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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!

David H
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    *Hint*. – Lucian Dec 31 '14 at 18:29
  • @Lucian I thank you for that, but I seem to be missing something. How is there a nonzero imaginary component? – David H Dec 31 '14 at 19:12
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    There isn't. :-$)$ The factor of i vanishes, and the answer is simply $\dfrac\pi2\ln2$. – Lucian Dec 31 '14 at 20:03
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    Apparently, $~\text{Li}_2\bigg[\dfrac{1-\sqrt2}2\bigg]+\text{Li}_2\bigg[\Big(1-\sqrt2\Big)^2\bigg]+\dfrac12\cdot\ln^2\bigg[\dfrac{1+\sqrt2}2\bigg]=0,~$ which is what one gets after simplifying the expression in question. – Lucian Dec 31 '14 at 20:17
  • @Lucian Doh! That probably confirms my suspicions that there's a vastly more elegant way to go about this problem lurking. Thanks again. – David H Dec 31 '14 at 21:11
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    Your conjecture is correct, both sides being equal with each other, and with $\dfrac\pi4\ln2$. – Lucian Dec 31 '14 at 21:38
  • I am very close to a proof and will post an answer when I have finished it. – user111187 Jan 01 '15 at 14:23
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    @user111187 No, your response below was quite satisfactory. To be honest, I simply forget about this question amidst holiday activities. Oops :) – David H Jan 05 '15 at 21:48
  • If you're watching this question, you may be interested in a recent similar question: https://math.stackexchange.com/questions/2395145 – Chris Culter Aug 18 '17 at 17:12

2 Answers2

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

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 $$ \boxed{ I = \operatorname{Im} J\left(\frac{1+i}{\sqrt 2}\right) = \frac \pi 2 \log 2.} $$

user111187
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$$\sinh^{-1}(\sqrt{\sin x}) = \sum\limits_{n=0}^{\infty} \frac{(-1)^n (2n)!}{2^{2n}(n!)^2}\frac{{(\sin x)}^{(2n+1)/2}}{2n+1}$$ 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 $$\sum\limits_{n=0}^{\infty} \frac{(-1)^n (2n)!}{2^{2n}(n!)^2(2n+1)}\frac{\Gamma\left(\dfrac{2n+3}{4}\right)}{\Gamma\left(\dfrac{2n+5}{4}\right)}=\sqrt\pi\ln2$$

user111187
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user170231
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