Who's up to the challenge? 27

this is a result of this ultimate generalized form

$\displaystyle\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ 1 }{ \left\{ \frac { x }{ y } \right\} ^{ a }\left\{ \frac { y }{ x } \right\} ^{ b }dxdy } } =\frac { a! }{ 2(b+1)! } \displaystyle\sum _{ n=1 }^{ \infty }{ \frac { (a+n)! }{ (b+n)! } (\zeta (a+n+1)-1) } +\frac { b! }{ 2(a+1)! } \displaystyle\sum _{ n=1 }^{ \infty }{ \frac { (b+n)! }{ (a+n)! } } (\zeta (b+n+1)-1)$

we let $\frac { x }{ y } =u$ then the integral converts to

$\displaystyle\int _{ 0 }^{ 1 }{ x\int _{ x }^{ \infty }{ \{ u\} ^{ a }\left\{ \frac { 1 }{ u } \right\} ^{ b }\frac { du }{ { u }^{ 2 } } } }$

integrating by parts we get

$\displaystyle\int _{ 0 }^{ 1 }{ x\int _{ x }^{ \infty }{ \{ u\} ^{ a }\left\{ \frac { 1 }{ u } \right\} ^{ b }\frac { du }{ { u }^{ 2 } } } } =\left( \frac { { x }^{ 2 } }{ 2 } \displaystyle\int _{ x }^{ \infty }{ \{ u\} ^{ a }\left\{ \frac { 1 }{ u } \right\} ^{ b }\frac { du }{ { u }^{ 2 } } } \right)\Big|_{x=0}^{x=1} +\frac { 1 }{ 2 } \displaystyle\int _{ 0 }^{ 1 }{ { x }^{ a }\left\{ \frac { 1 }{ x } \right\} ^{ b }dx } =\\ \frac { 1 }{ 2 } \displaystyle\int _{ 0 }^{ 1 }{ { x }^{ b }\left\{ \frac { 1 }{ x } \right\} ^{ a }dx } +\frac { 1 }{ 2 } \displaystyle\int _{ 0 }^{ 1 }{ { x }^{ a }\left\{ \frac { 1 }{ x } \right\} ^{ b }dx }$

see this for the evaluation of the above integrals

putting that in we get

$\displaystyle\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ 1 }{ \left\{ \frac { x }{ y } \right\} ^{ a }\left\{ \frac { y }{ x } \right\} ^{ b }dxdy } } =\frac { a! }{ 2(b+1)! } \displaystyle\sum _{ n=1 }^{ \infty }{ \frac { (a+n)! }{ (b+n)! } (\zeta (a+n+1)-1) } +\frac { b! }{ 2(a+1)! } \displaystyle\sum _{ n=1 }^{ \infty }{ \frac { (b+n)! }{ (a+n)! } } (\zeta (b+n+1)-1)$

putting $a=b=1$ we get $\boxed{ \large \displaystyle\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ 1 }{ \left \{ \frac { x }{ y } \right\} \left \{ \frac { y }{ x } \right \} \, dxdy=1-\dfrac { \zeta (2) }{ 2 } } } }$