7

$$I = \int_{0}^{\pi/2} \arccos\left(\frac{\cos x}{1 + \cos x}\right)\,\mathrm dx$$

This integral is very similar to Coxeter’s integral. I used integration by parts and obtained $$ I = \frac{\pi^2}{4} - \int_{0}^{\pi/2}\frac{x \sin x}{\sqrt{1 + 2\cos x}(1 + \cos x)}\,\mathrm dx $$ From then on, all my attempts were fruitless.

user170231
  • 25,320
Mathsource
  • 5,633

2 Answers2

3

If we use the series expanion of the arc cosine function, we have $$\frac {\pi^2}4-\sum_{n=0}^\infty \frac{4^{-n}\, (2 n)!}{(2 n+1)\, (n!)^2}\int_0^{\frac \pi 2}\left(\frac{\cos (x)}{\cos (x)+1}\right)^{2 n+1}$$ I suppose that $$J_n= \int_0^{\frac \pi 2}\left(\frac{\cos (x)}{\cos (x)+1}\right)^{2 n+1}$$ is given in terms of hypergeometric functions but computed one at the time, they are not bad.

They write $$J_n=\frac \pi 2 - a_n$$ where the first $a_n$ are $$\left\{1,\frac{22}{15},\frac{488}{315}, \frac{70544}{45045},\frac{1202048}{7 65765},\frac{22850816}{14549535},\frac {2628156416}{1673196525},\cdots\right\}$$ Alas, nothing in $OEIS$ for these numbers.

Limited to the given coefficients, this gives, as an approximation, $$I \sim \frac {\pi^2}4+\frac{23223252203341}{15073827493725}-\frac{62566171 }{92252160}\pi= \color{red}{1.877381}28$$ The summation converges quite fast. Adding the next term, we have $\color{red}{1.8773810}99$

Claude Leibovici
  • 289,558
3

For second integral I use substitution $t=\tan \left(\frac{x}{2}\right)$:

$$\int_0^{\frac{\pi }{2}} \cos ^{-1}\left(\frac{\cos (x)}{1+\cos (x)}\right) \, dx=\\\frac{\pi ^2}{4}-\int_0^{\frac{\pi }{2}} \frac{x \sin (x)}{\sqrt{1+2 \cos (x)} (1+\cos (x))} \, dx=\\\frac{\pi ^2}{4}-4 \int_0^1 \frac{ t \tan ^{-1}(t)}{\sqrt{3+2 t^2-t^4}} \, dt=\\\frac{\pi ^2}{4}-4 \int_0^1 \frac{\tan ^{-1}(t) t}{\sqrt{-t^2+3} \sqrt{t^2+1}} \, dt=\\\frac{\pi ^2}{4}-4 \int_0^1 \frac{t \left(\sum _{n=0}^{\infty } \frac{(-1)^n t^{1+2 n}}{1+2 n}\right) \sum _{m=0}^{\infty } t^{2 m} \binom{-\frac{1}{2}}{m}}{\sqrt{-t^2+3}} \, dt=\\\frac{\pi ^2}{4}-4 \sum _{m=0}^{\infty } \sum _{n=0}^{\infty } \int_0^1 \frac{t \left((-1)^n t^{1+2 n}\right) t^{2 m} \binom{-\frac{1}{2}}{m}}{\sqrt{-t^2+3} (1+2 n)} \, dt=\\\frac{\pi ^2}{4}-4 \sum _{m=0}^{\infty } \sum _{n=0}^{\infty } \frac{(-1)^n 3^{1+m+n} \sqrt{\pi } B_{\frac{1}{3}}\left(\frac{3}{2}+m+n,\frac{1}{2}\right)}{(2+4 n) \Gamma \left(\frac{1}{2}-m\right) \Gamma (1+m)}=\\\frac{\pi ^2}{4}-4 \sum _{m=0}^{\infty } \sum _{n=0}^{\infty } \frac{(-1)^n \sqrt{\frac{\pi }{3}} \, _2F_1\left(\frac{1}{2},\frac{3}{2}+m+n;\frac{5}{2}+m+n;\frac{1}{3}\right)}{\left(\frac{3}{2}+m+n\right) (2+4 n) \Gamma \left(\frac{1}{2}-m\right) \Gamma (1+m)}=\\\frac{\pi ^2}{4}-4 \sum _{m=0}^{\infty } \sum _{n=0}^{\infty } \frac{\left((-1)^n \sqrt{\frac{\pi }{3}}\right) \sum _{r=0}^{\infty } \frac{3^{-r} \left(\frac{1}{2}\right)_r \left(\frac{3}{2}+m+n\right)_r}{r! \left(\frac{5}{2}+m+n\right)_r}}{\left(\frac{3}{2}+m+n\right) (2+4 n) \Gamma \left(\frac{1}{2}-m\right) \Gamma (1+m)}=\\\frac{\pi ^2}{4}-4 \sum _{m=0}^{\infty } \sum _{n=0}^{\infty } \sum _{r=0}^{\infty } \frac{(-1)^{-m+n} 3^{-\frac{1}{2}-r} \Gamma \left(\frac{1}{2}+r\right) \left(\frac{1}{2}\right)_m}{(1+2 n) \sqrt{\pi } (3+2 m+2 n+2 r) \Gamma (1+m) \Gamma (1+r)}=\\\frac{\pi ^2}{4}-\sum _{m=0}^{\infty } \sum _{n=0}^{\infty } \sum _{r=0}^{\infty } \frac{4 \left(\left(\frac{1}{2}\right)_m \left(\frac{1}{2}\right)_n \left(\frac{1}{2}\right)_r (1)_n \left(\frac{3}{2}\right)_{m+n+r}\right) \left((-1)^m (-1)^n \left(\frac{1}{3}\right)^r\right)}{\left(3 \sqrt{3}\right) \left(\left(\frac{3}{2}\right)_n \left(\frac{5}{2}\right)_{m+n+r}\right) (m! n! r!)}$$

the last triple series can by expressed by: Srivastava - Daous multivariable hypergeometric function.

Mathematica code:

Pi^2/4 - N[Total@Total@
Total@Table[
  4/(3 Sqrt[3])*( 
   Pochhammer[1/2, m] Pochhammer[1/2, n] Pochhammer[1/2, 
     r] Pochhammer[1, n] Pochhammer[3/2, m + n + r])/(
   Pochhammer[3/2, n] Pochhammer[5/2, m + n + r])*((-1)^m (-1)^
    n (1/3)^r)/(m! n! r! ), {m, 0, 100}, {n, 0, 100}, {r, 0, 100}],20]

(1.87705047515715598394)


(Second code work faster,but needs Compiler C ! )


Q = Compile[{{z, _Integer}}, Pi^2/4 - 
Total@Total@
  Total@Total@
    Table[4/(3 Sqrt[3])( 
      Pochhammer[1/2, m] Pochhammer[1/2, n] Pochhammer[1/2, 
        r] Pochhammer[1, n] Pochhammer[3/2, m + n + r])/(
      Pochhammer[3/2, n] Pochhammer[5/2, m + n + r])((-1)^m 
(-1)^n (1/3)^r)/(m! n! r! ), {m, 0, z}, {n, 0, z}, {r, 0, z}], 
RuntimeAttributes -> {Listable}, Parallelization -> True, 
RuntimeOptions -> "Speed", CompilationTarget -> "C"];
Q[200] // AbsoluteTiming


(**)
Mariusz Iwaniuk
  • 6,292