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Question Number 138068 by Ñï= last updated on 09/Apr/21
∑∞n=1x2nζ(2n)=12−πx2cotπx
Answered by Dwaipayan Shikari last updated on 09/Apr/21
∑∞n=1∑∞k=1x2nk2n=∑∞k=1x2k2−x2=x2∑∞k=11k−x−1k+x=x2(ψ(x+1)−ψ(1−x))=x2(ψ(x)+1x−ψ(1−x))=12−xπ2cot(πx)ψ(x)−ψ(1−x)=−πcot(πx)
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