Today I Learned

April 17th

Today I learned the proof of the functional equation for the Dedekind zeta function $\zeta_K,$ mostly from here . The main idea is to use a global zeta integral $\zeta\left(g,|\bullet|^s\right)$ with a "mostly'' self-dual Gaussian $g.$ The exact specifics of our Gaussian comes down to local $\zeta$ computations, but we have to be careful about our measures.

To be explicit, we define $g_\nu(x_\nu):=e^{-\pi x_\nu^2}$ for real archimedean $\nu$ and $g_\nu:=e^{-2\pi z\overline z}$ for complex archimedean $\nu.$ Additionally, we take inspiration from our proof that the global zeta integrals converge for exponent greater than $1$ and set $g_\mf p=1_{\mathcal O_\mf p}$ for nonarchimedean $\mf p.$

Combining these local factors into $g=\prod_\nu g_\nu:\AA_K\to\CC$ gives a Shwartz-Bruhat function, practically by definition. (For the archimedean places, we don't prove that those Gaussians are Shwartz, but they are.) Now, the main claim is the following.

Expanding this out, we are simplifying\[\zeta\left(g,|\bullet|^s\right)=\int_{\AA_K^\times}g(x)|x|^s\,d^\times x=\prod_\nu\underbrace{\int_{K_\nu^\times}g_\nu(x_\nu)|x_\nu|_\nu^s\,d^\times x_\nu}_{\zeta_\nu\left(g_\nu,|\bullet|_\nu^s\right)}.\]Note that while $\AA_K^\times\not\cong\prod_\nu K_\nu^\times,$ this decomposition is legal because, after the projections, we are hitting all of those elements. In particular, all but finitely many (all the nonarchimedean places) of the $g_\nu$ are $1_{\mathcal O_\nu},$ so we are far from asserting that $\AA_K^\times$ is the product of the $K_\nu^\times.$

It remains to evaluate these integrals, but we need to discuss our measures first. We do this from the ground up because they introduce subtle factors. We are using the self-dual measure with respect to our standard character $\psi.$ What this means is that we have defined, for local fields $K_\nu$ with standard character $\psi_\nu,$\[\hat f(x):=\int_{K_\nu}f(x)\psi_\nu(xy)\,dx,\]which is legal because $\widehat{K_\nu}\cong K_\nu.$ However, this is a specific measure $d\chi$ on $\widehat{K_\nu}$ that makes the inverse Fourier transform well-behaved given a measure $dx$ on $K_\nu,$ and $d\chi$ might not match $\psi_\nu(xy)\,dx.$ Regardless, it is at least scaled by some $c$ (Haar measure is unique), so\[\hat{\hat f}(y)=\int_{K_\nu}f(y)\psi_\nu(xy)\,dy\]will be off from correctly inverting the Fourier transform by $c^2.$ Shifting $dy$ on $K_\nu$ to accommodate this $c^2,$ we can make $\hat{\hat f}(x)=f(-x)$ for all Shwartz-Bruhat $f:K_\nu\to\CC.$ In particular, it suffices to check a single function satisfies $\hat{\hat f}(x)=f(-x)$ to make sure we are using the self-dual measure.

We deal with our places in cases: real archimedean, complex archimedean, or nonarchimedean.

We check this with the Gaussian $g_\RR(x)=e^{-\pi x^2}.$ Our standard character is $\psi_\RR(x)=e^{-2\pi ix}.$ Now, we are evaluating\[\widehat{g_\RR}(y)=\int_\RR e^{-\pi x^2}e^{-2\pi ixy}\,dx.\]I steal the evaluation of this from this post : take the derivative under the integral sign so that\[\frac{\widehat{g_\RR}(y)}{dy}=\int_\RR e^{-\pi x^2}e^{-2\pi ixy}\cdot(-2\pi ix)\,dx.\]Now, $\frac d{dx}e^{-\pi x^2}=-2\pi xe^{-\pi x^2},$ so we may integrate by parts to get\[\frac{\widehat{g_\RR}(y)}{dy}=\underbrace{ie^{-\pi x^2}e^{-2\pi ixy}\bigg|_{-\infty}^\infty}_0-\underbrace{\int_\RR ie^{-\pi x^2}e^{-2\pi ixy}\cdot(-2\pi iy)\,dx}_{2\pi yg_\RR(y)}.\]So we have an ordinary differential equation $\frac{\widehat{g_\RR}(y)}{dy}=-2\pi yg_\RR(y),$ which we can solve to get solutions $\widehat{g_\RR}(y)=ce^{-\pi y^2}.$ We can evaluate $\widehat{g_\RR}(0)$ directly as $\int_\RR e^{-\pi x^2}\,dx=1,$ so we conclude $\widehat{g_{\RR}}=g_\RR.$ It follows that $g_\RR$ gives $\hat{\hat f}(x)=f(-x)$ because $g_\RR$ is also even. This finishes. $\blacksquare$

