Introduction:
In statistics, distributions in the exponential family are very natural to consider. One famous example is the Gaussian. It shows up in the heat equation, in probability theory, and in many other fields. While the Gaussian may be most important due to the central limit theorem, distributions in the exponential family have properties that may be connected to the properties of the Gaussian itself. So, consider the exponential family, the close cousin of the Gaussian, $f_s(x)=e^{sT(x)}$ for parameter $s$ and sufficient statistic $T(x)=\frac{1}{\log x}.$
Relating $f_s(x)$ to the Riemann zeta function:
Viewing $f_s(x)$ geometrically:
The function $f_s(x)$ is a particular solution to a linear parabolic diffusion equation (use an exponential ansatz):
$$s \frac{\partial ^2f(s,x)}{\partial s^2}=-x \frac{\partial f(s,x)}{\partial x} \tag{2}$$
and the Mellin transform can be used to transform that equation to (transform the solution $f_s(x)$ and use a CAS to verify):
$$r^2 \frac{\partial ^3\Psi(r,s)}{\partial r^3}=s^2 \frac{\partial \Psi(r,s)}{\partial s} \tag{3}$$
where a particular solution Bessel function, $K_1$ appears:
$$\Psi(r,s)=2 \sqrt{\frac{r}{s}}K_1(2\sqrt{r s})$$
We recover a connection with the Riemann zeta function and the Gamma function:
$$\frac{1}{\Gamma(w-1)\Gamma(w)}\int_0^\infty \sum_{r \in \Bbb N} \Psi(r,s) s^{w-1}~ds=\zeta(w-1) \tag{1}$$
convergent for $\Re(w)>2.$
We can recover the symmetric functional equation about the critical strip $\xi(s)=\xi(1-s)$ through this integral relationship and analytic continuation, although it is calculation intensive.
A more fundamental approach:
Another, perhaps more fundamental, way to obtain the symmetric functional equation is to start with any Schwartz function. Usually the self dual Gaussian is used to obtain the Gamma factor $\Gamma(s/2)\pi^{-s/2}.$ Here I ask for a closed form for the Gamma factor associated to the Schwartz class $f_s(x):$
Obtaining the correct Gamma factor for this Schwartz function.
We are looking for a functional equation of this form:
$$\Gamma(f,r,s)\cdot \zeta(s) \;=\; \Gamma(\hat{f},r,1-s)\cdot \zeta(1-s)\tag{4}$$
Here, $\Gamma$ is the Gamma factor and $\hat f$ denotes the Fourier transform. Note however, that the fundamental domain of $f$ is $(0,1)$ so the Fourier transform will act on this bounded domain.
Here I will show the calculation of $\Gamma(f,r,s):$
$$\Gamma(f,r,s)=\int_{\mathbb R^\times \cap ~(0,1)} |x|^r~f_s(x)~{dx\over |x|}=2 \sqrt{\frac{r}{s}}K_1(2\sqrt{r s})$$
for modified Bessel function of the second kind $K_1.$
Informal discussion and remarks:
In terms of geometric analysis $\Psi(r,s)$ has an interpretation as a $1$-parameter family of Riemannian metrics which can be shown using the Fisher methodology and information geometry, and this is how I initially started thinking about $\Psi(r,s)$. I initially thought of it as a Fisher metric.
With this linear geometric flow we have essentially summed discrete pieces of the metric, over a spectrum, in this case the natural numbers, and then taken the Mellin transform to recover the zeta function.
That is to say, $f_s(x)$ seems to be "weakly" tied to the Riemann zeta function. If you start with $f_s(x)$ and pretend that you can take two Mellin transforms in a row and sum over the natural numbers you recover the Riemann zeta i.e. formula $(1).$ Of course it doesn't make sense to take iterative Mellin transforms like this. But already, you can see that $f_s(x)$ is deeply linked to $\zeta(s)$. And this really isn't that surprising, given the modern viewpoint of things as in Iwasawa-Tate theory.
