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I am seeking a closed form for the function $$f(x)=\,_3F_2\left(\tfrac12,\tfrac12,\tfrac12;\tfrac32,\tfrac32;x\right)$$

I expect there to be one, because of this post and Wolfram. The Wolfram link produces closed forms involving $\mathrm{Li}_2$ for any value of $x$ that I've tried so far, so I can only assume that a general closed form exists.

I've started my attempts by noticing that $$f(x)=\frac12\int_0^1 \frac{_2F_1(\tfrac12,\tfrac12;\tfrac32;xt)}{\sqrt{t}}dt,$$ because $$\frac12\int_0^1 \frac{(xt)^n}{\sqrt{t}}dt=\frac{x^n}{2n+1}$$ which would introduce another factor of $$\frac{n+1/2}{n+3/2}$$ when computing the ratio of the terms. Similarly, $$_2F_1\left(\tfrac12,\tfrac12;\tfrac32;x\right)=\frac12\int_0^1 \frac{_1F_0(\tfrac12;;xt)}{\sqrt{t}}dt.$$

The last hypergeometric I was able to recognize as $$_1F_0\left(\tfrac12;;xt\right)=\frac1{\sqrt{1-xt}}.$$ So, all in all, $$f(x)=\frac14\int_0^1\int_0^1 \frac{1}{\sqrt{vu}\sqrt{1-xvu}}dvdu,$$ which looks like the Beta function's evil cousin.

I do not know how to turn this integral into something containing $\mathrm{Li}_2$ and I need some help. Thanks!

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    $\begingroup$ Just in case you don't know it : WA accepts many things as in Mathematica. You could also register at Wolfram Cloud (it is free and it is a complete version of Mathematica). Look at wolframalpha.com/input/… $\endgroup$ – Claude Leibovici Jul 2 at 7:49
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Starting from the integral representation, we let $v = w/u$ and change the order of integration to find \begin{align} f(x) &= \frac{1}{4} \int \limits_0^1 \int \limits_0^u \frac{1}{u \sqrt{w(1-xw)}} \, \mathrm{d} w \, \mathrm{d} u = \frac{1}{4} \int \limits_0^1 \frac{\mathrm{d} w}{\sqrt{w(1-xw)}} \int \limits_w^1 \frac{\mathrm{d} u}{u} \\ &= \frac{1}{4} \int \limits_0^1 \frac{-\log(w)}{\sqrt{w(1-xw)}} \, \mathrm{d} w \, . \end{align} The combination of the next few substitutions can be written as $w = \sin^2(t/2)/x$, which yields \begin{align} f(x) &= \frac{1}{2 \sqrt{x}} \int \limits_0^{2\arcsin(\sqrt{x})} - \log\left(\frac{\sin\left(\frac{t}{2}\right)}{\sqrt{x}}\right) \, \mathrm{d} t \\ &= \frac{\arcsin(\sqrt{x})}{\sqrt{x}} \log(2 \sqrt{x}) + \frac{1}{2\sqrt{x}} \int \limits_0^{2\arcsin(\sqrt{x})} - \log\left(2 \sin\left(\frac{t}{2}\right)\right) \, \mathrm{d} t \\ &= \frac{1}{\sqrt{x}} \left[\arcsin(\sqrt{x}) \log(2\sqrt{x}) + \frac{1}{2} \operatorname{Cl}_2(2\arcsin(\sqrt{x}))\right] \end{align} for $x \in (0,1]$, while $f(0) = 1$. The Clausen function is of course related to the dilogarithm. Interesting special values include $f(1) = \frac{\pi}{2} \log(2)$, $f\left(\frac{1}{2}\right) = \frac{1}{\sqrt{2}} \left[\frac{\pi}{4} \log(2) + \mathrm{G}\right]$ and $f\left(\frac{1}{4}\right) = \operatorname{Cl}_2\left(\frac{\pi}{3}\right)$ (see this question).

