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Se here, for instance. If one is going to define an analytic continuation by a given function, it must take the values that the function takes, even if they are infinite. So yes this function converges nicely at the negative odd numbers, but it will have simple poles at the negative evens, as well as all other zeroes of $\zeta$, leading to unremovable ...


The easiest method takes a slightly different tact after your first inequality: $$A\sum \frac{1}{n^s} \leq A\sum\frac{1}{|n^s|}\leq A\sum\frac{1}{n^{|R(s)|}}$$ The last inequality holds because $|z|\geq |R(z)|$ for any complex number $z$.


The Parseval identity (a.k.a. Plancherel's Theorem) for the Fourier transform is $$ \int_{-\infty}^{\infty}|f(t)|^2dt = \int_{-\infty}^{\infty}|\hat{f}(t)|^2 dt,\;\;\; f \in L^2(\mathbb{R}). $$ I assume you're familiar with this theorem. If you have a function $f\in L^2(\mathbb{R})$ that is supported in $[0,\infty)$ only, then the Fourier transform ...

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