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.

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  • $\ln (\zeta ((\sigma+it)(1-2^{1-\sigma-it})))$ has logarithmic singularities at RZ zeroes on the critical line say, while I do not see RHS of your expression having such, so I would suggest trying numerics for the first nontrivial zero with $\sigma =1/2, t=14.13....$; in particular the integral for that $1/2+it_0$ combination should diverge, while for nearby $1/2+it$ should be finite and I see no reason for that – Conrad Aug 16 '21 at 16:20
  • @Conrad those were some of the first things I tried, seems to hold there too- at least to Mathematica's precision, the integral explodes to $- \infty$. – KStarGamer Aug 16 '21 at 16:27
  • How is $\ln (\zeta ((\sigma+it)(1-2^{1-\sigma-it})))$ defined around $1/2+it_0$? – Conrad Aug 16 '21 at 16:34
  • The question is how do you define the logarithm of $f$ near a zero of $f$; it is definitely not possible to define it continuously so what choices do you make? At some point there is a $2\pi$ jump in the imaginary part – Conrad Aug 16 '21 at 16:39
  • Yeah, I see your point- not sure to be honest, but both sides of the representation give $-\infty$ at the zeros of $\zeta$, so not sure. – KStarGamer Aug 16 '21 at 16:44
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    Try just before the zero from all sides (in other words something like $1/2 \pm 1/100+it_0, 1/2 +i(t_0 \pm 1/100)$ and see the results in both imaginary and real parts, because the imaginary part jumps there - not sure how Mathematica computes the argument of zeta (hence the imaginary part of the logarithm) but there is a fairly standard way and tons of results about this quite mysterious function – Conrad Aug 16 '21 at 16:50
  • @KStarGamer Here is another integral representation using the [Abel Plana formula](https://en.m.wikipedia.org/wiki/Abel–Plana_formula). – Tyma Gaidash Aug 22 '21 at 19:25
  • Math SE is not a peer reviewed journal, nor a preprint service. We are not journal editors or peer reviewers here. Review of purported results (especially those which claim to resolve long standing open problems) is not within the scope of this website. Such questions are simply too broad. – Xander Henderson Aug 29 '21 at 00:08
  • @Xander Henderson I only included my attempted proof to give context to my conjectured representation. I’m not asking for verification of my proof, which I’m certain is likely wrong, I’m just asking for help with ideas on how to go about proving Conj 1.1 and 1.3, as well as if it worth pursuing any further. I felt that it would be more appropriate to give context to this representation, otherwise it wouldn’t be at all motivated, and no one would care about it. I can remove the attempted RH proof and just keep the conjectures and leave the question as help with proving them, if more suitable. – KStarGamer Sep 01 '21 at 20:01
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    @KStarGamer Your Conj 1.1 contains an intriguing integral. When you limit the question to only that, I believe it would be perfectly fine for MSE. Using the words "RH" and "proof" jointly in an OP, will almost automatically provoke the word "purported" when closed down :-) I've experimented quite a bit with the integral and numerically it appears to work fine (although I don't know why it does so). Note that it actually seems to work for Dirichlet series in particular (e.g. using the $\zeta$-function or the $\beta$-function yields the same results). – Agno Sep 01 '21 at 21:20
  • Ok, thanks for the comments @Agno I've removed all the unnecessary nonsense. I was aware that it seemed to work for other Dirichlet series, but I don't know why, and I cannot prove such a thing. – KStarGamer Sep 01 '21 at 21:40
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    Towards a resolution of this controversial question, note that it should come as no surprise that this formula holds, because it holds for any even function $f$. In particular a type of Fourier inversion formula indicates $f(t)=\frac{1}{2\pi}\int^{\infty}_{-\infty}\int^{\infty}_{-\infty} \cos(y(x-t)) f(x) dx dy$ which for even functions can be written $f(t)=\frac{2}{\pi}\int_0^{\infty}\int_0^{\infty}f(x)\cos(xy)\cos(yt)$. To obtain the desired results simply set $t\to \sqrt{t}e^{i\pi/4}$ and $f(x)=\zeta(\sigma+x^2)^b$ for example. – DinosaurEgg Sep 01 '21 at 21:57
  • @DinosaurEgg Doesn't Fourier inversion generally only work for functions $f:\mathbb{R} \to \mathbb{C}$? Maybe the Paley-Wiener theorem is at play or something related- not sure. It is also interesting that the $\sqrt{t} e^{i \pi/4}$ is required in particular for this to hold- anything else, it doesn't numerically work, so there is more to it I think than just simply setting $t \to \sqrt{t} e^{i \pi/4}$. – KStarGamer Sep 02 '21 at 15:07
  • @KStarGamer Your concerns are certainly valid, but can you elaborate on this last point? For example if I set $f(x)=\zeta(\sigma+x^4)^b$ and $t\to t^{1/4}e^{i\pi/8}$, will I not obtain the result $\zeta(\sigma+it)^b$? Is this what you mean by "nothing else works numerically"? – DinosaurEgg Sep 02 '21 at 16:15
  • @DinosaurEgg Yes, that is exactly what I had meant, and although I thought I was incorrect, I realised I had set $t=0$, I tried with a value for $t$, and now Mathematica gives me two different values between the integrals and $\zeta (\sigma + i t)^b$, and I ensured to change $y^2 \to y^4$ in the integral and $t \to t^{1/4} e^{i \pi/8}$. – KStarGamer Sep 02 '21 at 16:26

3 Answers3


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):


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:



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

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  • @KStarGamer Just one more observation: in the integral for $\eta(\sigma)$, it appears that taking $\sigma \in \mathbb{C}$ works fine as well (but obviously you've now lost the nice separation between the real and imaginary parts). Still feels like some Fourier inversion mechanism is at work here. – Agno Sep 03 '21 at 22:05

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

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:

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

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

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

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

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

  1. 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).
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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 $|\Im(u)|\le \sqrt{2+\sigma}$.

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  • Thank you for managing to prove the equality. I have one question and that is if the equality is still valid if one takes the partial derivative with respect to $b$, then the limit as $b \to 0^+$? – KStarGamer Dec 31 '21 at 21:41
  • For $|\Im(u)|< \sqrt{2+\sigma}$ yes of course. – reuns Dec 31 '21 at 21:43
  • What if instead one just sets $b=0$ immediately after taking the partial derivative rather than a limit? $b$ would be an integer in that case meaning that the equality would hold $\forall u \in \mathbb{C}$ including at $\sigma = \frac{1}{2}$ where the function would not be holomorphic at $t=14.1…$ i.e the first zeta zero. – KStarGamer Dec 31 '21 at 22:12
  • It works the same way with $H(y)=\log \eta(\sigma+y^2)$ which is analytic and Schwartz on horizontal lines for $|\Im(y)|< \sqrt{2+\sigma}$ (at $y=i\sqrt{2+\sigma}$ it reaches the singularity of $\log \eta(s)$ at $s=-2$) and $(1)$ is valid for $|\Im(u)| < \sqrt{2+\sigma}$. It doesn't reach the non-trivial zeros. – reuns Dec 31 '21 at 22:16