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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.2012-03-30
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    Is $2it\pi it$ a typo? Where does it come from? Why don't you write $-2\pi t^2$?2012-03-30

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