February 24th
Today I learned about the $p$-adic Gaussian to derive the completed Riemann $\xi$ function, from Terrence Tao . We're going to synthetically derive the $\Gamma$ factor of the $\xi$ function, and then this will give us an idea of how to synthetically derive the entire $\xi$ function.
Fix $G_\infty(x)=e^{-\pi x^2}$ to be our Gaussian over $\RR=\QQ_\infty.$ Recall that we also have the Gaussian $G_p(x)=1_{\ZZ_p}$ on $\QQ_p$ as we showed two days ago. Additionally, we want our $\Gamma$ factor to be\[\Gamma_\infty(s):=\pi^{-s/2}\Gamma(s/2)=\pi^{-s/2}\int_0^\infty e^{-t}t^{s/2}\,\frac{dt}t.\]The correct to do here is to attempt to make $G_\infty$ appear in this expression. So we take $t\mapsto\pi t^2$ and $dt/t\mapsto 2dt/t,$ giving\[\Gamma_\infty(s)=2\int_0^\infty e^{-\pi t^2}t^s\,\frac{dt}t.\]Now this definition looks more motivated, for we get to write\[\Gamma_\infty(s)=\int_{\QQ_\infty^\times}G_\infty(t)|t|^s\,\frac{dt}{|t|},\]and all the artifacts of $\RR$ have disappeared: we can define the $\Gamma_\infty$ factor as (twice) the Mellin transform of the Gaussian! Further, note that $dt/|t|$ is a Haar measure on $\QQ_\infty^\times,$ so it feels a bit more natural (notationally) to integrate over $\QQ_\infty^\times$ instead of $\QQ_\infty\setminus\{0\}.$
Anyways, this suggests that we define\[\Gamma_p(s):=\int_{\QQ_p^\times}G_p(t)|t_p|_p^s\left(\frac p{p-1}\cdot\frac{dt_p}{|t_p|_p}\right).\]We remark that we choose our measure $\frac p{p-1}\cdot\frac{dt_p}{|t_p|_p}$ to be a multiplicative measure on $\QQ_p^\times,$ scaled so that the units $\ZZ_p^\times=\ZZ_p\setminus p\ZZ_p$ has measure $1.$ We show explicitly that $\ZZ_p\setminus p\ZZ_p$ has the desired measure, for it's a basic case of something we'll do shortly. Essentially, $t_p\in\ZZ_p\setminus p\ZZ_p$ will always have $|t_p|_p=1,$ so\[\int_{\ZZ_p\setminus p\ZZ_p}\frac{dt_p}{|t_p|_p}=\int_{\ZZ_p\setminus p\ZZ_p}dt_p.\]Now, $\ZZ_p\setminus p\ZZ_p$ is made up of $p-1$ cosets of $\ZZ_p/p\ZZ_p,$ of which there are $p.$ So with the $p$ total cosets, which all have the same measure because $dt_p$ is shift-invariant, partitioning $\ZZ_p$ of measure $1,$ it follows each coset has measure $\frac1p.$ Thus,\[\int_{\ZZ_p\setminus p\ZZ_p}\frac{dt_p}{|t_p|_p}=\frac{p-1}p,\]which is what we wanted.
This argument turns out to be helpful for the following claim, which is what we're really here for.
Proposition. We have that \[\Gamma_p(s)=\frac1{1-p^{-s}}.\]
This is a miracle. The $p$-adic $\Gamma$ factor turns out to be an Euler factor! We will make more comments on this later.