Intuitively, we are doubling the Lebesgue measure here because there are two embeddings associated with $\nu$ because conjugation gives the same place. However, we can't lose this size, so we have to double to accommodate.

We check this with the Gaussian $g_\CC(z)=e^{-2\pi z\overline z}.$ Our standard character is $\psi_\CC(z)=\psi_\RR(\op{Trace}_{\CC/\RR}z),$ which is $e^{-4\pi i\op{Re}z}.$ (This extra factor of $2$ from $\op{Trace}_{\CC/\RR}$ is the rigorization of the above intuition.) Our Fourier transform is\[\widehat{g_\CC}(y)=\int_\CC e^{-2\pi z\overline z}e^{-4\pi i\op{Re}(zy)}\,dz.\]Writing $\CC\cong\RR^2$ with $y=c+di$ and $dz=2da\,db$ (recall we are using twice the Lebesgue measure) gives\[\widehat{g_\CC}(c+di)=\iint_{\RR^2}e^{-2\pi\left(a^2+b^2\right)}e^{-4\pi i(ac-bd)}\,2da\,db.\]Now, this integral can be separated into\[\widehat{g_\CC}(c+di)=\left(\int_\RR e^{-2\pi a^2}e^{-4\pi iac}\,\sqrt2da\right)\left(\int_\RR e^{-2\pi b^2}e^{4\pi ibd}\,\sqrt2db\right).\]Taking $a\sqrt2\mapsto a$ and $b\sqrt2\mapsto-b$ gives\[\widehat{g_\CC}(c+di)=\left(\int_\RR e^{-\pi a^2}e^{-2\pi ia(c\sqrt2)}\,da\right)\left(\int_\RR e^{-\pi b^2}e^{-2\pi ib(d\sqrt2)}\,db\right),\]which are themselves Fourier transforms of $g_\RR(x)=e^{-\pi x^2},$ which we know how to do. In particular, this collapses to $\widehat{g_\RR}(c\sqrt2)\widehat{g_\RR}(d\sqrt2)=e^{-2\pi\left(c^2+d^2\right)}.$ In other words, we see $\widehat{g_\CC}(y)=g_\CC(y),$ so it follows $g_\CC$ witnesses a $\hat{\hat f}(x)=f(-x)$ because $g_\CC$ is also even. This finishes. $\blacksquare$

The different becomes involved because our standard character is $\psi_\mf p(x):=\psi_p(\op{Trace}_{K_\mf p/\QQ_p}x),$ where $p$ is the rational prime below $\mf p.$ Anyways, we check this with (surprise!) the Gaussian $g_\mf p=1_{\mathcal O_\mf p}.$ We have that\[\widehat{g_\mf p}(y)=\int_{K_\mf p}g_\mf p(x)\psi_\mf p(xy)\,dx=\int_{\mathcal O_\mf p}\psi_\mf p(xy)\,dx.\]If $\op{Trace}_{K_\mf p/\QQ_p}(y\mathcal O_\mf p)\in\ZZ_p$—i.e., if $y$ is in the dual lattice $\mathcal O_\mf p^\vee$—then $x\mapsto\psi_\mf p(xy)$ takes all $x$ to $1,$ and we get out $\op{Vol}(\mathcal O_\mf p).$ Otherwise, we note $\psi_\mf p(1y)$ doesn't go to $1,$ so $x\mapsto\psi_\mf p(xy)$ gives a nontrivial character over the compact group $\mathcal O_\mf p,$ so this integral is $0$ by orthogonality of characters. Thus, $\widehat{g_\mf p}(y)=\op{Vol}(\mathcal O_\mf p)\cdot1_{\mathcal O_\mf p^\vee}.$

It will be worth our time to briefly talk about $\mathcal O_\mf p^\vee.$ We note that $\mathcal O_\mf p^\vee$ is closed under addition (the trace is additive) and $y\in\mathcal O_\mf p^\vee$ means that $y\mathcal O_\mf p\subseteq\mathcal O_\mf p^\vee,$ so $\mathcal O_\mf p^\vee$ is an $\mathcal O_\mf p$-ideal. Because $\mathcal O_\mf p$ is also a discrete valuation ring, we can then assert, for some positive integer $d,$\[\mathcal O_\mf p^\vee=\varpi^d\mathcal O_\mf p,\]where $\varpi$ is the uniformizer of $K_\mf p.$ In particular, we get the following properties.

