A question that has been puzzling me for quite some time now:
Why is the value of the Riemann Zeta function equal to $0$ for every even negative number?
I assume that even negative refers to the real part of the number, while its imaginary part is $0$.
So consider $-2$ for example:
$f(-2) =
\sum_{n=1}^{\infty}\frac{1}{n^{-2}} =
\frac{1}{1^{-2}}+\frac{1}{2^{-2}}+\frac{1}{3^{-2}}+\dots =
1^2+2^2+3^2+\dots =
\infty$
What am I missing here?
Best Answer
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 $$ \bbox[5px,border:2px solid #C0A000]{\zeta(s)\frac{\Gamma(s/2)}{\pi^{s/2}} =\zeta(1-s)\frac{\Gamma((1-s)/2)}{\pi^{(1-s)/2}}}\tag{17} $$