9
$\begingroup$

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 $\cos x+\cos y=1$

$\endgroup$
  • 1
    $\begingroup$ I've made some edits for instant comparison. I hope it's ok. $\endgroup$ – Tito Piezas III Aug 16 '17 at 12:59
  • $\begingroup$ @TitoPiezasIII Thanks! I edited it again to present both forms, for completeness. $\endgroup$ – Chris Culter Aug 16 '17 at 17:33
  • 1
    $\begingroup$ 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. $\endgroup$ – Tito Piezas III Aug 17 '17 at 6:29
  • 1
    $\begingroup$ 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}}} $ $\endgroup$ – user90369 Aug 21 '17 at 16:37
  • 1
    $\begingroup$ Another possible closed-form is given by $$\int_0^1 \frac{4\arccos x}{\sqrt{2x-x^2}} \, dx \stackrel{?}{=} 2\pi^2-4\pi\,\Re\left({_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,\tfrac34,\tfrac54\\1,\tfrac32,\tfrac32\end{array}\middle|\,4\right)\right).$$ $\endgroup$ – user153012 Aug 22 '17 at 21:52
9
+200
$\begingroup$

About the formula in the headline given by Mathematica.

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.

$k\to k+1$ :

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.