• We have that $\mathcal O_\mf p^{\vee\vee}$ is the set of elements $y\in K_\mf p$ such that $\op{Trace}_{K_\mf p/\QQ_p}(y\mathcal O^\vee)\in\ZZ_p,$ which really means $y$ needs to have a nonnegative number of factors of $\varpi,$ or $y\in\mathcal O_\mf p.$

• We have that $\op{Vol}(\mathcal O_\mf p^\vee)=p^{-d}\op{Vol}(\mathcal O_\mf p)$ because $\mathcal O_\mf p/\varpi^d\mathcal O_\mf p$ has $p^d$ elements.

• This also implies $\mathcal D_\mf p:=(\mathcal O_\mf p^\vee)^{-1}=\varpi^{-d}\mathcal O_\mf p$ has norm $p^{-d}.$

Now, taking the next Fourier transform, we see\[\widehat{\widehat{g_\mf p}}(y)=\int_{K_\mf p}g_\mf p(x)\overline{\psi_\mf p(xy)}\,dx=\op{Vol}(\mathcal O_\mf p)\int_{\mathcal O_\mf p^\vee}\overline{\psi_\mf p(xy)}\,dx.\]Again, if $y\in\mathcal O_\mf p^{\vee\vee}=\mathcal O_\mf p,$ then $x\mapsto\overline{\psi_\mf p(xy)}$ is the trivial character, so we get out $\op{Vol}(\mathcal O_\mf p)\op{Vol}(\mathcal O_\mf p^\vee).$ Otherwise, $x\mapsto\overline{\psi_\mf p(xy)}$ is a nontrivial character of the compact subgroup $\mathcal O_\mf p^\vee=\varpi^d\mathcal O_\mf p,$ so this evaluates to $0$ by orthogonality of characters. Thus,\[\widehat{\widehat{g_\mf p}}=\op{Vol}(\mathcal O_\mf p)\op{Vol}(\mathcal O_\mf p^\vee)1_{\mathcal O_\mf p}=\op{Vol}(\mathcal O_\mf p)^2\op N(\mathcal D_\mf p)1_{\mathcal O_\mf p}.\]We would like this to be $g_\mf p,$ so we set $\op{Vol}(\mathcal O_\mf p)=\op N(\mathcal D_\mf p)^{-1/2}$ (as prescribed in the lemma) so that this term cancels out properly. It follows that $g_\mf p$ again witnesses an $f$ with $\hat{\hat f}(x)=f(-x)$ because $g_\mf p$ is even. $\blacksquare$

Having properly established all of our measures, it remains to actually compute the integrals of interest. We begin have three cases: $\nu$ is real archimedean, complex archimedean, or nonarchimedean. We do these one at a time.

If $\nu$ is real archimedean, then $K_\nu^\times\cong\RR^\times,$ so we are evaluating\[\zeta_\nu\left(g_\nu,|\bullet|_\nu^s\right)=\int_{\RR^\times}e^{-\pi x^2}|x|^s\,d^\times x=2\int_0^\infty e^{-\pi x^2}x^s\,d^\times x.\]We want this to become a $\Gamma_\RR(s)$ factor, so we want to coerce this to look like $\Gamma(s)=\int_0^\infty e^{-x}x^s\,d^\times x.$ We let $y:=\pi x^2$ so that $d^\times y=2d^\times x.$ Our integral now looks like\[\zeta_\nu\left(g_\nu,|\bullet|_\nu^s\right)=\int_0^\infty e^{-y}\pi^{-s/2}y^{s/2} d^\times y.\]This is $\Gamma_\RR(s)=\pi^{-s/2}\Gamma(s/2),$ which is what we wanted. $\blacksquare$

