Double integral separating real and imaginary parts of $\zeta (\sigma+i t)$

Recently, I stumbled upon what I believe to be a new representation of $\zeta(\sigma+i t)^b$ by chance, and thus have no proof of it, and I am wondering if it is possible to prove.

Let $\eta (s) = \zeta (s) (1-2^{1-s})$ denote the Dirichlet eta function, where $\zeta$ is the Riemann zeta function and $s = \sigma + i t$.

Motivation: To motivative the potential usefulness of this representation, notice that $\zeta$ is transformed from taking a complex argument, to a real argument, allowing safer manipulations with it, as log singularities are no longer as major of an issue. I've had Mathematica numerically verify the conjecture on a few hundred combinations of $\sigma, t$, and $b$, each time producing the correct values- as far as I can tell, it even seems to behave appropriately near $\zeta$ zeros, from approaching the first few zeros, above and below the real line, from all directions.

$\forall \sigma, t, b \in \mathbb{R}_{\geq 0}$: $$\eta (s)^b-1=\frac{2}{\pi}\int_{0}^{\infty}\int_{0}^{\infty}\cos(\sqrt{t}x e^{i \pi/4})\cos(x y)\left(\eta(\sigma+y^2)^b-1\right)\,dy\,dx$$

Is it possible to prove/disprove this representation?

A suggestion by @DinosaurEgg was the fact that a Fourier inversion formula indicates: $$f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \cos \left(y\left(x-t\right)\right) f(x) \, dx \, dy$$ which for even functions can be written as $$f(t) = \frac{2}{\pi} \int_{0}^{\infty}\int_{0}^{\infty} \cos \left( x y \right) \cos \left( y t\right) f(x) \, dx \, dy$$

However, to my knowledge, Fourier inversion only generally holds for functions $f : \mathbb{R} \to \mathbb{C}$ so perhaps the Paley-Wiener theorem is at play here to allow my $\eta$ representation to hold, however, I’m not sure how to apply it here.

This is not an answer, but just to share some numerical observations about the Conj 1.1 integral (ignoring the $b$-power for now).

Take the Dirichlet function: $\eta(s) = \left(1-2^{1-s}\right)\zeta(s)$ and $s=\sigma + ti$, then we have your integral as:

$$\eta(s)=1+\frac{2}{\pi}\int_{0}^{\infty}\int_{0}^{\infty}\left(\eta(\sigma+y^2)-1\right)\cos\left(\sqrt{t}x e^{i \pi/4}\right)\,\cos(xy)\,dy\,dx$$

in which surprisingly, the real and imaginary parts of $s$ are 'separated'.

Numerical evidence suggests the double integral could be simplified further into a limit valid for all $s \in \mathbb{C}$:

$$\eta(s)=\lim_{V \to \infty} 1+\frac{1}{\pi}\int_{-V}^{V}\left(\eta(\sigma+y^2)-1\right)\frac{\sin\left(2 V\left(\sqrt{t} e^{i \pi/4}+y\right)\right)}{\left(\sqrt{t} e^{i \pi/4}+y\right)}dy$$

that for instance for $V = 33$ and $s=1/2+14.13472514173i$, i.e. a value relatively near to the first non-trivial zero, yields (correct value in the second line):

-5.3689806676380118318307396342475069763270134673799135834687318395168... E-13 - 8.82065894132048736467139905332032769853018880872300364479887520047395... E-12*I
-5.3689806676380118318307396342475069763270134673799135834687318395210... E-13 - 8.82065894132048736467139905332032769853018880872300364479887520047404... E-12*I

Focusing on the real part only, we get:

$$\eta(\sigma)=\lim_{V \to \infty} 1+\frac{2}{\pi}\int_{0}^{V}\left(\eta(\sigma+y^2)-1\right)\frac{\sin\left(2Vy\right)}{y}dy$$

that seems valid for all $\sigma \in \mathbb{R}$. With $V=20$ and $\sigma=2$ the result is (correct value $\pi^2/12$ in the second line):

0.82246703342411321823620758332301259460947495060339921886777911468500373520203498...
0.82246703342411321823620758332301259460947495060339921886777911468500373520160044...

