# Prove $\int_0^1 \frac{4\cos^{-1}x}{\sqrt{2x-x^2}}\,dx=\frac{8}{9\sqrt{\pi}}\left(9\Gamma(3/4)^2{}_4F_3(\cdots)+\Gamma(5/4)^2{}_4F_3(\cdots)\right)$

Mathematica gives the following. But how?!

$$\small{\int_0^1 \dfrac{4\cos^{-1}x}{\sqrt{2x-x^2}}\,dx=\frac{8}{9\sqrt{\pi}}\left( 9\Gamma\left(\tfrac{3}{4}\right)^2{}_4F_3\left( \begin{array}{c}\tfrac14\,\tfrac14\,\tfrac34\,\tfrac34\\\tfrac12\,\tfrac54\,\tfrac54\end{array};\tfrac14\right) +\Gamma\left(\tfrac{5}{4}\right)^2{}_4F_3\left( \begin{array}{c}\tfrac34\,\tfrac34\,\tfrac54\,\tfrac54\\\tfrac32\,\tfrac74\,\tfrac74\end{array};\tfrac14\right) \right)}$$

Is there some clever trig substitution that leads to the right-hand side? Or a integral on the complex plane? Can we make it look like a Mahler measure?

To aid comparison, we can rewrite this as:

$$\small{\int_0^1 \dfrac{4\cos^{-1}x}{\sqrt{2x-x^2}}\,dx =\frac{8\,\Gamma{\left(\frac34\right)}^2}{\sqrt{\pi}}\,{_4F_3}{\left(\frac14,\frac14,\frac34,\frac34;\frac12,\frac54,\frac54;\frac14\right)}\color{red}+\frac{\Gamma{\left(\frac14\right)}^2}{18\sqrt{\pi}}{_4F_3}{\left(\frac34,\frac34,\frac54,\frac54;\frac32,\frac74,\frac74;\frac14\right)}}=\color{red}{\,??}$$

This seems to have a similar flavor to Closed form for integral of inverse hyperbolic function in terms of ${_4F_3}$,

$$\small{\int_{0}^{1}\frac{x\sinh^{-1}{x}}{\sqrt{1-x^4}}\,\mathrm{d}x =\frac{\Gamma{\left(\frac34\right)}^2}{\sqrt{2\pi}}\,{_4F_3}{\left(\frac14,\frac14,\frac34,\frac34;\frac12,\frac54,\frac54;1\right)}\color{red}-\frac{\Gamma{\left(\frac14\right)}^2}{72\sqrt{2\pi}}{_4F_3}{\left(\frac34,\frac34,\frac54,\frac54;\frac32,\frac74,\frac74;1\right)}}=\color{red}{\frac{\pi}4\,\ln2}$$

Although in my case, it seems like there is less hope for a dramatic simplification. (Feel free to prove me wrong about that!) At least, I haven't found a simpler form in the Inverse Symbolic Calculator, even after playing with factors of $\ln2$, $\ln3$, $\pi$, and $\sqrt\pi$. Conversely, is it possible to prove (or give a plausible argument) that this integral doesn't have a simpler form (in terms of the functions that we'd hope to see)?

