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This question already has an answer here:

WolframAlpha says that $\zeta(0) = - \frac{1}{2}$ but I can't seem to get that result.

I found that for $\Re(s) < 1 $, \begin{equation}\label{1} \zeta(s) = 2^s \pi^{s-1}\sin\Bigl(\frac{s\pi}{2}\Bigr)\Biggl[\int_{0}^\infty e^{-y}y^{-s}\,\, dy \Biggr]\zeta(1-s),\tag{1} \end{equation}

And for $\Re(s) > 0 $, $$\zeta(s) = \frac{1}{1-2^{1-s}} \cdot \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}.\tag{2} $$

If I want to calculate $\zeta(0)$, using (1), then on the RHS I get $\zeta(1)$ which is undefined.

So how would I calculate $\zeta(0)$?

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marked as duplicate by Mark Viola complex-analysis Feb 13 at 20:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ What's reference 1? $\endgroup$ – J. W. Tanner Feb 13 at 20:06
  • $\begingroup$ @J.W.Tanner supposed to refer to the first equation, labelled (1). $\endgroup$ – Gurjinder Feb 13 at 20:08
  • $\begingroup$ It looks better now. Thank you for fixing $\endgroup$ – J. W. Tanner Feb 13 at 20:12
  • $\begingroup$ @J.W.Tanner no problem. $\endgroup$ – Gurjinder Feb 13 at 20:22
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Use $$\Gamma(s)\zeta(s)=\int_0^\infty\frac{x^{s-1}}{e^x-1}\,dx.$$ Now $$\frac1{e^x-1}=\frac1x-\frac12+f(x)$$ where $f(x)=O(x)$ as $x\to0^+$. Then $$\Gamma(s)\zeta(s)=\frac1{s-1}-\frac1{2s}+\int_0^\infty x^{s-1}f(x)\,dx.$$ As $f(x)=O(x)$ near zero, this last integral is holomorphic for $\text{Re }s>-1$. Now consider the residue at the pole at zero.

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