Maybe for a start, it would be possible to prove something simple like:

$$\eta(0)=\lim_{V \to \infty} 1+\frac{2}{\pi}\int_{0}^{V}\left(\eta(y^2)-1\right)\frac{\sin\left(2Vy\right)}{y}dy =\frac12$$

which at $V=20$ computes to:

0.50000000000000000000000000000000000000000000000000000000000000000000000000617584...
0.50000000000000000000000000000000000000000000000000000000000000000000000000000000...

P.S.:

The factor $2$ before the $V$ is a constant I've injected experimentally to maximise the speed of convergence towards the correct value.

With $\sigma > -2$, $H(y)=\eta(\sigma+y^2)^b-1$ is Schwartz on the real line so we can look at its inverse Fourier transform, which is Schwartz again $$h(x)=\frac1{2\pi}\int_{-\infty}^\infty e^{i xy} H(y)dy$$

And for all $u\in \Bbb{R}$

$$H(u) = \int_{-\infty}^\infty e^{-i x u}h(x)dx\tag{1}$$

When $\color{red}{b\text{ is an integer}}$, $H$ is entire and Schwartz on every horizontal line. The Cauchy integral theorem gives that for any $r$ $$h(x)=\frac1{2\pi}\int_{-\infty}^\infty e^{i x(y+ir)} H(y+ir)dy$$ from which $$|h(x)|\le e^{-r |x|} C_r, \qquad C_r=\int_{-\infty}^\infty (|H(y+ir)|+|H(y-ir)|)dy$$ This implies that there won't be any problem in continuing $(1)$ analytically to $\color{red}{\text{ every } u\in \Bbb{C}}$, obtaining $$\forall u\in \Bbb{C}, \qquad H(u)=\int_{-\infty}^\infty e^{-i x u}h(x)dx$$ With $u=\sqrt{it}$ it gives the formula in your question.

When $b$ is not an integer, we can only do the Cauchy integral theorem up to $r = \sqrt{2+\sigma}$, so that $(1)$ stays convergent and valid only for $|\Im(u)|\le \sqrt{2+\sigma}$.

$\DeclareMathOperator{\sech}{sech}$ $\DeclareMathOperator{\Si}{Si}$ $\DeclareMathOperator{\erf}{erf}$

Below is an attempt to prove your equation for $b=1$.

Take $s=\sigma+ti$ with $\sigma,t \in \mathbb{R}$ and $\eta(s) = \left(1-2^{1-s}\right)\zeta(s)$.

Note that:

$$\eta(s) = \frac12 \int_{-\infty}^{+\infty} \frac{\sech(\pi u)}{\left(\frac12+ui\right)^s} du \qquad s \in \mathbb{C} \tag{1}$$

The function to prove is:

$$\eta(\sigma+ti)=\frac{1}{\pi}\int_{0}^{\infty}\int_{0}^{\infty}\left(\eta(\sigma+y^2)-1\right)\cos\left(x\,\sqrt[4]{-1}\sqrt{t}\right)\,\cos(xy)\,dx\,dy +1\tag{2}$$

Let's start with the real part ($t=0$) which gives:

$$\eta(\sigma)=\frac{2}{\pi}\int_{0}^{\infty}\left(\eta(\sigma+y^2)-1\right)\,\int_{0}^{\infty}\cos(xy)\,dx\,dy +1\tag{3}$$

Simplifiying the integral over $x$ and expanding the domain over $y$ gives:

$$\eta(\sigma)=\lim_{v\to\infty} \frac{1}{\pi}\int_{-v}^{v}\left(\eta(\sigma+y^2)-1\right)\,\frac{\sin(vy)}{y}\,dy+1 \tag{4}$$

Let's remove the $-1$ by observing that $\frac{1}{\pi}\int_{-v}^{v}-\frac{\sin(vy)}{y} dy = -\frac{2\Si(v^2)}{\pi}$ where Si = Sine Integral:

$$\eta(\sigma)=\lim_{v\to\infty} \frac{1}{\pi}\int_{-v}^{v}\eta(\sigma+y^2)\,\frac{\sin(vy)}{y}\,dy-\frac{2\Si(v^2)}{\pi} +1 \tag{5}$$

Now inject integral (1) for $\eta(s)$:

$$\eta(\sigma)=\lim_{v\to\infty} \frac{1}{2\pi}\int_{-v}^{v} \int_{-v}^{v}\frac{\sech(\pi u)}{\left(\frac12+ui\right)^{\sigma+y^2}}\,\frac{\sin(vy)}{y}\,dy\,du -\frac{2\Si(v^2)}{\pi} +1 \tag{6}$$

