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


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

to find this?

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

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 $\zeta(0)=-\frac{1}{2}$.

  • $\begingroup$ 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. $\endgroup$ – GarouDan Nov 6 '11 at 22:25
  • 3
    $\begingroup$ @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 $$ \int_0^\infty\frac{1}{e^t+1}xt^{x-1}\;\mathrm{d}t=\left.\frac{t^x}{e^t+1}\right|_0^\infty+\int_0^\infty t^x\frac{e^t\;\mathrm{d}t}{(e^t+1)^2} $$ $\endgroup$ – robjohn Nov 7 '11 at 3:12
  • 2
    $\begingroup$ @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 $s=e^t$. $\endgroup$ – robjohn Nov 7 '11 at 3:12
  • $\begingroup$ Thx a lot @robjohn, this clarified everything. Nice answer^^ $\endgroup$ – GarouDan Nov 8 '11 at 1:28
  • $\begingroup$ @robjohn, the second part of your integral if x=0 then the integral is zero at that point (including zero )unless is (0 , infinity) and so the answer should be zero $\endgroup$ – user12878 Mar 30 '12 at 4:00

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.

  • $\begingroup$ Why can we consider $\eta(0)=1-1+1-1+1....=\frac{1}{2}$ in this case? $\endgroup$ – GarouDan Nov 5 '11 at 13:14
  • $\begingroup$ 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. $\endgroup$ – J. M. is a poor mathematician Nov 5 '11 at 13:21
  • $\begingroup$ 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? $\endgroup$ – Gottfried Helms Nov 5 '11 at 13:54
  • $\begingroup$ Nice use of the Euler transformation (one of my favorite tools). (+1) $\endgroup$ – robjohn Nov 5 '11 at 16:58

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