). \end{equation}

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You can convert the hypergeometric function to the Meijer G-function and use the formula for the integral of a product of two linear G-functions: $$\alpha \int_0^\infty {_2F_1}(1 - \alpha, 1; 2; -t) e^{-z t} dt = \\ \frac \alpha {\Gamma(1 - \alpha)} \int_0^\infty G_{2, 2}^{1, 2} \left(t \middle| {0, \alpha \atop 0, -1} \right) G_{0, 1}^{1, 0} \left(z t \middle| {- \atop 0} \right) dt = \\ \frac \alpha {\Gamma(1 - \alpha)} G_{2, 3}^{3, 1} \left(z \middle| {0, 1 \atop 0, 0, -\alpha} \right).$$ In terms of more elementary function, it'll be $$\frac z {1 + \alpha} \,{_2F_2}(1, 1; 2, 2 + \alpha; z) + \Gamma(\alpha + 1) z^{-\alpha} (-z)^\alpha \Gamma(-\alpha, -z) + \\ \ln z + \pi \csc (\pi \alpha) z^{-\alpha} (-z)^\alpha - \psi(\alpha) - \pi \cot \pi \alpha - \frac 1 \alpha,$$ $\alpha>0, \operatorname{Re} z > 0$. For integral $\alpha_0$, this is still valid as a limit, with the singular terms yielding $$\lim_{\alpha \to \alpha_0}( \pi \csc (\pi \alpha) z^{-\alpha} (-z)^\alpha - \pi \cot \pi \alpha) = \ln(-z) - \ln z.$$

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  • $\begingroup$ How did you convert the G function to more elementary functions? Was it done in mathematica? Or do you have an identity to reference? $\endgroup$ – Aaron Hendrickson Jun 17 '18 at 23:38
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    $\begingroup$ To derive this identity, write the G-function by definition as the integral of $\Gamma(1+y) \Gamma(-y)^2 \Gamma(-\alpha-y) \Gamma(1-y)^{-1} z^y$. The integral has to be taken over the right loop. When $\alpha$ isn't an integer, the sum of the residues over the simple poles at $y=-\alpha+k$ will give ${_1F_1}(-\alpha;-\alpha+1;z)$, which simplifies to the incomplete gamma function; the sum over the simple poles at $y=k$ will give ${_2F_2}(1,1;2,2+\alpha;z)$; and the double pole at $y=0$ will give the logarithmic and the polygamma terms. $\endgroup$ – Maxim Jun 18 '18 at 23:37
  • $\begingroup$ Thanks. I will take a look at it. $\endgroup$ – Aaron Hendrickson Jun 19 '18 at 0:11
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Wolfram Language all the way:

LaplaceTransform[Hypergeometric2F1[1-a,1,2,-t],t,z]

produces

$$\frac{-\mathcal{L}_t\left[\frac{1}{t}\right](z)+\frac{z \, _2F_2(1,1;2,a+2;z)}{a+1}+\frac{\pi z^{-a} \csc (\pi a) \left((-1)^{a+1} z^a \Gamma (1-a,-z)+e^z\right)}{\Gamma (1-a)}+\pi e^{i \pi a} \csc (\pi a)-\psi ^{(0)}(-a)-\gamma }{a}$$ Here $\gamma$ is the Euler constant, and $\psi^{(0)}(-\alpha)$ is the PolyGamma function. You still have an LT in here, unfortunately, but it seems reduced compared with the original.

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  • $\begingroup$ Thank you. I believe I have found a more compact way to express the solution to the integral. If the solution can be converted to MeijerG that would be accepted. $\endgroup$ – Aaron Hendrickson Apr 10 '18 at 13:37

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