2n-2n-n-Z-

Question Number 3156 by prakash jain last updated on 06/Dec/15

$$\Gamma\left(−\mathrm{2}{n}\right)\zeta\left(−\mathrm{2}{n}\right)=? \\ $$$${n}\in\mathbb{Z}^{+} \\ $$

Commented by prakash jain last updated on 06/Dec/15

Wolfram alpha Γ(−2)ζ(−2) is undefined. ζ(s)=2^s π^(s−1) sin (((πs)/2))Γ(1−s)ζ(1−s) ζ(3)=2^3 π^2 sin ((3π)/2)Γ(−2)ζ(−2) How is ζ(3)=1.2020 defined?

$$\mathrm{Wolfram}\:\mathrm{alpha} \\ $$$$\Gamma\left(−\mathrm{2}\right)\zeta\left(−\mathrm{2}\right)\:\mathrm{is}\:\mathrm{undefined}. \\ $$$$\zeta\left({s}\right)=\mathrm{2}^{{s}} \pi^{{s}−\mathrm{1}} \mathrm{sin}\:\left(\frac{\pi{s}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−{s}\right)\zeta\left(\mathrm{1}−{s}\right) \\ $$$$\zeta\left(\mathrm{3}\right)=\mathrm{2}^{\mathrm{3}} \pi^{\mathrm{2}} \mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{2}}\Gamma\left(−\mathrm{2}\right)\zeta\left(−\mathrm{2}\right) \\ $$$$\mathrm{How}\:\mathrm{is}\:\zeta\left(\mathrm{3}\right)=\mathrm{1}.\mathrm{2020}\:\mathrm{defined}? \\ $$

Commented by 123456 last updated on 06/Dec/15

$$\zeta\left(\mathrm{3}\right)=\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} } \\ $$

Commented by prakash jain last updated on 06/Dec/15

Functional equation should give the same result? I am trying to find out how to conclude that we get some regular result from functional equation.

$$\mathrm{Functional}\:\mathrm{equation}\:\mathrm{should}\:\mathrm{give}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{result}?\:\mathrm{I}\:\mathrm{am}\:\mathrm{trying}\:\mathrm{to}\:\mathrm{find}\:\mathrm{out} \\ $$$$\mathrm{how}\:\mathrm{to}\:\mathrm{conclude}\:\mathrm{that}\:\mathrm{we}\:\mathrm{get}\:\mathrm{some}\:\mathrm{regular} \\ $$$$\mathrm{result}\:\mathrm{from}\:\mathrm{functional}\:\mathrm{equation}. \\ $$

Commented by 123456 last updated on 06/Dec/15

Γ(z) has simple poles at z∈{0,−1,−2,...} so there some poblem in using it direct because this he dont have ζ(0)=0, so this ′trivial′ zero is anuled

$$\Gamma\left({z}\right)\:\mathrm{has}\:\mathrm{simple}\:\mathrm{poles}\:\mathrm{at}\: \\ $$$${z}\in\left\{\mathrm{0},−\mathrm{1},−\mathrm{2},…\right\} \\ $$$$\mathrm{so}\:\mathrm{there}\:\mathrm{some}\:\mathrm{poblem}\:\mathrm{in}\:\mathrm{using}\:\mathrm{it}\:\mathrm{direct} \\ $$$$\mathrm{because}\:\mathrm{this}\:\mathrm{he}\:\mathrm{dont}\:\mathrm{have} \\ $$$$\zeta\left(\mathrm{0}\right)=\mathrm{0},\:\mathrm{so}\:\mathrm{this}\:'\mathrm{trivial}'\:\mathrm{zero}\:\mathrm{is}\:\mathrm{anuled} \\ $$

Commented by prakash jain last updated on 06/Dec/15

I expected that since Γ has simple pole at −2n and ζ =0 at −2n. But Γ(−2n)ζ(−2n) could be regular value. Now considering that ζ(3)=x Why not Γ(−2)ζ(−2)=(x/(2^3 π^2 sin ((3π)/2))) why is it undefined.

$$\mathrm{I}\:\mathrm{expected}\:\mathrm{that}\:\mathrm{since}\:\Gamma\:\mathrm{has}\:\mathrm{simple}\:\mathrm{pole} \\ $$$$\mathrm{at}\:−\mathrm{2}{n}\:\mathrm{and}\:\zeta\:=\mathrm{0}\:\mathrm{at}\:−\mathrm{2}{n}.\:\mathrm{But}\:\Gamma\left(−\mathrm{2}{n}\right)\zeta\left(−\mathrm{2}{n}\right) \\ $$$$\mathrm{could}\:\mathrm{be}\:\mathrm{regular}\:\mathrm{value}. \\ $$$$\mathrm{Now}\:\mathrm{considering}\:\mathrm{that}\:\zeta\left(\mathrm{3}\right)={x} \\ $$$$\mathrm{Why}\:\mathrm{not}\:\Gamma\left(−\mathrm{2}\right)\zeta\left(−\mathrm{2}\right)=\frac{{x}}{\mathrm{2}^{\mathrm{3}} \pi^{\mathrm{2}} \mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{2}}} \\ $$$$\mathrm{why}\:\mathrm{is}\:\mathrm{it}\:\mathrm{undefined}. \\ $$

Commented by 123456 last updated on 06/Dec/15

$$\mathrm{i}\:\mathrm{think}\:\mathrm{that}\:\mathrm{the}\:\mathrm{key}\:\mathrm{is}\:\mathrm{limit} \\ $$

Commented by prakash jain last updated on 06/Dec/15

Thanks a lot. Now I realize my mistake. lim_(x→−2) Γ(x)ζ(x) = ((ζ(3))/(2^3 π^2 sin ((3π)/2)))

$$\mathrm{Thanks}\:\mathrm{a}\:\mathrm{lot}. \\ $$$$\mathrm{Now}\:\mathrm{I}\:\mathrm{realize}\:\mathrm{my}\:\mathrm{mistake}.\: \\ $$$$\underset{{x}\rightarrow−\mathrm{2}} {\mathrm{lim}}\Gamma\left({x}\right)\zeta\left({x}\right)\:=\:\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{2}^{\mathrm{3}} \pi^{\mathrm{2}} \mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{2}}} \\ $$

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