Who's up to the challenge? 26

Let $\frac { 1 }{ x } =u$ then the above integral is

$\displaystyle\int _{ 1 }^{ \infty }{ \frac { \left\{ u(-1)^{ \left\lfloor u \right\rfloor } \right\} }{ { u }^{ 2 } } du }$ $= \displaystyle\sum _{ n=1 }^{ \infty }{ \displaystyle\int _{ n }^{ n+1 }{ \frac { \left\{ (-1)^{ \left\lfloor u \right\rfloor }u \right\} }{ { u }^{ 2 } } du } }$ $= \displaystyle\sum _{ n=1 }^{ \infty }{ \int _{ n }^{ n+1 }{ \frac { \left\{ (-1)^{ n }u \right\} }{ { u }^{ 2 } } du } }$ $= \displaystyle\sum _{ n=1 }^{ \infty }{ \displaystyle\int _{ 2n-1 }^{ 2n }{ \frac { \{ -u\} }{ { u }^{ 2 } } du } } +\displaystyle\sum _{ n=1 }^{ \infty }{ \int _{ 2n }^{ 2n+1 }{ \frac { \{ u\} }{ { u }^{ 2 } } du } }$ $= \displaystyle\sum _{ n=1 }^{ \infty }{ \displaystyle\int _{ 2n-1 }^{ 2n }{ \frac { 1-\{ u\} }{ { u }^{ 2 } } du } } +\displaystyle\sum _{ n=1 }^{ \infty } \int _{ 2n }^{ 2n+1 }{ { \frac { u-2n }{ u^{ 2 } } du } }$ $= \displaystyle\sum _{ n=1 }^{ \infty }{ \int _{ 2n-1 }^{ 2n }{ \frac { 2n-u }{ { u }^{ 2 } } du } } +\displaystyle\sum _{ n=1 }^{ \infty }{ (\ln { \left( \frac { 2n+1 }{ 2n } \right) } -\frac { 1 }{ 2n+1 } )}$ $= \displaystyle\sum _{ n=1 }^{ \infty }{ (\frac { 1 }{ 2n-1 } -\ln { \left( \frac { 2n }{ 2n-1 } \right) } ) } +\displaystyle\sum _{ n=1 }^{ \infty }{ (\ln { \left( \frac { 2n+1 }{ 2n } \right) } -\frac { 1 }{ 2n+1 } )}$ $= \displaystyle\sum _{ n=1 }^{ \infty }{ \ln { \left( \frac { (2n+1)(2n-1) }{ 4{ n }^{ 2 } } \right) } }$ $=1+\ln { \left[ \displaystyle\prod _{ n=1 }^{ \infty }{ \frac { (2n+1)(2n-1) }{ 4{ n }^{ 2 } } } \right] }$ $=\boxed{1+\ln { \left( \frac { 2 }{ \pi } \right) } }$