Now I claimed above, that $f_s(x)$ was Schwartz $\forall s >0.$ You may be very skeptical about this and rightfully so. This is a delicate situation and I'll explain why. The Schwartz space is classically defined for functions on the real line and so there must be some functional analysis work done to show that $f_s(x)$ can legitimately be thought of as a kind of Schwartz function. I believe things will work out just fine, under the appropriate modifications, but I personally do not know enough functional analysis to tie everything up neatly here.
Question:
At this point I'd like to state my question. As the title says I am looking to extend the two PDE's $(2)$ and $(3)$ to the critical strip as defined through $(4).$
How do you extend $(2)$ and $(3)$ to the critical strip as defined by $(4)?$
Explorations of the literature:
Here are some other attempts of mine to educate myself on the approaches that have already been developed. Most relevant I think is the de Bruijn constant:
I've read Tao's article about freezing and vaporizing the Riemann zeta function. Also see de Bruijn constant. Here is Tao's presentation:
https://terrytao.files.wordpress.com/2018/08/webinar.pdf.
This seems related since $f_s(x)$ obeys a strange kind of "backwards heat equation" $$s \frac{\partial ^2f(s,x)}{\partial s^2}=-x \frac{\partial f(s,x)}{\partial x}$$
I do know that the Mellin transform by definition maps some function $\psi : (0,\infty) \to \mathbb{C}$ with $\mathcal{M}(\psi)(s) = \int_{0}^{\infty} \psi(x) x^s \frac{dx}{x}$
that's absolutely convergent on the vertical strip $\lbrace s \in \mathbb{C} : a < \Re(s) < b \rbrace$.
So it seems to me that this sum of the metric over a particular spectrum of the naturals, is being converted from a real $s$ to a complex $w$ through the Mellin transform. So, I guess that's where I'm stuck. I don't know how to connect that complex object to the $1$-parameter family of real metrics.
It seems closely related to this too: https://mathoverflow.net/q/139863/123449, but I don't know if the argument applies since the operator of my PDE is not a second order p.d.o. (it's third order) among other things. I mean I guess it could be reduced to second order through a Fourier transform but I don't know how to proceed exactly.
Final remarks:
$(A).~f_s(x)$ is a $1$-parameter class, and therefore we should have a varying, or $1$-parameter class of functional equations.
$(B).~$The Gamma factor is also a $1$-parameter class.
$(C).~$I think there is a way to unify $(A)$ and $(B)$ on the critical strip, using existing mathematical methods.
$(D).~$Any argument showing that $f_s(x)$ cannot be used to obtain $(4)$ would also answer this question, essentially providing a counterexample.
I believe the function
$$g_r(x)=r^2 \frac{\partial^3\Psi(r,x)}{\partial r^3}=x^2 \frac{\partial \Psi(r,x)}{\partial x}=-2 r \sqrt{\frac{1}{x}} x^{3/2} K_2\left(2 \sqrt{r x}\right),\quad r>0\tag{1}$$
is only Schwartz in the right-half plane, so consider the even function $h_r(x)=g_r(|x|)$ instead for which
$$\Gamma\left[h_r,s\right]=\int\limits_{-\infty}^\infty h_r(x)\, |x|^{s-1}\, dx=2 \int\limits_{-\infty}^\infty h_r(x)\, x^{s-1}\, dx$$ $$=-2\, r^{-s}\, \Gamma(s)\, \Gamma(s+2),\quad r>0\land\Re(s)>0\tag{2}$$
and
$$\Gamma\left[\hat{h_r},1-s\right]=\frac{\zeta(s)}{\zeta(1-s)}\, \Gamma\left[h_r,s\right]=2^s\, \pi^{s-1}\, \sin\left(\frac{\pi s}{2}\right)\, \Gamma(1-s)\, \Gamma\left[h_r,s\right]$$ $$=-(2 \pi)^s \left(\frac{1}{r}\right)^s \sec\left(\frac{\pi s}{2}\right) \Gamma(s+2)\tag{3}$$
where $\hat{h_r}(w)=\mathcal{F}_x[h_r(x)](w)$ is the Fourier transform of $h_r(x)$.
