Closed form for integral of inverse hyperbolic function in terms of ${_4F_3}$

Question

While attempting to evaluate the integral $\int_{0}^{\frac{\pi}{2}}\sinh^{-1}{\left(\sqrt{\sin{x}}\right)}\,\mathrm{d}x$, I stumbled upon the following representation for a related integral in terms of hypergeometric functions:

$$\small{\int_{0}^{1}\frac{x\sinh^{-1}{x}}{\sqrt{1-x^4}}\,\mathrm{d}x\stackrel{?}{=}\frac{\Gamma{\left(\frac34\right)}^2}{\sqrt{2\pi}}\,{_4F_3}{\left(\frac14,\frac14,\frac34,\frac34;\frac12,\frac54,\frac54;1\right)}-\frac{\Gamma{\left(\frac14\right)}^2}{72\sqrt{2\pi}}{_4F_3}{\left(\frac34,\frac34,\frac54,\frac54;\frac32,\frac74,\frac74;1\right)}}.$$

I'm having some trouble wading through the algebraic muckity-muck, so I'd like help confirming the above conjectured identity. More importantly, can these hypergeometrics be simplified in any significant way? The "niceness" of the parameters really makes me suspect it can be...

Any thoughts or suggestions would be appreciated. Cheers!

Answer 1

$$ \newcommand{\as}{\sinh^{-1}} \newcommand{\at}{\tan^{-1}} \begin{align} I &:= 2 \int_0^{\pi/2} \frac{x \as x}{\sqrt{1-x^4}} \\&= \int_0^{\pi/2} \as\sqrt{ \sin x } \\&= \frac 1 2\int_0^{\pi} \as\sqrt{ \sin x } \\&= \frac 1 2\int_0^{\infty} \frac{2}{1+t^2}\as\sqrt{ \frac{2t}{1+t^2}} \\&= \left.\at x \as\sqrt{ \frac{2t}{1+t^2}}\right\lvert _0^\infty - \int_0^{\infty} \at t \frac{1-t^2}{\sqrt{2t}(1+t^2)^{3/2}} \frac{1}{\sqrt{1+\frac{2t}{1+t^2}}} \\&= \int_0^{\infty} \at t \frac{t-1}{\sqrt{2t}(1+t^2)} \\&= \sqrt 2\int_0^{\infty} \frac{x^2-1}{1+x^4} \at x^2 \end{align} $$

Let $$ J(a) = \sqrt 2\int_0^{\infty} \frac{x^2-1}{1+x^4} \log(1+ a^2 x^2), $$ so $J(0) = 0$ and $$ \begin{align} J'(a) &= \sqrt 2\int_0^{\infty} \frac{x^2-1}{1+x^4} \frac{2a x^2}{1+a^2x^2} \\&= \sqrt 2\int_0^{\infty} \frac{2 \left(a^3 x^2-a^3-a x^2-a\right)}{\left(a^4+1\right) \left(x^4+1\right)}+\frac{2 a \left(a^2+1\right)}{\left(a^4+1\right) \left(a^2 x^2+1\right)} \\&= \frac{ \pi\sqrt{2} }{a^2+\sqrt{2} a+1} \\&= i \pi \left[\frac{1}{a+ \frac{1+i}{\sqrt 2}}-\frac{1}{a+\frac{1-i}{\sqrt 2}} \right]. \end{align} $$ This implies $$ J\left(\frac{1+i}{\sqrt 2}\right) = i \pi\left[\log\left(\sqrt 2 (1+i) \right) - \log \sqrt 2 \right], $$ whence $$ \boxed{ I = \operatorname{Im} J\left(\frac{1+i}{\sqrt 2}\right) = \frac \pi 2 \log 2.} $$

Answer 2

$$\sinh^{-1}(\sqrt{\sin x}) = \sum\limits_{n=0}^{\infty} \frac{(-1)^n (2n)!}{2^{2n}(n!)^2}\frac{{(\sin x)}^{(2n+1)/2}}{2n+1}$$ so the integral is equivalent to $$\sum\limits_{n=0}^{\infty} \frac{(-1)^n (2n)!}{2^{2n}(n!)^2(2n+1)}\int_0^{\pi/2}(\sin x)^{(2n+1)/2}\,dx$$ You have $$\int_0^{\pi/2}(\sin x)^{(2n+1)/2}\,dx=\frac{\sqrt\pi}{2}\frac{\Gamma\left(\dfrac{2n+3}{4}\right)}{\Gamma\left(\dfrac{2n+5}{4}\right)}$$ (I haven't done the work myself, but I've seen the derivation somewhere...) Now it suffices to show that $$\sum\limits_{n=0}^{\infty} \frac{(-1)^n (2n)!}{2^{2n}(n!)^2(2n+1)}\frac{\Gamma\left(\dfrac{2n+3}{4}\right)}{\Gamma\left(\dfrac{2n+5}{4}\right)}=\sqrt\pi\ln2$$