How is the Riemann zeta function zero at the negative even integers? [duplicate]

Solution 1:

For $\mathrm{Re}(s)\gt1$, $$ \zeta(s)=\sum_{n=1}^\infty\frac1{n^s}\tag{1} $$ For $\mathrm{Re}(s)\le1$, we need to use Analytic Continuation and another formula for $\zeta(s)$ since $(1)$ does not converge for $\mathrm{Re}(s)\le1$.

One formula that can be used is the Functional Equation for $\zeta$ which says $$ \zeta(s)\frac{\Gamma(s/2)}{\pi^{s/2}}=\zeta(1-s)\frac{\Gamma((1-s)/2)}{\pi^{(1-s)/2}}\tag{2} $$ The symmetric form in $(2)$ is given as $(14)$ on MathWorld.

Since the recurrence for $\Gamma$ says that for $n\in\mathbb{Z}$ and $n\le0$, $\frac1{\Gamma(n)}=0$, we get that $\zeta(2n)=0$ for $n\in\mathbb{Z}$ and $n\lt0$.

Analytic Continuation of $\boldsymbol{\zeta}$ Using $\boldsymbol{\eta}$ and Integration by Parts

Define $\eta$, the alternating $\zeta$ function, as $$ \begin{align} \eta(s) &=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}\\ &=\sum_{n=1}^\infty\frac1{n^s}-2\sum_{n=1}^\infty\frac1{(2n)^s}\\[6pt] &=\zeta(s)\left(1-2^{1-s}\right)\tag{3} \end{align} $$ Formula $(3)$ follows since the terms of an alternating series are the terms of the non-alternating series minus twice the even terms.

Formula $(3)$ increases the domain to $\operatorname{Re}(s)\gt0$. $\eta$ also comes in handy to define $$ \eta(s)\,\Gamma(s)=\int_0^\infty\frac{t^{s-1}}{e^t+1}\,\mathrm{d}t\tag{4} $$ which also converges for $\operatorname{Re}(s)\gt0$. However, we can integrate $(4)$ by parts $k$ times to get $$ \bbox[5px,border:2px solid #C0A000]{\eta(s)\,\Gamma(s)=\frac{(-1)^k}{s(s+1)\cdots(s+k-1)}\int_0^\infty t^{s+k-1}\frac{\mathrm{d}^k}{\mathrm{d}t^k}\frac1{e^t+1}\,\mathrm{d}t}\tag{5} $$ $(5)$ agrees with $(4)$ for $\operatorname{Re}(s)\gt0$ and converges for $\operatorname{Re}(s)\gt-k$. Thus, $(5)$ gives an analytic continuation of $\zeta(s)$ for $\operatorname{Re}(s)\gt-k$.

Using this method, $\zeta(0)=-\frac12$ is computed in this answer and $\zeta(-1)=-\frac1{12}$ is computed in this answer.

Proving $\boldsymbol{\zeta(-2n)=0}$

Note that $(5)$ can be rewritten as $$ \eta(s)\,\Gamma(s+k)=(-1)^k\int_0^\infty t^{s+k-1}\frac{\mathrm{d}^k}{\mathrm{d}t^k}\left(\frac12-\frac12\frac{e^{t/2}-e^{-t/2}}{e^{t/2}+e^{-t/2}}\right)\,\mathrm{d}t\tag{6} $$ Setting $s=1-k$ in $(6)$ gives $$ \begin{align} \eta(1-k) &=(-1)^k\int_0^\infty\frac{\mathrm{d}^k}{\mathrm{d}t^k}\left(\frac12-\frac12\frac{e^{t/2}-e^{-t/2}}{e^{t/2}+e^{-t/2}}\right)\,\mathrm{d}t\\[6pt] &=(-1)^{k-1}\frac{\mathrm{d}^{k-1}}{\mathrm{d}t^{k-1}}\left.\left(\frac12-\frac12\frac{e^{t/2}-e^{-t/2}}{e^{t/2}+e^{-t/2}}\right)\right|_{t=0}\tag{7} \end{align} $$ Set $k=2n+1$ and we get $$ \eta(-2n) =\frac{\mathrm{d}^{2n}}{\mathrm{d}t^{2n}}\left.\left(\frac12-\frac12\frac{e^{t/2}-e^{-t/2}}{e^{t/2}+e^{-t/2}}\right)\right|_{t=0}\tag{8} $$ For $n\ge1$, the right side of $(8)$ is an odd function, so evaluating at $t=0$, and applying $(3)$, yields $$ \bbox[5px,border:2px solid #C0A000]{\zeta(-2n)=0}\tag{9} $$