Motivation for this question: Area bounded by $\cos x+\cos y=1$

• I've made some edits for instant comparison. I hope it's ok. Aug 16 '17 at 12:59
• @TitoPiezasIII Thanks! I edited it again to present both forms, for completeness. Aug 16 '17 at 17:33
• I corrected a sign typo I made. Mea culpa. I guess this may be a relevant difference: your integral is a sum of two hypergeometrics, while the second is a difference of two hypergeometrics presumably of form $a\pm\frac{\pi}{8}\,\ln 2$. But I can't find a closed-form for $a$ though. Aug 17 '17 at 6:29
• I looked for a (more or less) closed form but only got: $\enspace\displaystyle \int\limits_0^1 \frac{\arccos x}{\sqrt{2x-x^2}}dx = \int\limits_0^{\pi/2} \frac{t\sin t }{\sqrt{2\cos t-(\cos t)^2}}dt =$ $\displaystyle =2\int\limits_0^{\pi/2} \arcsin(\sqrt{\frac{\cos t}{2}})dt = \sqrt{2}\sum\limits_{k=0}^\infty \binom {2k} k \frac{1}{8^k(2k+1)} \int\limits_0^{\pi/2} \sqrt{\cos t}^{2k+1}dt$ $\displaystyle =\sqrt{2}\sum\limits_{k=0}^\infty \frac{1}{8^k(2k+1)} \frac{\binom {2k} k}{\binom {\frac{k}{2}+\frac{1}{4}} {\frac{1}{2}}}$ Aug 21 '17 at 16:37
• Another possible closed-form is given by $$\int_0^1 \frac{4\arccos x}{\sqrt{2x-x^2}} \, dx \stackrel{?}{=} 2\pi^2-4\pi\,\Re\left({_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,\tfrac34,\tfrac54\\1,\tfrac32,\tfrac32\end{array}\middle|\,4\right)\right).$$ Aug 22 '17 at 21:52

For the proof of the formula in the headline we have to know that for $\,\displaystyle |x|\leq \frac{1}{16}\,$ exists :

$\displaystyle f(x):=\sum\limits_{k=0}^\infty \frac{x^k}{4k+1}{\binom {4k} {2k}} / {\binom {k+\frac{1}{4}} {\frac{1}{2}} }={ \binom {\frac{1}{4}} {\frac{1}{2}}}^{-1} {}_4F_3\left( \begin{array}{c}\tfrac14\,\tfrac14\,\tfrac34\,\tfrac34\\\tfrac12\,\tfrac54\,\tfrac54\end{array};16x\right)$

$\displaystyle g(x):=\sum\limits_{k=0}^\infty \frac{x^k}{4k+3}{\binom {4k+2} {2k+1}} / {\binom {k+\frac{3}{4}} {\frac{1}{2}} }=\frac{2}{3} { \binom {\frac{3}{4}} {\frac{1}{2}}}^{-1} {}_4F_3\left( \begin{array}{c}\tfrac34\,\tfrac34\,\tfrac54\,\tfrac54\\\tfrac32\,\tfrac74\,\tfrac74\end{array};16x\right)$

On the other hand we have (for the original integral divided by $\,4\,$):

$\enspace\displaystyle \int\limits_0^1 \frac{\arccos x}{\sqrt{2x-x^2}}dx = \int\limits_0^{\pi/2} \frac{t\sin t }{\sqrt{2\cos t-(\cos t)^2}}dt =$

$\displaystyle =2\int\limits_0^{\pi/2} \arcsin(\sqrt{\frac{\cos t}{2}})dt = \sqrt{2}\sum\limits_{k=0}^\infty \binom {2k} k \frac{1}{8^k(2k+1)} \int\limits_0^{\pi/2} \sqrt{\cos t}^{2k+1}dt$

$\displaystyle =\sqrt{2}\sum\limits_{k=0}^\infty \frac{1}{8^k(2k+1)} {\binom {2k} k}/{\binom {\frac{k}{2}+\frac{1}{4}} {\frac{1}{2}}}=\sqrt{2}\Big(f\big(\frac{1}{8^2}\big)+\frac{1}{8} g\big(\frac{1}{8^2}\big)\Big)$

Since $\,\,\displaystyle \frac{1}{\sqrt{2}} {\binom {\frac{1}{4}} {\frac{1}{2}}}^{-1} = \frac{1}{\sqrt{\pi}} \Gamma\left(\frac{3}{4}\right)^2\,$ and $\,\,\displaystyle \frac{1}{2\sqrt{2}} {\binom {\frac{3}{4}} {\frac{1}{2}}}^{-1} = \frac{2}{3\sqrt{\pi}} \Gamma\left(\frac{5}{4}\right)^2\,$ ,

we get what Mathematica creates.

