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Question Number 138068 by Ñï= last updated on 09/Apr/21

                  Σ_(n=1) ^∞ x^(2n) ζ(2n)=(1/2)−((πx)/2)cot πx

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{x}^{\mathrm{2}{n}} \zeta\left(\mathrm{2}{n}\right)=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\pi{x}}{\mathrm{2}}\mathrm{cot}\:\pi{x} \\ $$

Answered by Dwaipayan Shikari last updated on 09/Apr/21

Σ_(n=1) ^∞ Σ_(k=1) ^∞ (x^(2n) /k^(2n) )=Σ_(k=1) ^∞ (x^2 /(k^2 −x^2 ))=(x/2)Σ_(k=1) ^∞ (1/(k−x))−(1/(k+x))  =(x/2)(ψ(x+1)−ψ(1−x))=(x/2)(ψ(x)+(1/x)−ψ(1−x))  =(1/2)−((xπ)/2)cot(πx)               ψ(x)−ψ(1−x)=−πcot(πx)

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}} }{{k}^{\mathrm{2}{n}} }=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}} }{{k}^{\mathrm{2}} −{x}^{\mathrm{2}} }=\frac{{x}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}−{x}}−\frac{\mathrm{1}}{{k}+{x}} \\ $$$$=\frac{{x}}{\mathrm{2}}\left(\psi\left({x}+\mathrm{1}\right)−\psi\left(\mathrm{1}−{x}\right)\right)=\frac{{x}}{\mathrm{2}}\left(\psi\left({x}\right)+\frac{\mathrm{1}}{{x}}−\psi\left(\mathrm{1}−{x}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}−\frac{{x}\pi}{\mathrm{2}}{cot}\left(\pi{x}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\psi\left({x}\right)−\psi\left(\mathrm{1}−{x}\right)=−\pi{cot}\left(\pi{x}\right) \\ $$

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