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!
Best Answer
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: $$J=\int_1^{\sqrt{1+y}-\sqrt y}\ln\left(\frac{2t}{1-t^2}\right)\frac{dt}{t}=\left(\frac12\ln^2(2t) +\frac12 \operatorname{Li}_2(t^2)\right)\bigg|_1^{\sqrt{1+y}-\sqrt y}$$ $$=\frac12\left( \ln^2[2\sqrt{1+y}-2\sqrt y]-\ln^2 2 +\operatorname{Li}_2[(\sqrt{1+y}-\sqrt y)^2]-\frac{\pi^2}{6}\right)$$ $$\small \Rightarrow \boxed{f(-y)=\frac{1}{\sqrt y} \ln \sqrt y \operatorname{arcsinh}\sqrt y+\frac{1}{2\sqrt y}\left(\ln^2 2 -\ln^2[2\sqrt{1+y}-2\sqrt y] - \operatorname{Li}_2[(\sqrt{1+y}-\sqrt y)^2]+\frac{\pi^2}{6}\right)}$$