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
$$
\bbox[5px,border:2px solid #C0A000]{\zeta(s)\frac{\Gamma(s/2)}{\pi^{s/2}}
=\zeta(1-s)\frac{\Gamma((1-s)/2)}{\pi^{(1-s)/2}}}\tag{17}
$$
Best Answer
Raziwill has a paper about the moments of the Riemann Zeta function:
The 4.36-th moment of the Riemann Zeta function
Both Ingham's paper of 1926 and Hardy-Littlewood's paper of 1918 are in the references.
Radom matrix theory (and quantum billiard) is known to be related to the spacings between the non-trivial roots of the Riemann zeta function:
Is there an equivalent statement of Riemann Hypothesis in term of Random Matrix or physics theory?
Edit: To address the question why the moments of the Riemann zeta function are important - this is really involved with a lot of analytic number theory and the Riemann hypothesis. In fact, the quality of the asymptotic formula depends on RH. Since the latter is one of the most important conjectures in number theory, also the moments and other properties of the zeta function are important. The exact connections are a bit technical. The introduction of the paper Moments of the Riemann zeta function by Soundararajan give a good survey on this. There is also a connection to random matrix theory, i.e., to the link above. This makes it important, too.