If $\nu$ is complex archimedean, then $K_\nu^\times\cong\CC^\times,$ so we are evaluating\[\zeta_\nu\left(g_\nu,|\bullet|_\nu^s\right)=\int_{\CC^\times}e^{-2\pi|z|^2}|z|_\CC^s\,d^\times z,\]where $d^\times z/|z|_\CC$ with $dz$ twice the Lebesgue measure and $|\bullet|_\CC$ is the square of the typical magnitude. (We square so that multiplication corresponds to an actual scaling of space.) Transforming to polar coordinates, we get $dz=2dx\,dy=2rdr\,d\theta,$ and $d^\times z=(2/r)dr\,d\theta=2d^\times r\,d\theta,$ so\[\zeta_\nu\left(g_\nu,|\bullet|_\nu^s\right)=\int_0^{2\pi}\int_0^\infty e^{-2\pi r^2}r^{2s}\,2d^\times r\,d\theta.\]Note that we can remove the $d\theta$ intergral to get a $2\pi$ out. Again, we would like $\Gamma$ to appear somewhere here, so we integrate over $x:=2\pi r^2$ so that $d^\times x=2d^\times r,$ which gives\[\zeta_\nu\left(g_\nu,|\bullet|_\nu^s\right)=2\pi\int_0^\infty e^{-x}x^s(2\pi)^{-s}\,d^\times x.\]The integral is now cleanly $(2\pi)^{-s}\Gamma(s),$ so indeed we see everything collapses to $\pi\cdot2(2\pi)^{-s}\Gamma(s)=\pi\Gamma_\CC(s) \lt $ as desired. $\blacksquare$

We are evaluating\[\zeta_\mf p\left(g_\mf p,|\bullet|_\mf p^s\right)=\int_{K_\mf p}g_\mf p(x)|x|_\mf p^s\,d^\times x=\int_{\mathcal O_\mf p}|x|_\mf p^s\,d^\times x\]because $g_\mf p=1_{\mathcal O_\mf p}.$ Because $\mathcal O_\mf p$ is defined by $|x|_\mf p\le1,$ we can evaluate this as a geometric series by writing\[\zeta_\mf p\left(g_\mf p,|\bullet|_\mf p^s\right)=\sum_{k=0}^\infty\int_{|x|_\mf p=\op N(\mf p)^{-k}}\op N(\mf p)^{-ks}\,d^\times x.\]Because we are dealing with a multiplicative measure, we can transform the integral from $|x|_\mf p=\op N(\mf p)^{-k}$ to $|x|_\mf p=1$ (say, by multiplying by an appropriate power of the inverse uniformizer $\varpi^{-1}$). This means\[\zeta_\mf p\left(g_\mf p,|\bullet|_\mf p^s\right)=\left(\sum_{k=0}^\infty\op N(\mf p)^{-ks}\right)\int_{\mathcal O_\mf p^\times}d^\times x.\]The infinite sum collapses to $\left(1-\op N(\mf p)^{-s}\right)^{-1},$ so the integral is where the $\op N(\mathcal D_\mf p)^{-1/2}$ will come from.

The measure $d^\times x$ can technically be defined in lots of ways, but we need $\mathcal O_\mf p^\times$ needs to have measure $1$ have all but finitely many $\mf p$ so that open sets of $\mathcal A_K^\times$ (where all but finitely many of the components are $\mathcal O_\mf p^\times$) can actually have measure. To do this naturally, our definition is $d^\times x=\frac1{1-\op N(\mf p)^{-1}}\cdot\frac{dx}{|x|_\mf p}$ so that\[\int_{\mathcal O_\mf p^\times}d^\times x=\frac1{1-\op N(\mf p)^{-1}}\int_{\mathcal O_\mf p\setminus\mf p}dx=\op N(\mathcal D_\mf p)^{-1/2}.\]In particular, the last inequality holds because $\mf p$ is one of $\op N(\mf p)$ cosets of $\mathcal O_\mf p.$ Anyways, we see this is $1$ for unramified primes $\mf p,$ so our wish is granted.

