Theorem: Mean Distance between Two Random Points in Unit Cube

The mean distance $R$ between $2$ points chosen at random from the interior of a unit cube is given by:
{{begin-eqn}}
{{eqn | l = R
      | r = \frac {4 + 17 \sqrt 2 - 6 \sqrt3 - 7 \pi} {105} + \frac {\map \ln {1 + \sqrt 2 } } 5 + \frac {2 \, \map \ln {2 + \sqrt 3} } 5
}}
{{eqn | o = \approx
      | r = 0 \cdotp 66170 \, 71822 \, 67176 \, 23515 \, 582 \ldots
      | c = 
}}
{{end-eqn}}
The value $R$ is known as the Robbins constant.

Proof:
From Mean Distance between Two Random Points in Cuboid:
{{:Mean Distance between Two Random Points in Cuboid}}
The result follows by setting $a = b = c = \dfrac 1 2$.
Hence we have:
{{begin-eqn}}
{{eqn | l = r
      | r = \sqrt {\dfrac 1 {2^2} + \dfrac 1 {2^2} + \dfrac 1 {2^2} }
      | c = 
}}
{{eqn | r = \sqrt {\dfrac 3 4}
      | c = 
}}
{{eqn | r = \dfrac {\sqrt 3} 2
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = r_1 = r_2 = r_3
      | r = \sqrt {\dfrac 1 {2^2} + \dfrac 1 {2^2} }
      | c = 
}}
{{eqn | r = \sqrt {\dfrac 2 4}
      | c = 
}}
{{eqn | r = \dfrac {\sqrt 2} 2
      | c = 
}}
{{end-eqn}}
So:
:$r - r_1 = r - r_2 = r - r_3 = \dfrac {\sqrt 3 - \sqrt 2} 2$
Thus:
{{begin-eqn}}
{{eqn | l = E
      | r = \dfrac 2 {15} \dfrac {\sqrt 3} 2 - \dfrac 7 {45} \paren {\dfrac {\sqrt 3 - \sqrt 2} 2 \paren {\dfrac {\sqrt 2} {2 a} }^2 + \dfrac {\sqrt 3 - \sqrt 2} 2 \paren {\dfrac {\sqrt 2} {2 b} }^2 + \dfrac {\sqrt 3 - \sqrt 2} 2 \paren {\dfrac {\sqrt 2} {2 c} }^2}
      | c = substituting for all instances of $r$, $r_1$ etc.
}}
{{eqn | o =
      | ro= +
      | r = \dfrac 8 {315 a^2 b^2 c^2} \paren {a^7 + b^7 + c^7 - 3 \paren {\dfrac {\sqrt 2} 2}^7 + \paren {\dfrac {\sqrt 3} 2}^7}
      | c = 
}}
{{eqn | o =
      | ro= +
      | r = \dfrac 1 {15 a b^2 c^2} \paren {b^6 \sinh^{-1} \dfrac a b + c^6 \sinh^{-1} \dfrac a c -  \paren {\dfrac {\sqrt 2} 2}^2 \paren { \paren {\dfrac {\sqrt 2} 2}^4 - 8 b^2 c^2} \sinh^{-1} \dfrac {2 a} {\sqrt 2} }
      | c = 
}}
{{eqn | o =
      | ro= +
      | r = \dfrac 1 {15 a^2 b c^2} \paren {c^6 \sinh^{-1} \dfrac b c + a^6 \sinh^{-1} \dfrac b a -  \paren {\dfrac {\sqrt 2} 2}^2 \paren { \paren {\dfrac {\sqrt 2} 2}^4 - 8 c^2 a^2} \sinh^{-1} \dfrac {2 b} {\sqrt 2} }
      | c = 
}}
{{eqn | o =
      | ro= +
      | r = \dfrac 1 {15 a^2 b^2 c} \paren {a^6 \sinh^{-1} \dfrac c a + b^6 \sinh^{-1} \dfrac c b -  \paren {\dfrac {\sqrt 2} 2}^2 \paren { \paren {\dfrac {\sqrt 2} 2}^4 - 8 a^2 b^2} \sinh^{-1} \dfrac {2 c} {\sqrt 2} }
