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Can anybody see how to deduce an asymptotic formula for the hypergeometric function $$ _3F_2\left(\frac{1}{2},x,x;x+\frac{1}{2},x+\frac{1}{2} \hskip2pt\bigg|\hskip2pt 1\right), \quad\mbox{ as } x\to\infty?$$

For the standard definition of the hypergeometric series, see here.

Remark: I've tried to combine the numerous transformations and integral representations which $_3F_2$-functions fulfill. I've also tried to get inspired by existing works on asymptotic expansions of similar (but simpler) $_2F_1$-functions but never ended up with the desired result.

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2 Answers 2

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$\newcommand\Ga\Gamma$Let $f(x)$ denote your hypergeometric expression. Then $$f(x)\sim\sqrt{\pi x}$$ (as $x\to\infty$).

Indeed, $$f(x)=\sum_{k\ge0}\frac{(1/2)_k}{k!}r_k(x)^2,$$ where $(a)_k:=a(a+1)\cdots(a+k-1)=\Ga(a+k)/\Ga(a)$ and $$r_k(x):=\frac{(x)_k}{(x+1/2)_k}=\frac{\Ga(x+k)}{\Ga(x)}\Big/\frac{\Ga(x+1/2+k)}{\Ga(x+1/2)} \sim\sqrt{\frac x{x+k}}$$ uniformly in $k\ge0$ (as $x\to\infty$). So, \begin{equation*} f(x)\sim x\,\sum_{k\ge0}\frac{(1/2)_k}{k!} \frac1{x+k} =\frac{\sqrt{\pi }\, \Ga(x+1)}{\Ga(x+1/2)} \sim\sqrt{\pi x} \tag{1}\label{1} \end{equation*} (as $x\to\infty$).


Details on the equality in \eqref{1}: \begin{equation*} \begin{aligned} \sum_{k\ge0}\frac{(1/2)_k}{k!} \frac1{x+k} & =\sum_{k\ge0}\frac{(1/2)_k}{k!} \int_0^1 dt\,t^{x+k-1} \\ & =\int_0^1 dt\,t^{x-1}\sum_{k\ge0}\frac{(1/2)_k}{k!} t^k \\ & =\int_0^1 dt\,t^{x-1}(1-t)^{-1/2}, \end{aligned} \end{equation*} in view of the Maclaurin series for $(1-t)^{-1/2}$. Therefore, \begin{equation*} \begin{aligned} x\,\sum_{k\ge0}\frac{(1/2)_k}{k!} \frac1{x+k} =x\,B(x,1/2)= x\,\frac{\Ga(1/2)\, \Ga(x)}{\Ga(x+1/2)} =\frac{\sqrt{\pi }\, \Ga(x+1)}{\Ga(x+1/2)}, \end{aligned} \end{equation*} as claimed.

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Use the identity $$_3F_2\left(\frac{1}{2},x,x;x+\frac{1}{2},x+\frac{1}{2} \bigg|1\right)=\frac{\sqrt{\pi }\, \Gamma \left(x+\frac{1}{2}\right)}{\Gamma (x)}\,_3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};1,x+\frac{1}{2};1\right),$$ and the large-$x$ limits $$_3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};1,x+\frac{1}{2};1\right)\rightarrow 1,$$ $$\frac{\sqrt{\pi }\, \Gamma \left(x+\frac{1}{2}\right)}{\Gamma (x)}\rightarrow\sqrt{\pi x},$$ to conclude that $$_3F_2\left(\frac{1}{2},x,x;x+\frac{1}{2},x+\frac{1}{2} \bigg|1\right)\rightarrow \sqrt{\pi x}.$$

A numerical test shows that the $\sqrt{\pi x}$ asymptote (gold) is nearly indistinguishable from the exact function (blue) for $x$ larger than 4.

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  • $\begingroup$ oh, I only saw your answer, Iosif, when I posted mine, apologies; the calculation is slightly different, so I'll just leave it here for the record. $\endgroup$ Dec 28, 2022 at 18:57
  • $\begingroup$ Yes, Carlo, your approach is different. Can you give a reference on the displayed identity and details on $\to1$? $\endgroup$ Dec 28, 2022 at 19:23
  • $\begingroup$ It is, for example, [Eq. 7.4.1.1] in Prudnikov, Brychkov, Marichev Vol. 3 $\endgroup$
    – Twi
    Dec 28, 2022 at 19:31
  • $\begingroup$ Alas, I don't see this identity there. Here is what I see there instead: u.pcloud.link/publink/… $\endgroup$ Dec 28, 2022 at 19:53
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    $\begingroup$ Thefunctional identity for "$\rightarrow 1$" can be found here: functions.wolfram.com/07.27.17.0034.01 $\endgroup$ Dec 30, 2022 at 9:24

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