The Fourier transform of $h_r(x)$ is
$$\hat{h_r}(w)=\mathcal{F}_x[h_r(x)](w)=\int\limits_{-\infty }^{\infty} h_r(x)\, e^{-i 2 \pi w x} \, dx=\frac{i r e^{-\frac{i r}{2 \pi w}}}{8 \pi^3 w^3} \left(4 i \pi w e^{\frac{i r}{2 \pi w}}+r e^{\frac{i r}{\pi w}}\, \Gamma\left(0,\frac{i r}{2 \pi w}\right)-r\, \Gamma\left(0,-\frac{i r}{2 \pi w}\right)\right),\quad r>0\tag{4}$$
and $\Gamma\left[\hat{h_r},1-s\right]$ evaluated using formula (4) for $\hat{h_r}(w)$ above is
$$\Gamma\left[\hat{h_r},1-s\right]=\int\limits_{-\infty}^\infty \hat{h_r}(w)\, |w|^{s-1}\, dw=2 \int\limits_{-\infty}^\infty \hat{h_r}(w)\, w^{s-1}\, dw=\frac{i r}{4 \pi^3} \left(8 i \pi^2 \delta(-i (s+1))+\frac{2^{s+2} \pi^{s+3} \left(\left(\frac{i}{r}\right)^s-\left(-\frac{i}{r}\right)^s\right) \csc(\pi s) \Gamma(s+2)}{r}\right)\tag{5}$$
which is equivalent to formula (3) for $\Gamma\left[\hat{h_r},1-s\right]$ above when ignoring the $8 i \pi^2 \delta(-i (s+1))$ term in formula (5) above (which corresponds to a pole of $\Gamma\left[\hat{h_r},1-s\right]$ at $s=-1$).
I believe the function
$$u_r(x)=r \frac{\partial ^2}{\partial r^2}e^{\frac{r}{\log (x)}}=-x \frac{\partial }{\partial x}e^{\frac{r}{\log (x)}}=\frac{r\, e^{\frac{r}{\log (x)}}}{\log ^2(x)}\tag{6}$$
may perhaps be a Schwartz function for $x\in\mathbb{R}$, but instead consider the simpler even function $v_r(x)=u(|x|)$ over the interval $x\in(-1,1)$ for which
$$\Gamma\left[v_r,s\right]=\int\limits_{-1}^1 v_r(x)\, |x|^{s-1}\, dx=2 \int\limits_{-1}^1 v_r(x)\, x^{s-1}\, dx$$ $$=4 \sqrt{r s}\, K_1\left(2 \sqrt{r s}\right),\quad r>0\land\Re(s)>0\tag{7}$$
and
$$\Gamma\left[\hat{v_r},1-s\right]=\frac{\zeta(s)}{\zeta(1-s)}\, \Gamma\left[v_r,s\right]=2^s\, \pi^{s-1}\, \sin\left(\frac{\pi s}{2}\right)\, \Gamma(1-s)\, \Gamma\left[v_r,s\right]\tag{8}$$
where $\hat{v_r}(w)=\mathcal{F}_x[v_r(x)](w)$ is the Fourier transform of $v_r(x)$.
I don't have a closed form for $\hat{v_r}(w)$ but $v_1(x)$, $\hat{v_1}(w)$ and $\Gamma\left[\hat{v_1},1-s\right]$ can all be approximated by series representations analogous to formulas (4), (7), and (9) of this answer to a related question where in this case the Fourier coefficients
$$a_k=\int\limits_{-1}^1 v_1(x)\, \cos(\pi k x)\, dx\tag{9}$$
can again be approximated via numeric integration.
Figure (1) below illustrates the series representation of $\Gamma\left[\hat{v_1},1-s\right]$ in orange overlaid on the closed form for $\Gamma\left[\hat{v_1},1-s\right]$ defined in formula (8) above in blue.
Figure (1): Illustration of the series representation of $\Gamma\left[\hat{v_1},1-s\right]$ (orange) overlaid on formula (8) for $\Gamma\left[\hat{v_1},1-s\right]$ (blue)