Notes:

The formula for $f$ (above) is based on

$\displaystyle \frac{1}{4k+1} \binom {4k} {2k} / \binom {k+\frac{1}{4}} {\frac{1}{2}} = {\binom {\frac{1}{4}} {\frac{1}{2}} }^{-1} \frac{ ( \frac{1}{4} )_k (\frac{1}{4})_k (\frac{3}{4})_k (\frac{3}{4})_k }{(\frac{1}{2})_k (\frac{5}{4})_k (\frac{5}{4})_k } \frac{16^k}{k!}$

e.g. proofed by induction :

$k=0$ : o.k.

$k\to k+1$ :

$\displaystyle\frac{ \frac{1}{4k+3} \binom {4k+4} {2k+2} / \binom {k+\frac{5}{4}} {\frac{1}{2}} }{\frac{1}{4k+1} \binom {4k} {2k} / \binom {k+\frac{1}{4}} {\frac{1}{2}}} = \frac{2 (4k+1)^2 (4k+3)^2}{(k+1) (2k+1) (4k+5)^2} = \frac{{\binom {\frac{1}{4}} {\frac{1}{2}} }^{-1} \frac{ ( \frac{1}{4} )_{k+1} (\frac{1}{4})_{k+1} (\frac{3}{4})_{k+1} (\frac{3}{4})_{k+1} }{(\frac{1}{2})_{k+1} (\frac{5}{4})_{k+1} (\frac{5}{4})_{k+1} } \frac{16^{k+1} }{(k+1)!}}{ {\binom {\frac{1}{4}} {\frac{1}{2}} }^{-1} \frac{ ( \frac{1}{4} )_k (\frac{1}{4})_k (\frac{3}{4})_k (\frac{3}{4})_k }{(\frac{1}{2})_k (\frac{5}{4})_k (\frac{5}{4})_k } \frac{16^k}{k!}}$

The formula for $g$ (above) is based on

$\displaystyle \frac{1}{4k+3} \binom {4k+2} {2k+1} / \binom {k+\frac{3}{4}} {\frac{1}{2}} = \frac{2}{3}{\binom {\frac{3}{4}} {\frac{1}{2}} }^{-1} \frac{ ( \frac{3}{4} )_k (\frac{3}{4})_k (\frac{5}{4})_k (\frac{5}{4})_k }{(\frac{3}{2})_k (\frac{7}{4})_k (\frac{7}{4})_k } \frac{16^k}{k!}$

e.g. proofed by induction :

$k=0$ : o.k.

$k\to k+1$ :

$\displaystyle\frac{ \frac{1}{4k+7} \binom {4k+6} {2k+3} / \binom {k+\frac{7}{4}} {\frac{1}{2}} }{\frac{1}{4k+3} \binom {4k+2} {2k+1} / \binom {k+\frac{3}{4}} {\frac{1}{2}}} = \frac{2 (4k+3)^2 (4k+5)^2}{(k+1) (2k+3) (4k+7)^2 } = \frac{\frac{2}{3} {\binom {\frac{3}{4}} {\frac{1}{2}} }^{-1} \frac{ ( \frac{3}{4} )_{k+1} (\frac{3}{4})_{k+1} (\frac{5}{4})_{k+1} (\frac{5}{4})_{k+1} }{(\frac{3}{2})_{k+1} (\frac{7}{4})_{k+1} (\frac{7}{4})_{k+1} } \frac{16^{k+1} }{(k+1)!}}{ \frac{2}{3} {\binom {\frac{3}{4}} {\frac{1}{2}} }^{-1} \frac{ ( \frac{3}{4} )_k (\frac{3}{4})_k (\frac{5}{4})_k (\frac{5}{4})_k }{(\frac{3}{2})_k (\frac{7}{4})_k (\frac{7}{4})_k } \frac{16^k}{k!}}$