$\xi(s)=\xi(1-s)$ functional equation from $\Gamma(s)\Gamma(s+1)\zeta(s)=\int_0^\infty t^{s-1}\sum_{\alpha \in \Bbb N} \Phi_{\alpha}(t)~ dt$

Question

The following came up for me while doing an exercise in geometry.

I considered a smooth foliation $\Phi_{\alpha}(t)=2\sqrt{\alpha t}K_1(2\sqrt{\alpha t})$ of the manifold $(0,\infty)\times (0,1).$ I then invoked the following relationship between $\Phi_{\alpha}(t)$ and $\zeta(s),$ the Riemann zeta function:

$$ \zeta(s)=\frac{1}{Z}M_s \bigg [\sum_{\alpha \in \Bbb N} \Phi_\alpha(t)\bigg],~~~~~ \Re(s)>1. $$

Here $Z$ is a Gamma factor, $M_s$ is a Mellin transform, $K_1$ is a modified Bessel function, and the sum is taken over the natural leaves of the foliation.

How can this be extended to the critical strip with symmetry about $\Re(s)=1/2?$ In other words what are the steps to obtain the classic $\xi(s)=\xi(1-s)$ equation for $\zeta(s)$ starting from my integral representation?

Well, for $$ \zeta(s)\Gamma(s)=\int_0^\infty\frac{x^{s-1}}{e^x-1}dx$$

the idea is to introduce a branch cut for $x^{s-1}$ along the positive real axis, and to replace the above integral by one running from $+\infty$ along the bottom of the positive real axis, around the origin, and back to $+\infty$ along the top of the real axis. This introduces an extra factor $1-e^{2\pi i s}$. Now start expanding the circle around the origin, taking account of the poles of the integrand along the imaginary axis as we go, and end up with

$$\Gamma(s)\zeta(s)=(2\pi)^{s-1}\Gamma(1-s)\sin(\tfrac12\pi s)\zeta(1-s).$$

Of course more work must be done to put this in the correct functional equation form but it can be done.

But in this case we have:

$$ \zeta(s)=\frac{1}{Z}M_s \bigg [\sum_{\alpha \in \Bbb N} \Phi_\alpha(t)\bigg]$$

and equivalently:

$$\Gamma(s)\Gamma(s+1)\zeta(s)=\int_0^\infty t^{s-1}\sum_{\alpha \in \Bbb N} \Phi_{\alpha}(t)~ dt$$

I'm unsure if the approach I described above will work here.

Answer 1

A related integral representation of $\zeta(s)$ which extends convergence to the critical strip is

$$\zeta (s)=\frac{1}{\left(1-2^{1-s}\right) \Gamma(s)\, \Gamma(s+1)} \int\limits_0^\infty f(x)\, x^{s-1} \, dx,\quad\Re(s)>0\tag{1}$$

where

$$f(x)=\underset{N\to \infty }{\text{lim}}\left(\sum\limits_{n=1}^N (-1)^{n+1}\, 2\, \sqrt{n\, x}\, K_1\left(2 \sqrt{n\, x}\right)\right)\tag{2}$$.