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}}