Who's Up To the Challenge?82

On substituting $x=t+2\pi n$ ,(since we know $cos\left( t+2\pi n \right) =cos(t)$ )we can write $f\left( 2\pi n \right) =\int _{ 2\pi n }^{ \infty }{ \frac { cosx }{ x } } dx=\int _{ 0 }^{ \infty }{ \frac { cos\left( t \right) }{ \left( t+2\pi n \right) } dt }$ we know that $\int _{ 0 }^{ \infty }{ \frac { cos\left( t \right) }{ \left( t+2\pi n \right) } dt } =\int _{ 0 }^{ \infty }{ L \left\{ cos(t) \right\} } { L }^{ -1 }\left\{ \frac { 1 }{ t+2\pi n } \right\} ds=\int _{ 0 }^{ \infty }{ \left( \frac { s }{ { s }^{ 2 }+1 } \right) { e }^{ -2n\pi s } } ds$ on substituting the integral representation of $f\left( 2\pi n \right)$ into the summation we'll get $\sum _{ n=1 }^{ \infty }{ \int _{ 0 }^{ \infty }{ \left( \frac { s }{ { s }^{ 2 }+1 } \right) { e }^{ -2n\pi s } } ds } =\int _{ 0 }^{ \infty }{ \left( \frac { s }{ { s }^{ 2 }+1 } \right) \sum _{ n=1 }^{ \infty }{ { e }^{ -2\pi ns }ds } }$ by applying infinite geometric series approach we can compute the following sum to be ${ \sum _{ n=1 }^{ \infty }{ { e }^{ -2\pi ns }ds } =\frac { 1 }{ { e }^{ 2\pi s }-1 } }$ on rearranging the third equation of this solution part , we can write it as $\int _{ 0 }^{ \infty }{ \frac { s }{ \left( { s }^{ 2 }+1 \right) \left( { e }^{ 2\pi s }-1 \right) } } ds=\frac { -\psi \left( 1 \right) }{ 2 } -\frac { 1 }{ 4 }$ (binets second integral ) formula for digamma function) we know that $\psi \left( x \right) =-\gamma +\sum _{ n=0 }^{ \infty }{ \frac { \left( x-1 \right) }{ \left( n+1 \right) \left( n+x \right) } }$ therefore $\psi \left( 1 \right) =-\gamma$ and so our final result will be $\sum _{ n=1 }^{ \infty }{ f\left( 2\pi n \right) } =\frac { \gamma }{ 2 } -\frac { 1 }{ 4 }$