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Is there any chance to find a closed form for this integral? $$I=\int_0^\infty{_3F_2}\left(\begin{array}c\tfrac58,\tfrac58,\tfrac98\\\tfrac12,\tfrac{13}8\end{array}\middle|\ {-x}\right)^2\frac{dx}{\sqrt x}$$

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  • $\begingroup$ Mathematica can give a result in terms of hypergeometric functions (see here), but I have no idea how to prove it. $\endgroup$ Mar 15, 2014 at 4:08
  • $\begingroup$ @Vladimir Reshetnikov. Could you post you Mathematica syntax here ? Thanks. $\endgroup$ Mar 15, 2014 at 4:55
  • $\begingroup$ @ClaudeLeibovici 25 π/144 HypergeometricPFQ[{1/2, 3/4, 3/4}, {7/4, 2}, 1] + 25 Gamma[1/4]^2/(24 √2 √π) HypergeometricPFQ[{-1/2, -1/4, 1/2}, {3/4, 3/4}, 1] - 25 Gamma[3/4]^2/(2016 √π) (135 HypergeometricPFQ[{1, 1, 7/4, 7/4, 9/4}, {3/2, 2, 2, 11/4}, 1] - 672 Log[2] + 448) $\endgroup$ Mar 15, 2014 at 6:28
  • $\begingroup$ @ClaudeLeibovici ...and I got it by FullSimplify[FunctionExpand[Integrate[HypergeometricPFQ[{5/8, 5/8, 9/8}, {1/2, 13/8}, -x]^2/√x, {x, 0, ∞}]]] and some manual simplification. $\endgroup$ Mar 15, 2014 at 6:42
  • $\begingroup$ @VladimirReshetnikov. Thanks ! I suppose that you had fun with the manual simplifications. Cheers. $\endgroup$ Mar 15, 2014 at 6:59

3 Answers 3

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$$I=\frac{50\,\pi^{3/2}}{3\,\Gamma^2\left(\frac14\right)}\Big(\ln\left(3+\sqrt8\right)-\sqrt2\Big)$$

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Looks like we have a more general result: $$I(a)=\int_0^\infty{_3F_2}\left(\begin{array}ca,a,a+\tfrac12\\\tfrac12,a+1\end{array}\middle|\ {-x}\right)^2\frac{dx}{\sqrt x}\\=\frac{2^{4a-2}a^2}{\Gamma(2a)^2}G^{3~3}_{4~4}\left(1\middle|\begin{array}c\tfrac12,1,1;2a+\tfrac12\\2a-\tfrac12,2a-\tfrac12,2a;0\end{array}\right)\\ =\frac{2^{4a-2}a^2}{\Gamma(2a)^2}G^{4~4}_{6~6}\left(1\middle|\begin{array}c\tfrac12,\tfrac12,1,1;2a,2a+\tfrac12\\2a-\tfrac12,2a-\tfrac12,2a,2a;0,\tfrac12\end{array}\right)\\=\frac{4\pi a^2}{\Gamma(2a)^2}G^{2~2}_{3~3}\left(1\middle|\begin{array}c1,1;4a\\4a-1,4a-1;0\end{array}\right).$$

To prove Cleo's form is correct, we need to prove that $$G^{2~2}_{3~3}\left(1\middle|\begin{array}c1,1;\tfrac52\\\tfrac32,\tfrac32;0\end{array}\right)\stackrel?=\frac{2\sqrt\pi}{3}(2\log(1+\sqrt2)-\sqrt2).$$

Here Mathematica gives $$G^{2~2}_{3~3}\left(z\middle|\begin{array}c1,1;\tfrac52\\\tfrac32,\tfrac32;0\end{array}\right)=\frac{2\sqrt\pi}{3}\left(\log(\sqrt z+\sqrt{z+1})+z^{3/2}\log(1+\sqrt{z+1})-z^{3/2}\log(\sqrt z)-\sqrt{z}\sqrt{z+1})\right).$$

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I know that I'm late to post here, but here is an alternative proof of @Cleo's answer.

