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$$\zeta(it)=2it\pi it−1\sin(i\pi t/2)\Gamma(1−it)\zeta(1−it).$$ Everything on the RHS is never zero,

Does that means LHS has no zeros, since $\sin(s)$ has a simple zero at $s=0$ while $\zeta(1−s)$ has a simple pole at $s=0$ (the Laurent expansion), so the product $\sin(i\pi t/2)\zeta(1−it)$ is finite and nonzero at $t=0$.

My question is that from functional equation one can't get direct answer $\zeta(0)=−1/2$ unless using alternate infinite series !

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For a question, that looks an awful lot like an assertion. –  Gerry Myerson Mar 30 '12 at 6:06
    
Is $2it\pi it$ a typo? Where does it come from? Why don't you write $-2\pi t^2$? –  draks ... Mar 30 '12 at 7:00

2 Answers 2

Do you mean the functional equation

$$\zeta(s) = 2(2\pi)^{s-1} \Gamma(1-s) \zeta(1-s) \sin (\frac{1}{2} \pi s)$$

?

which is valid for all $s \neq 1$ and $s \neq 0$.

Even though you cannot directly substitute $s =0$ or $s=1$, what you can do is multiply by $s-1$ to get

$$[\zeta(s)(s-1)] = 2(2\pi)^{s-1} [\Gamma(1-s) (s-1)] \zeta(1-s) \sin (\frac{1}{2} \pi s) \tag{1}$$

$\zeta$ has a simple pole at $1$ with Residue $1$, and $\Gamma$ has a simple pole at $0$ with Residue $1$.

Taking $s \to 1$ in 1) now gives you the result that $\displaystyle \zeta(0) = -\frac{1}{2}$.

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The LHS has no zeroes and therein lies the proof of the Prime Number Theorem.

You can get $\zeta(0)=-1/2$ from the functional equation; though you must be comfortable with limit manipulation as the poster above rightly states that the functional equation is not valid for $s=0,1$. The derivation at the end of this page is simple enough.

http://planetmath.org/valueoftheriemannzetafunctionats0.

Regards

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