Analytic Continuation of $\boldsymbol{\zeta}$ Using the Euler-Maclaurin Sum Formula

As described in this answer, this answer, and this answer, for $\operatorname{Re}(s)\gt-2m-1$, $\zeta(s)$ can be represented as $$ \hspace{-12pt}\bbox[5px,border:2px solid #C0A000]{\zeta(s)=\lim_{n\to\infty}\left[\sum_{k=1}^n\frac{1}{k^s}-\left(\frac{1}{1-s}n^{1-s}+\frac12n^{-s}-\sum_{k=1}^m\frac{B_{2k}}{2k}\binom{s+2k-2}{2k-1}n^{-s-2k+1}\right)\right]}\tag{10} $$ where the formula in the parentheses is obtained from the Euler-Maclaurin Sum Formula.

Proving $\boldsymbol{\zeta(-2m)=0}$

If we plug $s=-2m$ into $(10)$, for $m\ge1$, the formula in parentheses exactly matches the sum outside the parentheses by Faulhaber's Formula. In fact, the Euler-Maclaurin Sum Formula is one way to prove Faulhaber's Formula. This means that $$ \bbox[5px,border:2px solid #C0A000]{\zeta(-2m)=0}\tag{11} $$ We can also plug $s=-2m+1$, for $m\ge1$, into $(10)$ and note that Faulhaber's Formula does not include the constant term. Thus, we get $$ \zeta(-2m+1)=-\frac{B_{2m}}{2m}\tag{12} $$

Functional Equation for $\boldsymbol{\zeta}$

The Fourier Transform of $e^{-\pi x^2t}$ is $$ \begin{align} \int_{-\infty}^\infty e^{-\pi x^2t}e^{-2\pi ix\xi}\,\mathrm{d}x &=\int_{-\infty}^\infty e^{-\pi(x-i\xi/t)^2t}e^{-\pi\xi^2/t}\,\mathrm{d}x\\ &=\frac1{\sqrt{t}}e^{-\pi\xi^2/t}\tag{13} \end{align} $$ Applying the Poisson Summation Formula to $(13)$ says that $$ 1+2\sum_{n=1}^\infty e^{-\pi n^2t} =\frac1{\sqrt{t}}+\frac2{\sqrt{t}}\sum_{n=1}^\infty e^{-\pi n^2/t}\tag{14} $$ Note that for $s\gt1$, $$ \begin{align} &\zeta(s)\frac{\Gamma(s/2)}{\pi^{s/2}}\\ &=\sum_{n=1}^\infty\frac1{n^s}\int_0^\infty e^{-\pi t}t^{\frac{\large s}2}\,\frac{\mathrm{d}t}t\\ &=\int_0^\infty\left(\sum_{n=1}^\infty e^{-\pi n^2t}\right)t^{\frac{\large s}2}\,\frac{\mathrm{d}t}t\\ &=\int_0^1\left(\frac1{2\sqrt{t}}-\frac12+\frac1{\sqrt{t}}\sum_{n=1}^\infty e^{-\pi n^2/t}\right)t^{\frac{\large s}2}\,\frac{\mathrm{d}t}t +\int_1^\infty\left(\sum_{n=1}^\infty e^{-\pi n^2t}\right)t^{\frac{\large s}2}\,\frac{\mathrm{d}t}t\\ &=-\frac1{1-s}-\frac1s+\int_1^\infty\left(\sum_{n=1}^\infty e^{-\pi n^2t}\right)t^{\frac{1-\large s}2}\,\frac{\mathrm{d}t}t +\int_1^\infty\left(\sum_{n=1}^\infty e^{-\pi n^2t}\right)t^{\frac{\large s}2}\,\frac{\mathrm{d}t}t\\ &=-\frac1{1-s}-\frac1s+\int_1^\infty\left(\sum_{n=1}^\infty e^{-\pi n^2t}\right)\left(t^{\frac{1-\large s}2}+t^{\frac{\large s}2}\right)\,\frac{\mathrm{d}t}t\tag{15} \end{align} $$ The following integral is increasing in $\alpha$. For $\alpha\ge0$, we have $$ \begin{align} \int_1^\infty\left(\sum_{n=1}^\infty e^{-\pi n^2t}\right)t^\alpha\,\mathrm{d}t &=\sum_{n=1}^\infty\frac1{\pi^{\alpha+1}n^{2\alpha+2}}\int_{\pi n^2}^\infty e^{-t}\,t^\alpha\,\mathrm{d}t\\ &\le\sum_{n=1}^\infty\frac1{\pi^{\alpha+1}n^{2\alpha+2}}\left(\int_{\pi n^2}^\infty e^{-t}\,\mathrm{d}t\right)^{1/2}\left(\int_{\pi n^2}^\infty e^{-t}t^{2\alpha}\,\mathrm{d}t\right)^{1/2}\\ &\le\frac{\Gamma(2\alpha+1)^{1/2}}{\pi^{\alpha+1}}\sum_{n=1}^\infty\frac{e^{-\pi n^2/2}}{n^{2+2\alpha}}\\ &\le\frac{\Gamma(2\alpha+1)^{1/2}}{\pi^{\alpha+1}}\sum_{n=1}^\infty\frac2{\pi n^{4+2\alpha}}\\ &=\frac{2\,\Gamma(2\alpha+1)^{1/2}}{\pi^{\alpha+2}}\zeta(4+2\alpha)\tag{16} \end{align} $$ Thus, the last integral in $(15)$ defines an entire function. Therefore $(15)$ defines $\zeta(s)\frac{\Gamma(s/2)}{\pi^{s/2}}$ for all $s\in\mathbb{C}$. Since $(15)$ is invariant under $s\leftrightarrow1-s$, we have $$ \bbox[5px,border:2px solid #C0A000]{\zeta(s)\frac{\Gamma(s/2)}{\pi^{s/2}} =\zeta(1-s)\frac{\Gamma((1-s)/2)}{\pi^{(1-s)/2}}}\tag{17} $$