Preliminaries:

  1. Ramanujan's Master Theorem: $$\int_{0}^{\infty}f(x)x^{s-1}dx=\int_{0}^{\infty}\left(\sum_{n=0}^{\infty}\frac{\left(-x\right)^n}{n!}\varphi(n)\right)x^{s-1}dx=\Gamma(s)\varphi(-s) $$
  2. Plancherel's Theorem for Mellin Transform: $$\int_{0}^{\infty}\left|f(x)\right|^2x^{2c-1}dx=\frac{1}{2\pi}\int_{\infty}^{\infty}\left|\hat{f}(c+iz)\right|^2dz $$
  3. Series representation of the Hyper Geometric Function: $${_3F_2}\left(\begin{array}c\frac{5}{8},\frac{5}{8},\frac{9}{8}\\\tfrac12,\frac{13}{8}\end{array}\middle|\ {-x}\right)=\frac{5\sqrt{\pi}}{2\Gamma\left(\frac{1}{4}\right)}\sum_{n=0}^{\infty}\frac{\left(-x\right)^n}{n!}\frac{2^{-2n}\Gamma\left(\frac{5}{4}+2n\right)}{\Gamma\left(\frac{1}{2}+n\right)\left(\frac{5}{8}+n\right)} $$
  4. The Mellin Transform of the Hyper Geometric Function: $$\mathcal{M}\left[{_3F_2}\left(\begin{array}c\frac{5}{8},\frac{5}{8},\frac{9}{8}\\\tfrac12,\frac{13}{8}\end{array}\middle|\ {-x}\right)\right](s)=\frac{5\sqrt{\pi}}{2\Gamma\left(\frac{1}{4}\right)}\frac{2^{2s}\Gamma\left(\frac{5}{4}-2s\right)}{\Gamma\left(\frac{1}{2}-s\right)\left(\frac{5}{8}-s\right)} $$

Computation

Therefore, the desired integral can be rewritten as:

$$I=\int_{0}^{\infty}{_3F_2}\left(\begin{array}c\frac{5}{8},\frac{5}{8},\frac{9}{8}\\\tfrac12,\frac{13}{8}\end{array}\middle|\ {-x}\right)^2x^{2/4-1}dx\\ =\frac{25}{8\Gamma^2\left(\frac{1}{4}\right)}\int_{-\infty}^{\infty}\left|\frac{2^{1/2+2iz}\Gamma\left(\frac{3}{4}-2iz\right)\Gamma\left(\frac{1}{4}+iz\right)}{\Gamma\left(\frac{1}{4}-iz\right)\left(\frac{3}{8}-iz\right)}\right|^2dz\\ =\frac{25}{8\Gamma^2\left(\frac{1}{4}\right)}\int_{-\infty}^{\infty}\frac{2\Gamma\left(\frac{3}{4}-2iz\right)\Gamma\left(\frac{3}{4}+2iz\right)\color{red}{\Gamma\left(\frac{1}{4}+iz\right)\Gamma\left(\frac{1}{4}-iz\right)}}{\color{red}{\Gamma\left(\frac{1}{4}+iz\right)\Gamma\left(\frac{1}{4}-iz\right)}\left(\frac{9}{64}+z^2\right)}dz;\ 2z\rightarrow z \\ =\frac{25}{2\Gamma^2\left(\frac{1}{4}\right)}\int_{-\infty}^{\infty}\frac{\Gamma\left(\frac{3}{4}-iz\right)\Gamma\left(\frac{3}{4}+iz\right)}{\frac{9}{16}+z^2}dz=\frac{25\Gamma\left(\frac{3}{2}\right)}{2\Gamma^2\left(\frac{1}{4}\right)}\int_{-\infty}^{\infty}\frac{\mathfrak{B}\left(\frac{3}{4}-iz,\frac{3}{4}+iz\right)}{\frac{9}{16}+z^2}dz\\ =\frac{25\sqrt{\pi}}{4\Gamma^2\left(\frac{1}{4}\right)}\int_{0}^{1}\frac{d\omega}{\left[\omega(1-\omega)\right]^{1/4}}\int_{-\infty}^{\infty}\frac{\left(\frac{1-\omega}{\omega}\right)^{iz}}{\frac{9}{16}+z^2}dz $$ Some critical information before we proceed with the evaluation:

