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Question Number 134662 by mnjuly1970 last updated on 06/Mar/21

....nice calculus... prove that:: 𝛗= Σ_(n=1) ^∞ ((ζ(2n))/((n+1)(2n+1)))=(1/2) ...m.n...

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....{nice}\:\:{calculus}... \\ $$$$\:\:\:\:{prove}\:\:{that}:: \\ $$$$\:\:\:\boldsymbol{\phi}=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\zeta\left(\mathrm{2}{n}\right)}{\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:...{m}.{n}... \\ $$

Answered by Dwaipayan Shikari last updated on 06/Mar/21

Σ_(n=1) ^∞ ((2ζ(2n))/(2n+1))−((ζ(2n))/(n+1)) =∫_0 ^1 2Σ_(n=1) ^∞ Σ_(k=1) ^∞ (x^(2n) /k^(2n) )−Σ_(n=1) ^∞ Σ_(k=1) ^∞ (x^n /k^(2n) ) =∫_0 ^1 Σ_(k=1) ^∞ ((2x^2 )/(k^2 −x^2 ))−(x/(k^2 −x))dx =∫_0 ^1 xΣ_(k=1) ^∞ (1/(k−x))−(1/(k+x))−(1/2)(√x) Σ_(k=1) ^∞ (1/(k−(√x)))−(1/(k+(√x)))dx =∫_0 ^1 xψ(1+x)−xψ(1−x)−(1/2)(√x) ψ(1+(√x))+(√x)ψ(1−(√x))dx =1−∫_0 ^1 xπcot(πx)dx−(1/2)+(1/2)∫_0 ^1 (√x)πcot(π(√x))dx =(1/2)−∫_0 ^1 xπcot(πx)dx+∫_0 ^1 tπcot(πt)dt=(1/2)

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}\zeta\left(\mathrm{2}{n}\right)}{\mathrm{2}{n}+\mathrm{1}}−\frac{\zeta\left(\mathrm{2}{n}\right)}{{n}+\mathrm{1}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}} }{{k}^{\mathrm{2}{n}} }−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{k}^{\mathrm{2}{n}} } \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{x}^{\mathrm{2}} }{{k}^{\mathrm{2}} −{x}^{\mathrm{2}} }−\frac{{x}}{{k}^{\mathrm{2}} −{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {x}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}−{x}}−\frac{\mathrm{1}}{{k}+{x}}−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{{x}}\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}−\sqrt{{x}}}−\frac{\mathrm{1}}{{k}+\sqrt{{x}}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {x}\psi\left(\mathrm{1}+{x}\right)−{x}\psi\left(\mathrm{1}−{x}\right)−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{{x}}\:\psi\left(\mathrm{1}+\sqrt{{x}}\right)+\sqrt{{x}}\psi\left(\mathrm{1}−\sqrt{{x}}\right){dx} \\ $$$$=\mathrm{1}−\int_{\mathrm{0}} ^{\mathrm{1}} {x}\pi{cot}\left(\pi{x}\right){dx}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{{x}}\pi{cot}\left(\pi\sqrt{{x}}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}−\int_{\mathrm{0}} ^{\mathrm{1}} {x}\pi{cot}\left(\pi{x}\right){dx}+\int_{\mathrm{0}} ^{\mathrm{1}} {t}\pi{cot}\left(\pi{t}\right){dt}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by mnjuly1970 last updated on 06/Mar/21

tayebballah...sir payan... your solution is very nice and better than that me.. mercey...

$${tayebballah}...{sir}\:\:{payan}... \\ $$$${your}\:{solution}\:{is}\:{very}\:{nice} \\ $$$$\:{and}\:{better}\:{than}\:{that}\:{me}.. \\ $$$${mercey}... \\ $$

Commented by mnjuly1970 last updated on 06/Mar/21

(1/2)∫_0 ^( 1) (√x) πcot(π((√x) ))dx=^(WHY??) ∫_0 ^( 1) tπcot(πt)dt please explain....

$$\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \sqrt{{x}}\:\pi{cot}\left(\pi\left(\sqrt{{x}}\:\right)\right){dx}\overset{{WHY}??} {=}\int_{\mathrm{0}} ^{\:\mathrm{1}} {t}\pi{cot}\left(\pi{t}\right){dt} \\ $$$$\:\:\:\:\:{please}\:\:{explain}.... \\ $$

Answered by mnjuly1970 last updated on 07/Mar/21

Answered by mnjuly1970 last updated on 07/Mar/21

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