Combining these, we see\[\zeta_\mf p\left(g_\mf p,|\bullet|_\mf p^s\right)=\frac1{1-\op N(\mf p)^{-s}}\cdot\op N(\mathcal D_\mf p)^{-1/2},\]which is what we wanted. $\blacksquare$

We are finally ready to complete the proof of the proposition. We recall\[\zeta\left(g,|\bullet|^s\right)=\prod_\nu\zeta_\nu\left(g_\nu,|\bullet|_\nu^s\right).\]Because $K$ has signature $(r_1,r_2),$ we know there are $r_1$ real archimedean places and $r_2$ complex archimedean places. We've evaluated these $\zeta_\nu$ elements, so our archimedean places contribute a total of $\Gamma_\RR(s)^{r_1}\cdot(\pi\Gamma_\CC(s))^{r_2}.$

As for our archimedean places, our evaluation of the $\zeta_\mf p$ tells us we have\[\prod_\mf p\zeta_\mf p\left(g_\mf p,|\bullet|_\mf p^s\right)=\left(\prod_\mf p\op N(\mathcal D_\mf p)^{-1/2}\right)\left(\prod_\mf p\frac1{1-\op N(\mf p)^{-s}}\right).\]It happens that the individual $\mathcal D_\mf p$ multiply together to give the full different $\mathcal D_K,$ which has norm $|\op{disc}\mathcal O_K|,$ so the first factor contributes $|\op{disc}\mathcal O_K|.$ The second factor is the Euler product for $\zeta_K(s).$

Bringing this all together, we see that\[\prod_\mf p\zeta_\mf p\left(g_\mf p,|\bullet|_\mf p^s\right)=\Gamma_\RR(s)^{r_1}\cdot(\pi\Gamma_\CC(s))^{r_2}\cdot|\op{disc}\mathcal O_K|^{-1/2}\zeta_K(s).\]This completes the proof of the proposition. $\blacksquare$

We remark that the proposition provides an meromorphic continuation of $\zeta_K(s)$ because $\zeta\left(g,|\bullet|^s\right)$ and the $\Gamma$ factors all have meromorphic continuations. We do not talk in detail about what this looks like here; we're moving towards a real functional equation of $\zeta_K$ anyways.

There is also a sister of the proposition we need to establish before we can actually apply the global $\zeta$ functional equation.

We have actually done most of the work for this already. Starting the same way, we are evaluating\[\zeta\left(\hat g,|\bullet|^s\right)=\prod_\nu\zeta_\nu\left(\widehat{g_\nu},|\bullet|_\nu^s\right).\]All of our archimedean arguments now carry through verbatim because $\widehat{g_\nu}=g_\nu,$ which we found when looking for our self-dual nonarchimedean measures. We have left to compute the local $\zeta$ integral when $\nu$ is nonarchimedean, which is easier the second time.

Observe that when trying to establish our self-dual archimedean measures, we found $\widehat{g_\mf p}=\op N(\mathcal D_\mf p)^{-1/2}1_{\mathcal O_\mf p^\vee}.$ Fixing $d$ so that $\mathcal O_\mf p^\vee=\varpi^d\mathcal O_\mf p,$ we see we want\[\zeta_\mf p\left(\widehat{g_\mf p},|\bullet|_\mf p^s\right)=\int_{K_\mf p}\widehat{g_\mf p}(x)|x|_\mf p^s\,d^\times x=\op N(\mathcal D_\mf p)^{-1/2}\int_{\varpi^d\mathcal O_\mf p}|x|_\mf p^s\,d^\times x.\]As before, we turn this into an infinite geometric series, but this time we are integrating over $|x|_\mf p\le\op N(\mf p)^{-d},$ so\[\zeta_\mf p\left(\widehat{g_\mf p},|\bullet|_\mf p^s\right)=\op N(\mathcal D_\mf p)^{-1/2}\sum_{k=d}^\infty\int_{|x|_\mf p=\op N(\mf p)^{-k}}\op N(\mf p)^{-ks}\,d^\times x.\]Rescaling multiplicatively the integral over $|x|_\mf p=\op N(\mf p)^{-k}$ to $\mathcal O_\mf p^\times,$ we see\[\zeta_\mf p\left(\widehat{g_\mf p},|\bullet|_\mf p^s\right)=\op N(\mathcal D_\mf p)^{-1/2}\left(\sum_{k=d}^\infty\op N(\mf p)^{-ks}\right)\int_{\mathcal O_\mf p^\times}d^\times x.\]Recalling that the measure of $\mathcal O_\mf p^\times$ is $\op N(\mf p)^{-1/2},$ we see this is\[\zeta_\mf p\left(\widehat{g_\mf p},|\bullet|_\mf p^s\right)=\op N(\mathcal D_\mf p)^{-1}\sum_{k=d}^\infty\op N(\mf p)^{-ks}.\]Finally, the infinite series collapses to $\op N(\mf p)^{-ds}\cdot\frac1{1-\op N(\mf p)^{-s}}.$ After noting that $\op N(\mf p)^{-d}=\op N(\mathcal D_\mf p),$ we see that the above rearranges into the desired. $\blacksquare$

