\[ \displaystyle\sum _{ n=1 }^{ \infty }{ \frac { \zeta (2n,x) }{ 4^{ n }(2n^{ 2 }+n) } } =(2x-1)\ln { (x-\frac { 1 }{ 2 }) } -2x+1+\ln { 2\pi } -2\ln { \Gamma (x) } \]

Prove the equation above, where \(\zeta(s,x)\) denotes the Hurwitz zeta function

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TopNewest-1/2 +1 what is that check your question... Again :/

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the 1/2 is supposed to be inside the ln,sorry for the confusion :)

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