      | c = 
}}
{{eqn | o =
      | ro= -
      | r = \dfrac 4 {15 a b c} \paren {a^4 \, \map \arcsin {2 b c} + b^4 \, \map \arcsin {2 a c} + c^4 \, \map \arcsin {2 a b} }
      | c = 
}}
{{eqn | r = \dfrac 2 {15} \dfrac {\sqrt 3} 2 - \dfrac 7 {45} \paren {\dfrac {\sqrt 3 - \sqrt 2} 2 \paren {\sqrt 2}^2 + \dfrac {\sqrt 3 - \sqrt 2} 2 \paren {\sqrt 2}^2 + \dfrac {\sqrt 3 - \sqrt 2} 2 \paren {\sqrt 2}^2}
      | c = substituting for all instances of $a = b = c = \dfrac 1 2$
}}
{{eqn | o =
      | ro= +
      | r = \dfrac 8 {315 \paren {\dfrac 1 2}^2 \paren {\dfrac 1 2}^2 \paren {\dfrac 1 2}^2} \paren {\paren {\dfrac 1 2}^7 + \paren {\dfrac 1 2}^7 + \paren {\dfrac 1 2}^7 - 3 \paren {\dfrac {\sqrt 2} 2}^7 + \paren {\dfrac {\sqrt 3} 2}^7}
      | c = 
}}
{{eqn | o =
      | ro= +
      | r = \dfrac 1 {15 \paren {\dfrac 1 2} \paren {\dfrac 1 2}^2 \paren {\dfrac 1 2}^2} \paren {\paren {\dfrac 1 2}^6 \sinh^{-1} 1 + \paren {\dfrac 1 2}^6 \sinh^{-1} 1 -  \paren {\dfrac {\sqrt 2} 2}^2 \paren { \paren {\dfrac {\sqrt 2} 2}^4 - 8 \paren {\dfrac 1 2}^2 \paren {\dfrac 1 2}^2} \sinh^{-1} \dfrac 1 {\sqrt 2} }
      | c = 
}}
{{eqn | o =
      | ro= +
      | r = \dfrac 1 {15 \paren {\dfrac 1 2}^2 \paren {\dfrac 1 2} \paren {\dfrac 1 2}^2} \paren {\paren {\dfrac 1 2}^6 \sinh^{-1} 1 + \paren {\dfrac 1 2}^6 \sinh^{-1} 1 -  \paren {\dfrac {\sqrt 2} 2}^2 \paren { \paren {\dfrac {\sqrt 2} 2}^4 - 8 \paren {\dfrac 1 2}^2 \paren {\dfrac 1 2}^2} \sinh^{-1} \dfrac 1 {\sqrt 2} }
      | c = 
}}
{{eqn | o =
      | ro= +
      | r = \dfrac 1 {15 \paren {\dfrac 1 2}^2 \paren {\dfrac 1 2}^2 \paren {\dfrac 1 2} } \paren {\paren {\dfrac 1 2}^6 \sinh^{-1} 1 + \paren {\dfrac 1 2}^6 \sinh^{-1} 1 -  \paren {\dfrac {\sqrt 2} 2}^2 \paren { \paren {\dfrac {\sqrt 2} 2}^4 - 8 \paren {\dfrac 1 2}^2 \paren {\dfrac 1 2}^2} \sinh^{-1} \dfrac 1 {\sqrt 2} }
      | c = 
}}
{{eqn | o =
      | ro= -
      | r = \dfrac 4 {15 \paren {\dfrac 1 2} \paren {\dfrac 1 2} \paren {\dfrac 1 2} } \paren {\paren {\dfrac 1 2}^4 \, \map \arcsin {2 \paren {\dfrac 1 2} \paren {\dfrac 1 2} } + \paren {\dfrac 1 2}^4 \, \map \arcsin {2 \paren {\dfrac 1 2} \paren {\dfrac 1 2} } + \paren {\dfrac 1 2}^4 \, \map \arcsin {2 \paren {\dfrac 1 2} \paren {\dfrac 1 2} } }
      | c = 
}}
{{eqn | r = \dfrac {\sqrt 3} {15} - \dfrac {21} {45} \paren {\sqrt 3 - \sqrt 2}
