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Jameson

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Thanks,

Jameson

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- #1

Jameson

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Thanks,

Jameson

- #2

benorin

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Which long expression do you mean? Do you mean the analytic continuation via functional equation?

Do you want the globally convergent (entire complex plane except@z=1) continuation of [tex]\zeta(z)[/tex]? It is obtained by first doing the usual continuation of [tex]\sum_{k=1}^{\infty}\frac{1}{k^z}[/tex] which converges for Re[z]>1 to [tex]\frac{1}{1-2^{1-z}}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^z}[/tex] which converges for Re[z]>0 :zzz: . This can be further continued to [tex]z\in \mathbb{C}\setminus\{1\}[/tex] applying Euler's series transformation to the prior series to obtain

[tex]\frac{1}{1-2^{1-z}}\sum_{n=0}^{\infty}\frac{1}{2^{n+1}}\sum_{k=0}^{n}(-1)^k\left(\begin{array}{cc}n\\k\end{array}\right)(k+1)^{-z}[/tex].

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- #3

Jameson

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[tex]\zeta(1-s)=2^{1-s}{\pi}^{-s}(\sin{\frac{(1-s)\pi}{2}}})(s-1)!\zeta(s)[/tex]

- #4

shmoe

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Riemann provided two methods of proving the functional equation in his original paper (though in a symmetric form, it's just minor fiddling to get the version you want). Here's a translation:

http://www.maths.tcd.ie/pub/HistMath/People/Riemann/Zeta/EZeta.pdf

Titchmarhs's "Theory of the Riemann Zeta Function" has several different proofs, as will any book with "Zeta function" in the title and many with "Analytic number theory" will as well.

http://www.maths.tcd.ie/pub/HistMath/People/Riemann/Zeta/EZeta.pdf

Titchmarhs's "Theory of the Riemann Zeta Function" has several different proofs, as will any book with "Zeta function" in the title and many with "Analytic number theory" will as well.

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- #5

benorin

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This one gives the symmetric form of the above, from there you can use your knowledge of the gamma function to obtain the desired form. And by this one, I do mean exercise #21 from pg. 51 of Andrews, G., Askey, R., & Roy, R. (2000). Encyclopedia of mathematics and its applications: special functions . 1st paperback ed. Cambridge, UK: Cambridge University Press., which reads as follows:Jameson said:

[tex]\zeta(1-s)=2^{1-s}{\pi}^{-s}(\sin{\frac{(1-s)\pi}{2}}})(s-1)!\zeta(s)[/tex]

Prove that [tex]\pi^{-\frac{s}{2}}\Gamma(\frac{s}{2})\zeta(s)=\pi^{-\frac{1-s}{2}}\Gamma(\frac{1-s}{2})\zeta(1-s)[/tex] as follows:

a) Observe that [tex]\sum_{n=0}^{\infty}\frac{\sin((2n+1)x)}{2n+1}=(-1)^{m}\frac{\pi}{4}[/tex] for [tex]m\pi<x<(m+1)\pi, m\in\mathbb{N}[/tex]

b) Multiply the equation by [tex]x^{s-1}[/tex] (0 < s < 1) and integrate over [tex](0,\infty)[/tex]. Show that the left side is [tex]\Gamma(s)\sin(\frac{s\pi}{2})(1-2^{-s-1})\zeta(s+1)[/tex] and that the right represents an analytic function for [tex]\Re(s)<1[/tex] and is equal to [tex]2(1-2^{s+1}) \zeta(1-s)[/tex] for [tex]\Re(s)<0[/tex].

c) Deduce the functional equation for the zeta function. (Hardy)

(I added the smiley, and note that the text had the sum in part (a) starting at n=1, which my prof. confirmed is a typo).

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