For $y > 0$ similar steps lead to \begin{align} f(-y) &= \frac{\operatorname{arsinh}(\sqrt{y})}{\sqrt{y}} \log(2 \sqrt{y}) + \frac{1}{2\sqrt{y}} \int \limits_0^{2\operatorname{arsinh}(\sqrt{y})} - \log\left(2 \sinh\left(\frac{t}{2}\right)\right) \, \mathrm{d} t \\ &= \frac{1}{\sqrt{y}} \left[\operatorname{arsinh}(\sqrt{y})\log(2\sqrt{y}) - \frac{1}{2} \operatorname{arsinh}^2(\sqrt{y}) + \frac{1}{2} \int \limits_0^{2\operatorname{arsinh}(\sqrt{y})} - \log\left(1 - \mathrm{e}^{-t}\right) \, \mathrm{d} t \right] \\ &= \frac{1}{\sqrt{y}} \left[\operatorname{arsinh}(\sqrt{y})\log(2\sqrt{y}) - \frac{1}{2} \operatorname{arsinh}^2(\sqrt{y}) + \frac{\pi^2}{12} - \frac{1}{2} \operatorname{Li}_2\left(\mathrm{e}^{-2\operatorname{arsinh}(\sqrt{y})}\right) \right] \\ &= \frac{1}{\sqrt{y}} \left[\operatorname{arsinh}(\sqrt{y})\log(2\sqrt{y}) - \frac{1}{2} \operatorname{arsinh}^2(\sqrt{y}) + \frac{\pi^2}{12} - \frac{1}{2} \operatorname{Li}_2\left[\left(\sqrt{1+y} - \sqrt{y}\right)^2\right] \right] , \end{align} which closely resembles Claude Leibovici's result (some dilogarithm identities should do the trick). Here $f\left(-\frac{1}{4}\right) = \frac{\pi^2}{10}$ looks rather nice, as does $f\left(-\frac{1}{8}\right) = \frac{\pi^2 - 3 \log^2(2)}{6 \sqrt{2}}$.

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  • $\begingroup$ There are so many great answers to choose from, but this one's gotta be the quickest. Thanks! $\endgroup$ – clathratus Jul 2 at 20:53
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For $x>0$ we have:$$f(x)=\frac14\int_0^1\int_0^1 \frac{1}{\sqrt{vu}\sqrt{1-xvu}}dvdu\overset{vu=t}=\frac14\int_0^1\frac{1}{u}\int_0^u \frac{1}{\sqrt{t}\sqrt{1-xt}}dtdu$$ $$=\frac12\int_0^1 \frac{1}{u}\frac{\arcsin \sqrt{xt}}{\sqrt{x}}\bigg|_0^udu=\frac1{2\sqrt x} \int_0^1 \frac{\arcsin\sqrt{xu}}{u}du\overset{xu=t}=\frac{1}{2\sqrt x}\int_0^x \frac{\arcsin \sqrt t}{t}dt$$ $$\overset{t=y^2}=\frac{1}{\sqrt x}\int_0^\sqrt x \frac{\arcsin y}{y}dy\overset{IBP}=\frac{1}{\sqrt x} \ln \sqrt x \arcsin\sqrt x-\frac{1}{\sqrt x}\int_0^\sqrt x \frac{\ln y}{\sqrt{1-y^2}}dy$$ $$\overset{y=\sin z}=\frac{1}{\sqrt x} \ln \sqrt x \arcsin\sqrt x-\frac{1}{\sqrt x}\int_0^{\arcsin \sqrt x} \ln(\sin z)dz$$ $$=\frac{1}{\sqrt x} \ln \sqrt x \arcsin\sqrt x+\frac{\arcsin \sqrt x}{\sqrt x}\ln 2+\frac{1}{2\sqrt x}\operatorname{Cl}_2(2\arcsin \sqrt x)$$ $$=\boxed{\frac{1}{\sqrt x} \left(\arcsin \sqrt x \ln(2\sqrt x) +\frac12\operatorname{Cl}_2(2\arcsin \sqrt x) \right)}$$ In terms of Clausen function, which of course is the imaginary part of some dilogarithms.


