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I would like to know how we can find the following result:

$$\zeta(0)=-\frac12$$

Is there a way, using the definition, $$\zeta(s)=\sum_{i=1}^{\infty}i^{-s}$$

to find this?

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2  
Shouldn't you let the index i begin at 1 ? –  Gottfried Helms Nov 5 '11 at 3:07
    
Yes. I'll edit. –  GarouDan Nov 5 '11 at 12:54
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2 Answers

up vote 13 down vote accepted

Consider the integral $$ \begin{align} \int_0^\infty\frac{xt^{x-1}}{e^t+1}\mathrm{d}t &=\int_0^\infty\frac{xt^{x-1}}{1+e^{-t}}e^{-t}\;\mathrm{d}t\\ &=x\sum_{k=1}^\infty(-1)^{k-1}\int_0^\infty t^{x-1}e^{-kt}\;\mathrm{d}t\\ &=x\sum_{k=1}^\infty(-1)^{k-1}k^{-x}\int_0^\infty t^{x-1}e^{-t}\;\mathrm{d}t\\ &=x\eta(x)\Gamma(x)\\ &=(1-2^{1-x})\zeta(x)\Gamma(x+1)\tag{1} \end{align} $$ Integrate by parts to get $$ \begin{align} \lim_{x\to0^+}\int_0^\infty\frac{xt^{x-1}}{e^t+1}\mathrm{d}t &=\lim_{x\to0^+}\int_0^\infty\frac{t^xe^t}{(e^t+1)^2}\mathrm{d}t\\ &=\int_1^\infty\frac{\mathrm{d}u}{(u+1)^2}\\ &=\frac{1}{2}\tag{2} \end{align} $$ Sending $x$ to $0$ in $(1)$ and combining with $(2)$, we get $\zeta(0)=-\frac{1}{2}$.

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+1. Cute. You're a master with the syntax :-). –  Jonas Teuwen Nov 5 '11 at 17:00
    
This was a bit hard to understand because you jumped some "trivial" steps, not so trivial to me =). I need to understand the part after Integrate by parts to get and then I'll embrace your answer. At the end I realized I'm more interested in the result, not find the result by definition and you did well. Thx. –  GarouDan Nov 6 '11 at 22:25
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@GarouDan: Let $u=\frac{1}{e^t+1}$ and $\mathrm{d}v=xt^{x-1}\;\mathrm{d}t$, then $v=t^x$ and $\mathrm{d}u=-\frac{e^t\;\mathrm{d}t}{(e^t+1)^2}$. Integration by Parts then says that $$ \int_0^\infty\frac{1}{e^t+1}xt^{x-1}\;\mathrm{d}t=\left.\frac{t^x}{e^t+1}\right|‌​_0^\infty+\int_0^\infty t^x\frac{e^t\;\mathrm{d}t}{(e^t+1)^2} $$ –  robjohn Nov 7 '11 at 3:12
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@GarouDan: When $x>0$, $\frac{t^x}{e^t+1}=0$ at both $t=0$ and $t=\infty$. Thus, when taking $\lim\limits_{x\to0^+}$ of both sides, the limit term (on the left of the sum) vanishes. Therefore, $$ \begin{align} \lim_{x\to0^+}\int_0^\infty\frac{1}{e^t+1}xt^{x-1}\;\mathrm{d}t &=\int_0^\infty\frac{e^t\;\mathrm{d}t}{(e^t+1)^2}\\ &=\int_1^\infty\frac{\mathrm{d}s}{(s+1)^2} \end{align} $$ where $s=e^t$. –  robjohn Nov 7 '11 at 3:12
    
Thx a lot @robjohn, this clarified everything. Nice answer^^ –  GarouDan Nov 8 '11 at 1:28
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The "defining sum" converges only for $\Re s > 1$. One can however use the related formula

$$\zeta(s)=\frac1{1-2^{1-s}}\sum_{n=0}^\infty \frac1{2^{n+1}}\sum_{k=0}^n (-1)^k \binom{n}{k} (k+1)^{-s}$$

for $s=0$. This complicated sum can be derived by applying the Euler transformation to the series for Dirichlet $\eta$, the alternating version of Riemann's function.

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Why can we consider $\eta(0)=1-1+1-1+1....=\frac{1}{2}$ in this case? –  GarouDan Nov 5 '11 at 13:14
    
You do what I indicated above: since the usual series for Dirichlet $\eta$ converges only for $\Re s > 0$, performing the Euler transformation is what's needed to have an analytic continuation. The other possibility is to do analytic continuation via Abel summation. See this article as well. –  J. M. Nov 5 '11 at 13:21
    
It is important always to keep uniqueness with respect to the single terms in the series, so if you want to derive a result for $\small \eta(0) $ you should write this as $\small \eta(0) = \lim_{s->0} \eta(s) = 1^{-s} - 2^{-s} + 3^{-s} - \ldots + \ldots $ so each of your ones above is uniquely identified. Only then you may operate algebraically until you have a form (or formula) which allows to read off the final value by neglecting residuals, which vanish as s goes to zero. I think the derivation for the $\small \eta(0) $ was given in the wikipedia? –  Gottfried Helms Nov 5 '11 at 13:54
    
Nice use of the Euler transformation (one of my favorite tools). (+1) –  robjohn Nov 5 '11 at 16:58
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