Riemann-zeta Function Evaluated at $\zeta(0)$ [duplicate]

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

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

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 $$\zeta(0)$$?

marked as duplicate by Mark Viola complex-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 13 at 20:05

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• What's reference 1? – J. W. Tanner Feb 13 at 20:06
• @J.W.Tanner supposed to refer to the first equation, labelled (1). – Gurjinder Feb 13 at 20:08
• It looks better now. Thank you for fixing – J. W. Tanner Feb 13 at 20:12
• @J.W.Tanner no problem. – Gurjinder Feb 13 at 20:22

1 Answer

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.