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

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!

• Hint. – Lucian Dec 31 '14 at 18:29
• There isn't. :-$)$ The factor of i vanishes, and the answer is simply $\dfrac\pi2\ln2$. – Lucian Dec 31 '14 at 20:03
• 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
• Your conjecture is correct, both sides being equal with each other, and with $\dfrac\pi4\ln2$. – Lucian Dec 31 '14 at 21:38
• @user111187 No, your response below was quite satisfactory. To be honest, I simply forget about this question amidst holiday activities. Oops :) – David H Jan 5 '15 at 21:48

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 $$\boxed{ I = \operatorname{Im} J\left(\frac{1+i}{\sqrt 2}\right) = \frac \pi 2 \log 2.}$$
$$\sinh^{-1}(\sqrt{\sin x}) = \sum\limits_{n=0}^{\infty} \frac{(-1)^n (2n)!}{2^{2n}(n!)^2}\frac{{(\sin x)}^{(2n+1)/2}}{2n+1}$$ 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 $$\sum\limits_{n=0}^{\infty} \frac{(-1)^n (2n)!}{2^{2n}(n!)^2(2n+1)}\frac{\Gamma\left(\dfrac{2n+3}{4}\right)}{\Gamma\left(\dfrac{2n+5}{4}\right)}=\sqrt\pi\ln2$$
• It should be $\frac{2n+5}{4}$ instead of $\frac{2n+3}{5}$. – user111187 Jan 2 '15 at 16:44
• Ah I copied it incorrectly, it's actually $\dfrac{2n+5}{4}$ but I wouldn't know for certain, I didn't derive it myself – user170231 Jan 2 '15 at 16:45
• It is proven using the Beta function identity $$\int_0^{\pi/2} dx 2 \sin^{2a+1}x \cos^{2b+1} x = B(a,b)$$ The trouble with this approach is that evaluating the sum is not trivial at all (the sum is basically identical to the hypergeometric function expression) – user111187 Jan 2 '15 at 16:47