$\zeta(0)$ and the limit of $(1-s)\zeta(s)$ as $s\to 1$

Question

I am hoping to compute $\zeta(0)$ where $\zeta$ is of course the Riemann zeta function. My first attempt was to use the functional equation which yields: $$\zeta(0) = \frac{1}{\pi}\cos\left(\frac{\pi}{2}\right)\zeta(1)~.$$

Now, since $\cos(\pi/2)=0$ and $\zeta(1)\to\pm\infty~,$ it looks like L'Hospital would be my best friend here, but alas: what on earth is $\zeta'(1)$? If anything this seems to make it worse.

I am aware that this question has been asked several times already so I tried to find a nice answer for it and I sort of did. How to Compute $\zeta (0)$? DonAntonio offers a very nice solution, but it relies on the equation: $$\lim_{s\to1}~(1-s)\zeta(s)=-1~.$$

Again, my only tool for limit evaluation, L'Hospital, causes the same problem as above. Is there a nice elementary way of computing either of those limits? Apparently this is related to something called residue which I don't really know what that is. I tried to look it up and there seems to be some pretty heavy theory involved. I hope somebody can provide a more elementary explanation for all of this.
Thaks a lot,
Alex

Answer 1

You may consider that $$ \zeta(s)=\sum_{n\geq 1}\frac{1}{n^s} = \sum_{n\geq 1}\frac{1}{\Gamma(s)}\int_{0}^{+\infty}x^{s-1}e^{-nx}\,dx =\frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{x^{s-1}}{e^x-1}\,dx$$ holds for any $s$ such that $\Re(s)>1$. Similarly $$ \eta(s) = \sum_{n\geq 1}\frac{(-1)^{n+1}}{n^s} = \frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{x^{s-1}}{e^x+1}\,dx $$ holds for any $s$ such that $\Re(s)>0$. For $\Re(s)>1$ we may define the $\zeta$-function in terms of the $\eta$-function via $$ \eta(s) = \zeta(s)-2\sum_{n\geq 1}\frac{1}{(2n)^s} = \left(1-\frac{2}{2^s}\right)\zeta(s)\qquad\Rightarrow\quad \zeta(s)=\frac{2^s}{2^s-2}\eta(s) $$ such that $$ \zeta(s) = \frac{2^s}{2^s-2}\cdot\frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{x^{s-1}}{e^x+1}\,dx. $$ The RHS is convergent for $\Re(s)>0$, so the previous line provides an analytic continuation for the $\zeta$-function over such region. On the other hand, by integration by parts, $$ \frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{x^{s-1}}{e^x+1}\,dx=\frac{1}{\Gamma(s+1)}\int_{0}^{+\infty}\frac{x^{s}e^x}{(e^x+1)^2}\,dx $$ with the RHS being convergent for $\Re(s)>-1$, and providing the following analytic continuation over such region: $$ \zeta(s) = \frac{2^s}{2^s-2}\cdot\frac{1}{\Gamma(s+1)}\int_{0}^{+\infty}\frac{x^s e^x}{(e^x+1)^2}\,dx. $$ By evaluating the RHS at $s=0$ we get $$ \zeta(0) = -\int_{0}^{+\infty}\frac{e^x}{(e^x+1)^2}\,dx \stackrel{e^x\mapsto t}{=} -\int_{1}^{+\infty}\frac{dt}{(t+1)^2}=\color{red}{-\frac{1}{2}}.$$

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We may notice that $$ \zeta(s) = \frac{2^s}{2^s-2}\cdot\frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{x^{s-1}}{e^x+1}\,dx\quad \text{for }\Re(s)>0 $$ implies that $\zeta(s)$ has a simple pole with residue $1$ at $s=1$. Indeed $$ \left.\frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{x^{s-1}}{e^x+1}\,dx\right|_{s=1}=\log(2) $$ and in a neighbourhood of $s=1$ $$ \frac{2^s}{2^s-2}=\frac{2^{s-1}}{2^{s-1}-1}\sim \frac{1}{(s-1)\log 2}.$$ This is enough to state $\lim_{s\to 1}(1-s)\zeta(s)=\color{red}{-1}.$

Answer 2

$\newcommand{\multichoose}[2]{{#1}^{[\!\underline{#2}\!]}}$

If you want to show $$\lim_{s\to 1}(s-1)\zeta(s)=1\text{,}$$ here's a way that doesn't invoke other special functions, built on estimating the zeta sum by certain integrals.

Start with the rectangle rule for integration:

$$\int_{n}^{n+1}f(x)\mathrm{d}x\approx f(n+1)\text{.}$$ If derive the error term using integration by parts, we get $$\int_{n}^{n+1}f(x)\mathrm{d}x=f(n+1)-\int_{n}^{n+1}(x-n)f'(x)\mathrm{d}x\text{.}$$ This is just as much a relation expressing the value of $f$ in terms of its integrals:

$$f(n+1) = \int_{n}^{n+1}f(x)\mathrm{d}x+\int_{n}^{n+1}(x-n)f'(x)\mathrm{d}x\text{.}$$ Summing over $n$, we get $$\sum_{n=1}^{\infty}f(n)=\int_{1}^{\infty}f(x)\mathrm{d}x +f(1) +\int_{1}^{\infty}(x-\lfloor x \rfloor)f'(x)\mathrm{d}x\text{.}$$ Let $f(k)=k^{-s}$. Then $$\sum_{n=1}^{\infty}\frac{1}{k^s}=\int_{1}^{\infty}\frac{\mathrm{d}x}{x^s} +1 -s\int_{1}^{\infty}\frac{(x-\lfloor x \rfloor)}{x^{s+1}}\mathrm{d}x\text{,}$$

i.e., $$\boxed{\zeta(s)=\frac{1}{s-1} +1 -s\int_{1}^{\infty}\frac{x-\lfloor x \rfloor}{x^{s+1}}\mathrm{d}x}$$

(DLMF 25.2.8). The right side of this equation is defined for all $\Re s>0$, $s\neq 1$.