Solution 2:

The zeta function is defined as a meromorphic function given by

$$\zeta(s):=1^{-s}+2^{-s}+3^{-s}+\dots=\sum_{n=1}^\infty\frac1{n^s}\ \forall\ \Re(s)>1$$

Though it is common misconception to apply this definition whenever $\Re(s)\le1$, since clearly,

$$1+2+3+\dots$$

doesn't make much sense, normally. To define things like $\zeta(0)$, we use something called analytic continuation (very important part), which allows us to make sense of things. For example,

$$\zeta(s)=\frac1{1-2^{1-s}}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s}\ \forall\ \Re(s)>0\tag!$$

which is not only true for $\Re(s)>1$ but also defined for $\Re(s)>0$. Here, we find that

$$\zeta(0)=\lim_{s\to0^+}\frac1{1-2^{1-s}}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s}=-\frac12$$

We also have the nice reflection formula

$$\zeta(-s)=2^{-s}\pi^{-(s+1)}\sin\left(\frac{-s\pi}2\right)\Gamma(s+1)\zeta(s+1)$$

which can be used to show that

$$\zeta(-2k)=0\ \forall\ k\in\{1,2,3,\dots\}$$

As a short little proof of $(!)$ above, see that if we have

$$\eta(s)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s}\tag{converges for $\Re(s)>0$}$$

then, by subtracting out the odd terms and adding in the even terms, we are left with double the even terms:

$$\zeta(s)-\eta(s)=\sum_{n=1}^\infty\frac2{(2n)^s}=2^{1-s}\sum_{n=1}^\infty\frac1{n^s}=2^{1-s}\zeta(s)$$

$$\zeta(s)-\eta(s)=2^{1-s}\zeta(s)$$

$$(1-2^{1-s})\zeta(s)=\eta(s)$$

$$\zeta(s)=\frac1{1-s^{1-s}}\eta(s)$$

And that is how one can expand the domain of the zeta function.

Remark: Deriving other representations of the zeta function are usually much more hairy than above and often times require more areas of mathematics than you might first think. However, if you are so inclined, taking the Taylor expansion of $f(s)=s\zeta(s+1)$ should provide a series form of the zeta function converging almost everywhere.