In light of the above lemma, we see that the nonarchimedean factors of $\zeta\left(\hat g,|\bullet|^s\right)$ contribute\[\prod_\mf p\zeta_\mf p\left(\widehat{g_\mf p},|\bullet|_\mf p^s\right)=\left(\prod_\mf p\op N(\mathcal D_\mf p)^{s-1}\right)\left(\prod_\mf p\frac1{1-\op N(\mf p)^{-s}}\right).\]As discussed, the product of $\op N(\mathcal D_\mf p)$ multiply out to $\op N(\mathcal D_K)=|\op{disc}\mathcal O_K|,$ so the left product is $|\op{disc}\mathcal O_K|^{s-1}.$ And again, the right-hand product is the Euler product of $\zeta_K(s).$

Combining the archimedean factors with the nonarchimedean ones, we see\[\zeta\left(\hat g,|\bullet|^s\right)=\Gamma_\RR(s)^{r_1}(\pi\Gamma_\CC(s))^{r_2}\cdot|\op{disc}\mathcal O_K|^{s-1}\zeta_K(s),\]which is what we wanted. $\blacksquare$

We are finally able to apply the global $\zeta$ functional equation. Namely, we know that\[\zeta\left(g,|\bullet|^s\right)=\zeta\left(\hat g,|\bullet|^{1-s}\right).\]Almost everything in the two propositions is symmetric except that $\zeta\left(g,|\bullet|^s\right)$ has a factor of $|\op{disc}\mathcal O_K|^{-1/2}$ where $\zeta\left(\widehat g,|\bullet|^{1-s}\right)$ has a factor of $|\op{disc}\mathcal O_K|^{-s}.$ To deal with this, we define\[\xi_K(s):=|\op{disc}\mathcal O_K|^{(s+1)/2}\zeta\left(g,|\bullet|^s\right)\]so that\[\xi_K(s)=\Gamma_\RR(s)^{r_1}(\pi\Gamma_\CC(s))^{r_2}\cdot|\op{disc}\mathcal O_K|^{s/2}\zeta_K(s),\]and\[\xi_K(1-s)=\Gamma_\RR(1-s)^{r_1}(\pi\Gamma_\CC(1-s))^{r_2}\cdot|\op{disc}\mathcal O_K|^{(1-s)/2}\zeta_K(1-s),\]which was configured to be $|\op{disc}\mathcal O_K|^{(s+1)/2}\zeta\left(\widehat g,|\bullet|^s\right).$ In particular, we know $\xi_K(s)=\xi_K(1-s).$ This gives the following.

This follows from the above discussion after expanding out our definition of $\xi_K.$ Namely, we know that\[\xi_K(s)=\Gamma_\RR(s)^{r_1}(\pi\Gamma_\CC(s))^{r_2}\cdot|\op{disc}\mathcal O_K|^{s/2}\zeta_K(s)\]satisfies $\xi_K(s)=\xi_K(1-s),$ so it remains to show that this equation for $\xi_K$ implies the one in the theorem. Plugging into the definitions $\Gamma_\RR(s)=\pi^{-s/2}\Gamma(s/2)$ and $\Gamma_\CC(s)=2(2\pi)^{-s}\Gamma(s),$ this is\[\xi_K(s)=\pi^{-r_1s/2+r_2-r_2s}\cdot2^{r_2-r_2s}\cdot\Gamma(s/2)^{r_1}\Gamma(s)^{r_2}\cdot|\op{disc}\mathcal O_K|^{s/2}\zeta_K(s).\]The factors $\pi^{r_2}$ and $2^{r_2}$ we may safely ignore because they do not depend on $s.$ Noting that $r_1+2r_2=n,$ we can simplify this a bit into\[\frac{\xi_K(s)}{2^{r_2}\pi^{r_2}}=\left(\frac{|\op{disc}\mathcal O_K|}{4^{r_2}\pi^n}\right)^{s/2}\Gamma(s/2)^{r_1}\Gamma(s)^{r_2}\zeta_K(s).\]The right-hand side is $\Lambda_K,$ so we conclude $\Lambda_K(s)=\Lambda_K(1-s).$ This finishes the proof of the theorem. $\blacksquare$