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.
 A: 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 $|\Im(u)|\le \sqrt{2+\sigma}$.
A: $\DeclareMathOperator{\sech}{sech}$
$\DeclareMathOperator{\Si}{Si}$
$\DeclareMathOperator{\erf}{erf}$
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:

*

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


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

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


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

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


*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).