  • $$\left(\frac{1-\omega}{\omega}\right)^{iz}=\exp\left(iz\log\left(\frac{1-\omega}{\omega}\right)\right)$$
  • $$\int_{-\infty}^{\infty}\frac{e^{iaz}}{b^2+z^2}dz=\int_{-\infty}^{\infty}\frac{\cos(az)}{b^2+z^2}dz=\frac{\pi e^{-|a|}}{|b|}$$
  • For $\omega\in\left(0,\frac{1}{2}\right),\log\left(\frac{1-\omega}{\omega}\right)>0$ and for $\omega\in\left(\frac{1}{2},1\right)\log\left(\frac{1-\omega}{\omega}\right)<0$

Thus: $$I=\frac{25\sqrt{\pi}}{4\Gamma^2\left(\frac{1}{4}\right)}\int_{0}^{1}\frac{d\omega}{\left[\omega(1-\omega)\right]^{1/4}}\int_{-\infty}^{\infty}\frac{\exp\left(iz\log\left(\frac{1-\omega}{\omega}\right)\right)}{\frac{9}{16}+z^2}dz\\ =\frac{25\pi^{3/2}}{3\Gamma^2\left(\frac{1}{4}\right)}\int_{0}^{1}\frac{\exp\left(-\frac{3}{4}\left|\log\left(\frac{1-\omega}{\omega}\right)\right|\right)}{\left[\omega(1-\omega)\right]^{1/4}}d\omega\\ =\frac{25\pi^{3/2}}{3\Gamma^2\left(\frac{1}{4}\right)}\int_{0}^{1/2}\frac{\left(\frac{\omega}{1-\omega}\right)^{3/4}}{\left[\omega(1-\omega)\right]^{1/4}}d\omega+\frac{25\pi^{3/2}}{3\Gamma^2\left(\frac{1}{4}\right)}\underbrace{\int_{1/2}^{1}\frac{\left(\frac{1-\omega}{\omega}\right)^{3/4}}{\left[\omega(1-\omega)\right]^{1/4}}d\omega}_{1-\omega\rightarrow \omega}\\ =\frac{50\pi^{3/2}}{3\Gamma^2\left(\frac{1}{4}\right)}\int_{0}^{1/2}\frac{\sqrt{\omega}}{1-\omega}d\omega=\frac{50\pi^{3/2}}{3\Gamma^2\left(\frac{1}{4}\right)}\left(2\log(\sqrt{2}+1)-\sqrt{2}\right) $$ Addendum: if my calculations are correct, the generalization provided by @Chen Wang can be expressed as the following after repeating the same procedures as above:

$$\int_{0}^{\infty}{_3F_2}\left(\begin{array}c a,a,a+\frac{1}{2}\\\tfrac12,a+1\end{array}\middle|\ {-x}\right)^2x^{2/4-1}dx$$ $$\boxed{=\frac{2^{4(1-a)}\pi}{\mathfrak{B}(2a,2a)}\left(\frac{a}{4a-1}\right)^2\Phi\left(\frac{1}{2},1,4a-1\right)}$$

$\Phi\left(z,s,\alpha\right)$ is the Lerch Transcendent

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    $\begingroup$ The $x^a\Phi(x,1,a)$ is an expression in terms of $\text B_z(a,0)$ with the incomplete beta function $\endgroup$ Sep 19, 2022 at 17:34
  • $\begingroup$ @P.Teruo Nagasava Please, Is there any chance to find a closed form for this integral? $\int_1^\infty t^2 \operatorname{ArcTanh} \left(\sqrt{\frac{t^2-1}{t^2}} \dfrac{k}{x}\right)\, e^{-a\,t} \, dt$ $\endgroup$
    – Gallagher
    Dec 24, 2022 at 15:07

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