We can push this method farther. Rewrite the integral as

$$\begin{split} \int_{1}^{\infty}\frac{x-\lfloor x \rfloor}{x^{s+1}}\mathrm{d}x&=\int_{1}^{\infty}\frac{x-\lfloor x \rfloor-\tfrac{1}{2}+\tfrac{1}{2}}{x^{s+1}}\mathrm{d}x\\ &=\frac{1}{2}\int_{1}^{\infty}\frac{\mathrm{d}x}{x^{s+1}}+\int_{1}^{\infty}\frac{x-\lfloor x \rfloor-\tfrac{1}{2}}{x^{s+1}}\mathrm{d}x \\ &=\frac{1}{2s}+(s+1)\int_{1}^{\infty}\frac{b_2(x-\lfloor x\rfloor)}{x^{s+2}}\mathrm{d}x \end{split}$$

where $b_2(u)=\tfrac{1}{2}(u^2-u)$. In these steps, we separated out the mean value of $x-\lfloor x \rfloor$ then integrated by parts. Therefore

$$\boxed{\zeta(s)=\frac{1}{s-1} +1 -\frac{1}{2} -s(s+1)\int_{1}^{\infty}\frac{b_2(x-\lfloor x\rfloor)}{x^{s+2}}\mathrm{d}x}\text{.}$$ The right side of this equation is defined for all $\Re s > -1$, $s\neq 1$.

Then $\lim_{s\to 1}(s-1)\zeta(s)=1$ and $\zeta(0)=-\tfrac{1}{2}$ follow by direct substitution into the boxed equations.

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The two expressions above are special cases of

$$\zeta(s)=\frac{1}{s-1}+1+\sum_{k=0}^{n-1}\multichoose{s}{k}\frac{B_{k+1}}{k+1}-\multichoose{s}{n+1}\int_1^{\infty}\frac{(B_{n+1}(x-\lfloor x \rfloor) -B_{n+1})\mathrm{d}x}{x^{s+n+1}}\text{,}$$ valid for $\Re s > -n$, $s\neq 1$ (DLMF 25.2.10); here $\multichoose{s}{k}$ are the multiset coefficients, $B_k(u)$ are the Bernoulli polynomials, and $B_k$ are the Bernoulli numbers:

$$\begin{align} (1-t)^{-s}&=\sum_{k=0}^{\infty}\multichoose{s}{k}t^k \\ \frac{t\mathrm{e}^{tu}}{\mathrm{e}^t-1}&=\sum_{k=0}^{\infty}B_k(u) \frac{t^k}{k!} \\ \frac{t}{\mathrm{e}^t-1}&=\sum_{k=0}^{\infty}B_k \frac{t^k}{k!}\text{.} \end{align}$$

Answer 3

Given your question you should prove carefully this theorem :

• For $\Re(s) > 1$ $$\zeta(s)-\frac1{s-1} = \sum_{n=1}^\infty(n^{-s}-\int_n^{n+1} x^{-s}dx)\tag{1}$$ Since $n^{-s}-\int_n^{n+1} x^{-s}dx=\int_n^{n+1}\int_n^x st^{-s-1}dtdx$ the RHS of $(1)$ converges absolutely and it is continuous and analytic for $\Re(s) > 0$,

  Thus we have found the analytic continuation of $\zeta(s)-\frac1{s-1}$ to $\Re(s) > 0$.

Since the RHS of (1) is continuous it is immediate that $$\lim_{s \to 1} (s-1)(\zeta(s)-\frac1{s-1}) = 0 \implies \lim_{s \to 1} (s-1)\zeta(s)=1$$

• To find $\zeta(0)$ we can continue following the same lines : $\sum_{n=1}^\infty \int_n^{n+1}\int_n^x (t^{-s-1}-n^{-s-1})dtdx$ converges and is analytic for $\Re(s) > -1$ thus $$\zeta(s)-\frac1{s-1}-\frac12 s\zeta(s+1)=\zeta(s)-\frac1{s-1}-s\sum_{n=1}^\infty \int_n^{n+1}(\int_n^x n^{-s-1}dt)dx\\=s\sum_{n=1}^\infty \int_n^{n+1}\int_n^x (t^{-s-1}-n^{-s-1})dtdx$$ is analytic for $\Re(s) > -1$ which means $$\zeta(0) = \lim_{s \to 0}\frac1{s-1}+\frac12 s\zeta(s+1)=-\frac12$$

• Going further leads to the 2nd, 3rd Bernouilli polynomials (K B Dave's answer) from which we have such an expression valid for $\Re(s) > -2,-3,..$ and so on.

Answer 4

See Hadamar Product for details: $$ \zeta\left(s\right) = \left.{\pi^{s/2}\prod_{\rho}\left(1 - s/\rho\right) \over 2\left(s - 1\right)\Gamma\left(1 + s/2\right)}\right\vert_{\ s\ =\ 0} =\ \bbox[#ffc,15px,border:1px solid navy]{-\,{1 \over 2}} $$