Which allows the $y^2$ to move to the right:

$$\eta(\sigma)=\lim_{v\to\infty} \frac{1}{2\pi}\int_{-v}^{v} \frac{\sech(\pi u)}{\left(\frac12+ui\right)^{\sigma}}\,\int_{-v}^{v}\frac{1}{\left(\frac12+ui\right)^{y^2}}\frac{\sin(vy)}{y}\,dy\,du -\frac{2\Si(v^2)}{\pi} +1 \tag{7}$$

The $y$-integral now nicely evaluates as: $\pi \erf\left(\frac{v}{2\log(1/2+ui)} \right)$, with erf = error function, which is always $\pi$ independent of $u$ when $v \rightarrow \infty$.

With $\displaystyle \lim_{v\to\infty}\frac{2\Si(v^2)}{\pi} = 1$ we then obtain the desired result:

$$\frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{\sech(\pi u)}{\left(\frac12+ui\right)^{\sigma}}\,\pi\,du -1 +1 = \eta(\sigma) \tag{8}$$

Side comment: equation (7) is also valid for $\sigma \in \mathbb{C}$.

For $t \ne 0$, the situation turns out to be a bit more complicated. Starting from equation (7):

$$\eta(\sigma,t)=\lim_{v\to\infty} \frac{1}{2\pi}\int_{-v}^{v} \frac{\sech(\pi u)}{\left(\frac12+ui\right)^{\sigma}}\,\int_{-v}^{v}\frac{1}{\left(\frac12+ui\right)^{y^2+ti}}\frac{\sin(vy)}{y}\,dy\,du -\frac{2\Si(v^2)}{\pi} +1 \tag{9}$$

which already gives a partial 'separation' between the real ($\sigma$) and imaginary ($t$) parts. Using the following relation for $\Re(z) > 0, a \in \mathbb{C}$ that I found numerically (hard proof required, asked here):

$$\lim_{v\to\infty} \int_{-v}^{v}\frac{1}{z^{y^2}}\frac{\sin(v\,(\sqrt{a}+y))}{\sqrt{a}+y}\,dy = \frac{\pi}{z^{a}} \tag{10}$$

and with $z=\frac12+ui, a = ti$ the final integral becomes:

$$\frac{1}{2\pi}\int_{-\infty}^{+\infty} \frac{\sech(\pi u)}{\left(\frac12+ui\right)^{\sigma}}\,\frac{\pi}{\left(\frac12+ui\right)^{ti}} \,du = \eta(\sigma,t) \tag{11}$$

which is the desired outcome (note $\sqrt{ti}=\sqrt[4]{-1}\sqrt{t}$).

Couple of observations:

1. The proof shows that $\sigma$ and $ti$ could also be 'swapped' as follows:

$$\eta (s)-1=\frac{2}{\pi}\int_{0}^{\infty}\int_{0}^{\infty}\left(\eta(ti+y^2)-1\right)\cos(\sqrt{\sigma}\,x)\cos(x y)\,dy\,dx \tag{12}$$

or even stretch it to:

$$\eta (s)-1=\frac{2}{\pi}\int_{0}^{\infty}\int_{0}^{\infty}\left(\eta(y^2)-1\right)\cos(\sqrt{s}\,x)\cos(x y)\,dy\,dx \tag{12}$$

2. Other combinations than $y^2$ and $\sqrt{ti}$ are allowed, e.g.:

$$\eta (s)-1=\frac{2}{\pi}\int_{0}^{\infty}\int_{0}^{\infty}\left(\eta(\sigma+y^4)-1\right)\cos(\sqrt[4]{ti}\,x)\cos(x y)\,dy\,dx \tag{13}$$

3. I believe the proof still works for $b \ne 1$ (the u-integral 'contracts' back to its original form):

$$\eta (s)^b-1=\frac{2}{\pi}\int_{0}^{\infty}\int_{0}^{\infty}\left(\eta(\sigma+y^2)^b-1\right)\cos(\sqrt{ti}\,x)\cos(x y)\,dy\,dx \tag{14}$$

4. The mechanism seems to work for a broader class of entire functions like $(s-1)\zeta(s), \eta(s), \beta(s), \frac{1}{\Gamma(s)}, \sin(s)$ (i.e. it is not specific for Dirichlet series).