At a high level, what makes this formula so nice is that our Gaussian $G_p$ is so nice. Because $G_p$ merely indicates $\ZZ_p,$ our integral over $\QQ_p^\times$ turns into an integral over $\ZZ_p\setminus\{0\}.$ So we see\[\Gamma_p(s)=\frac p{p-1}\int_{\ZZ_p\setminus\{0\}}|t_p|^s_p\,\frac{dt_p}{|t_p|_p}.\]Here we mimic the argument that $\ZZ_p\setminus p\ZZ_p$ has multiplicative maesure $1.$ We see $\ZZ_p$ is inherently granular, so we can make the integrand constant by summing over each possible $|x_p|_p.$ In particular, $x_p\in\ZZ_p\setminus\{0\}$ implies $|x_p|_p=p^{-\nu}$ for some nonnegative integer $\nu.$ So because $\ZZ_p\setminus\{0\}$ is countably partitioned by $\nu,$ we may write\[\Gamma_p(s)=\frac p{p-1}\sum_{\nu=0}^\infty\int_{|x_p|=p^{-\nu}}p^{-\nu s}\,\frac{dt_p}{p^{-\nu}}.\]Now, the integrand is constant with respect to $t_p,$ so we can move it out of the integral, giving\[\Gamma_p(s)=\frac p{p-1}\sum_{\nu=0}^\infty\frac{p^{-\nu s}}{p^{-\nu}}\int_{|x_p|=p^{-\nu}}dt_p.\]At this point, we can begin to see the $\left(1-p^{-s}\right)^{-1}$ coming from the infinite series. We note that $\left\{x_p:|x_p|=p^{-\nu}\right\}$ is the $p^\nu\ZZ_p\setminus p^{\nu+1}\ZZ_p,$ the set of elements divisible by $p^\nu$ but not $p^{\nu+1}.$ So our integral is\[\Gamma_p(s)=\frac p{p-1}\sum_{\nu=0}^\infty\frac{p^{-\nu s}}{p^{-\nu}}\int_{p^{\nu}\ZZ_p\setminus p^{\nu+1}\ZZ_p}dt_p.\]We can evaluate the integral as we did when showing $\ZZ_p\setminus p\ZZ_p$ has multiplicative measure $1.$ The coset $p^\nu\ZZ_p$ is one of $p^\nu$ in $\ZZ_p/p^\nu\ZZ_p,$ and all cosets have the same size, so the measure of $p^\nu\ZZ_p$ is $1/p^\nu.$ The same argument shows $p^{\nu+1}\ZZ_p$ has measure $1/p^{\nu+1},$ so\[\int_{p^{\nu}\ZZ_p\setminus p^{\nu+1}\ZZ_p}dt_p=\frac1{p^\nu}-\frac1{p^{\nu+1}}=\frac{p-1}{p^{\nu+1}}\]because $p^{\nu+1}\ZZ_p\subseteq p^\nu\ZZ_p.$ From here, we have\[\Gamma_p(s)=\frac p{p-1}\sum_{\nu=0}^\infty\frac{p^{-\nu s}}{p^{-\nu}}\cdot\frac{p-1}{p^{\nu+1}}.\]Simplifying, we see\[\Gamma_p(s)=\sum_{\nu=0}^\infty p^{-\nu s}=\frac1{1-p^{-s}},\]which is what we wanted. $\blacksquare$
Quickly, we remark that if I were more comfortable with $p$-adic integration, we could write $d^\times t_p:=\frac p{p-1}\cdot\frac{dt_p}{|t_p|_p}$ as our multiplicative measure so that\[\int_{|t_p|_p=p^{-\nu s}}d^\times t_p=\int_{|t_p|_p=1}d^\times t_p=\int_{\ZZ_p\setminus p\ZZ_p}d^\times t_p=1\]because we are integrating over that multiplicative (!) measure. This is faster, but the multiplicative measure scares me.
We now remark on the miracle that just happened. It would appear that we can define $\zeta$ synthetically (!) by\[\zeta(s)=\prod_{p \lt \infty}\Gamma_p(s),\]and this feels somewhat more natural than the the infinite sum definition. In particular, the analysis that we did above kind of implies that the "correct'' way to define our $\zeta$-functions is by an Euler product, which aligns with the way Artin $L$-functions are defined by Euler factors. Further,\[\xi(s)=\prod_\nu\Gamma_\nu(s),\]where $\nu$ ranges over all places (finite and infinite) of $\QQ.$ This presents a much more motivated view of the $\Gamma$ factor by placing it on equal footing with the rest of the Euler factors.