For $x<0$ we can work with $x=-y$, $y>0$ and using: $$\frac{\arcsin \sqrt{-z}}{\sqrt{-z}}=\frac{\operatorname{arcsinh} \sqrt{z}}{\sqrt{z}}$$ We will arrive at: $$f(-y)=\frac{1}{2\sqrt y} \int_0^1 \frac{\operatorname{arcsinh}\sqrt{yu}}{u}du\overset{yu=t}=\frac{1}{2\sqrt y}\int_0^y \frac{\operatorname{arcsinh}\sqrt{t}}{t}dt$$ $$\overset{t=v^2}=\frac{1}{\sqrt y} \int_0^\sqrt y \frac{\operatorname{arcsinh}v}{v}dv\overset{IBP}=\frac{1}{\sqrt y} \ln \sqrt y \operatorname{arcsinh}\sqrt y-\frac{1}{\sqrt y} \underbrace{\int_0^\sqrt y\frac{\ln v}{\sqrt{1+v^2}}dv}_{=J}$$ For $J$ put $v=\frac{1-t^2}{2t}$ then: $$J=\int_1^{\sqrt{1+y}-\sqrt y}\ln\left(\frac{2t}{1-t^2}\right)\frac{dt}{t}=\left(\frac12\ln^2(2t) +\frac12 \operatorname{Li}_2(t^2)\right)\bigg|_1^{\sqrt{1+y}-\sqrt y}$$ $$=\frac12\left( \ln^2[2\sqrt{1+y}-2\sqrt y]-\ln^2 2 +\operatorname{Li}_2[(\sqrt{1+y}-\sqrt y)^2]-\frac{\pi^2}{6}\right)$$ $$\small \Rightarrow \boxed{f(-y)=\frac{1}{\sqrt y} \ln \sqrt y \operatorname{arcsinh}\sqrt y+\frac{1}{2\sqrt y}\left(\ln^2 2 -\ln^2[2\sqrt{1+y}-2\sqrt y] - \operatorname{Li}_2[(\sqrt{1+y}-\sqrt y)^2]+\frac{\pi^2}{6}\right)}$$

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Here's the shortest closed-form:

$$\,_3F_2\left(\frac12,\frac12,\frac12;\frac32,\frac32;x\right) = -\,\frac{{\rm Li}_2(1-\beta^2)}{2\sqrt{-x}}-\frac{\ln^2 \beta}{2\sqrt{-x}}$$

where,

$$\beta = \sqrt{1-x}+\sqrt{-x}$$

Note: This was derived using WA's solution and the 4th identity here.

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  • $\begingroup$ Very nice! Thanks $\endgroup$ – clathratus Jul 2 at 21:12
  • $\begingroup$ Nice but you can make it shorter in terms of number of characters$$ -\,\frac{{\rm Li}_2(1-\beta^2)+\ln^2(\beta)}{2\sqrt{-x}}$$ $\endgroup$ – Claude Leibovici Jul 3 at 4:34
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    $\begingroup$ @ClaudeLeibovici: Ah, I should have written it like that! $\endgroup$ – Tito Piezas III Jul 3 at 14:24
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I hope that you will enjoy $$12 \sqrt{-x}\,\, f(x)=12 \text{Li}_2\left(-\sqrt{1-x}-\sqrt{-x}\right)-12 \text{Li}_2\left(-\sqrt{1-x}-\sqrt{-x}+1\right)+6 \sin ^{-1}\left(\sqrt{x}\right)^2+$$ $$12 \log \left(\sqrt{1-x}+\sqrt{-x}+1\right) \sinh ^{-1}\left(\sqrt{-x}\right)+\pi ^2$$

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  • $\begingroup$ Its beautiful! how'd you get it? $\endgroup$ – clathratus Jul 2 at 7:29
  • $\begingroup$ @clathratus. Guess ! Using a CAS for sure.Look at wolframalpha.com/input/… $\endgroup$ – Claude Leibovici Jul 2 at 7:33
  • $\begingroup$ Ahh I see. I was unaware of the FunctionExpand command. Thanks! $\endgroup$ – clathratus Jul 2 at 7:38
  • $\begingroup$ This can be condensed even more. Kindly see my answer. :) $\endgroup$ – Tito Piezas III Jul 2 at 17:51

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