      | c = simplification
}}
{{eqn | o =
      | ro= +
      | r = \dfrac {8 \times 2^6} {315} \paren {\dfrac 3 {2^7} - \dfrac {3 \sqrt 2^7} {2^7} + \dfrac {\sqrt 3^7} {2^7} }
      | c = 
}}
{{eqn | o =
      | ro= +
      | r = \dfrac {2^5} {15} \paren {\dfrac 1 {2^5} \sinh^{-1} 1 - \dfrac 1 2 \paren {\dfrac 1 4 - \dfrac 1 2} \sinh^{-1} \dfrac 1 {\sqrt 2} }
      | c = 
}}
{{eqn | o =
      | ro= +
      | r = \dfrac {2^5} {15} \paren {\dfrac 1 {2^5} \sinh^{-1} 1 - \dfrac 1 2 \paren {\dfrac 1 4 - \dfrac 1 2} \sinh^{-1} \dfrac 1 {\sqrt 2} }
      | c = 
}}
{{eqn | o =
      | ro= +
      | r = \dfrac {2^5} {15} \paren {\dfrac 1 {2^5} \sinh^{-1} 1 - \dfrac 1 2 \paren {\dfrac 1 4 - \dfrac 1 2} \sinh^{-1} \dfrac 1 {\sqrt 2} }
      | c = 
}}
{{eqn | o =
      | ro= -
      | r = \dfrac {4 \times 2^3} {15} \paren {\dfrac 3 {2^4} \arcsin \dfrac 1 2}
      | c = 
}}
{{eqn | r = \dfrac {3 \sqrt 3 - 21 \sqrt 3 + 21 \sqrt 2} {45} + \dfrac 4 {315} \paren {3 - 3 \sqrt 2^7 + \sqrt 3^7}
      | c = simplification
}}
{{eqn | o =
      | ro= +
      | r = \dfrac 1 5 \sinh^{-1} 1 + \dfrac 4 5 \sinh^{-1} \dfrac 1 {\sqrt 2} - \dfrac 2 5 \arcsin \dfrac 1 2
      | c = 
}}
{{eqn | r = \dfrac {21 \sqrt 2 - 18 \sqrt 3} {45} + \dfrac 4 {315} \paren {3 + 27 \sqrt 3 - 24 \sqrt 2}
      | c = simplification
}}
{{eqn | o =
      | ro= +
      | r = \dfrac 1 5 \map \ln {1 + \sqrt {1^2 + 1} } + \dfrac 4 5 \map \ln {\dfrac 1 {\sqrt 2} + \sqrt {\paren {\dfrac 1 {\sqrt 2} }^2 + 1} } - \dfrac 2 5 \arcsin \dfrac 1 2
      | c = Inverse Hyperbolic Sine Logarithmic Formulation
}}
{{eqn | r = \dfrac {7 \sqrt 2 - 6 \sqrt 3} {15} + \dfrac 4 {105} \paren {1 + 9 \sqrt 3 - 8 \sqrt 2}
      | c = simplification
}}
{{eqn | o =
      | ro= +
      | r = \dfrac 1 5 \map \ln {1 + \sqrt 2} + \dfrac 4 5 \map \ln {\dfrac {1 + \sqrt 3} {\sqrt 2} } - \dfrac 2 5 \dfrac \pi 6
      | c = Sine of $30 \degrees$
}}
{{eqn | r = \dfrac {49 \sqrt 2 - 42 \sqrt 3 + 4 + 36 \sqrt 3 - 32 \sqrt 2 - 7 \pi} {105}
      | c = common denominator
}}
{{eqn | o =
      | ro= +
      | r = \dfrac 1 5 \map \ln {1 + \sqrt 2} + \dfrac 2 5 \map \ln {\paren {\dfrac {1 + \sqrt 3} {\sqrt 2} }^2 }
      | c = Logarithm of Power
}}
{{eqn | r = \frac {4 + 17 \sqrt 2 - 6 \sqrt3 - 7 \pi} {105} + \dfrac {\map \ln {1 + \sqrt 2} } 5 + \dfrac {2 \, \map \ln {2 + \sqrt 3} } 5
      | c = tidying up
}}
{{end-eqn}}
{{qed}}