Background: I'm an undergraduate at an institution with no researchers in analytic number theory, and no ties to the analytic number theory community. I believe I have found what is, as far as I can tell after some googling, a new family of integral representations of $\zeta(2n+1)$. I was told I should post this here by an algebraic number theorist at my university, to see if it was a known result or not.

Feel free to look at Equations (4), (5), and (7), to see if you recognize them before reading the entire text. Most of this text is for the special case of $\zeta(3)$, but all results generalize quite readily to the case of $\zeta(2n+1)$, which I discuss at the end.

Beginning of setup

As a preliminary step, we will split the sum $\zeta(3)$ into even and odd parts. Observe that $$ \sum_{n=1}^\infty \frac{1}{(2n)^3} = \frac{1}{8} \sum_{n=1}^\infty \frac{1}{n^3} = \frac{1}{8}\zeta(3) $$ Therefore $$ \sum_{n=1}^\infty \frac{1}{(2n+1)^3} = \frac{7}{8}\zeta(3) \tag{1} $$ We now write $$ \frac{1}{(2n+1)^3} = \left(\int_0^1 x^{2n} dx\right)^3 = \int_0^1 \int_0^1 \int_0^1 x^{2n} y^{2n} z^{2n} \tag{2} dx dy dz $$ and plug this triple integral into Equation (1), obtaining $$ \zeta(3) = \frac{8}{7} \sum_{n=0}^\infty \left[ \int_0^1 \int_0^1 \int_0^1 x^{2n} y^{2n} z^{2n} dxdydz \right] $$ Using the absolute convergence of the geometric series on $(0,1)$, we can interchange the sum and integrals, obtaining $$ \zeta(3) = \frac{8}{7} \int_0^1 \int_0^1 \int_0^1 \frac{1}{1-x^2y^2z^2} dxdydz \tag{3} $$

End of setup

With this part out of the way, I will now describe the integral representations that form the meat of my question.

Consider the $u$-substitution for the integral given by Equation (3) $$ x = \frac{\sinh{u}}{\cosh{v}}, \qquad y = \frac{\sinh{v}}{\cosh{w}}, \qquad z = \frac{\sinh{w}}{\cosh{u}} $$ Some computation shows that $$ dxdydz = [1-(\tanh{u}\tanh{v}\tanh{w})^2]dudvdw $$ But dividing both sides by $1-(\tanh{u}\tanh{v}\tanh{w})^2$ and rewriting the left-hand side in terms of $(x,y,z)$ gives $$ \frac{1}{1-x^2y^2z^2}dxdydz = dudvdw $$ After changing limits, the transformed integral then reads $$ \zeta(3) = \frac{8}{7} \int_0^\infty \int_0^{\sinh^{-1}(\cosh(w))} \int_0^{\sinh^{-1}(\cosh(v))} dudvdw \\ = \frac{8}{7} \int_0^\infty \int_0^{\sinh^{-1}(\cosh(w))} \sinh^{-1}(\cosh(v)) dv dw \tag{4} $$

But the most interesting part of the above is the following generalization. Let $f : (0,\infty) \to (0,\infty)$ be a function satisfying the following three properties.

  1. $f$ is surjective, with $\lim\limits_{x\to 0} f(x) = 0$ and $\lim\limits_{x \to \infty} f(x) = \infty$
  2. $f$ is invertible
  3. $f$ is differentiable

Now consider the $u$-substitution $$ x = \frac{\sinh{f(u)}}{\cosh{f(v)}}, \qquad y = \frac{\sinh{f(v)}}{\cosh{f(w)}}, \qquad z = \frac{\sinh{f(w)}}{\cosh{f(u)}} $$ We then find that $$ dxdydz = f'(u)f'(v)f'(w)[1-(\tanh{f(u)}\tanh{f(v)}\tanh{f(w)})^2]dudvdw $$ and hence $$ \zeta(3) = \frac{8}{7} \int_0^\infty \int_0^{g(w)} \int_0^{g(v)} f'(u)f'(v)f'(w) dudvdw \tag{5} $$ where $g(x) = f^{-1}(\sinh^{-1}(\cosh(f(x))))$.

All of the above can be generalized to $\zeta(2n+1)$. Equation (3) becomes $$ \zeta(2n+1) = \frac{2^{2n+1}}{2^{2n+1}-1} \int_0^1 \int_0^1 \cdots \int_0^1 \frac{1}{1-x_1^2 x_2^2 \cdots x_{2n+1}^2} dx_1 dx_2 \cdots dx_{2n+1} \tag{6} $$ Our $u$-substitution becomes (where $u_{(2n+1)+1} = u_1$) $$ x_i = \frac{\sinh(f(u_i))}{\cosh(f(u_{i+1}))} $$ which transforms Equation (6) to $$ \zeta(2n+1) = \frac{2^{2n+1}}{2^{2n+1}-1} \int_0^\infty \int_0^{g(u_{2n})} \int_0^{g(u_{2n-1})} \cdots \int_0^{g(u_1)} f'(u_1)f'(u_2)\cdots f'(u_{2n+1}) du_1 du_2 \cdots du_{2n+1} \tag{7} $$

My questions are then:

  1. Are Equations (4), (5), and (7) known results?
  2. Equation (4) is a volume integral. Exploring the region of integration in Mathematica numerically, it looks like an octant of a hyperbolic cube with vertices at infinity. Is this in fact the case?
  • $\begingroup$ In equation (5), did you mean to put a $g$ in? $\endgroup$ – David Roberts Sep 17 '15 at 10:26
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    $\begingroup$ mathoverflow.net/questions/54237/… $\endgroup$ – Felipe Voloch Sep 17 '15 at 13:01
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    $\begingroup$ I assume you are aware of the following integral formula though: $\zeta(n) = \int_0^1\cdots\int_0^1 \frac{dx_1dx_2\cdots dx_n}{1-x_1x_2\cdots x_n}$? $\endgroup$ – Suvrit Sep 17 '15 at 13:36
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    $\begingroup$ Did you look at: arxiv.org/pdf/1003.3602v1.pdf --- that paper has very similar "Beukers-Kolk-Calabi" change of variables and very closely related integrals for $\zeta(3)$ and $\zeta(2n+1)$....see also the discussion in arxiv.org/pdf/1207.2055.pdf --- I think (7) is "essentially" proved in these papers. $\endgroup$ – Suvrit Sep 17 '15 at 17:41
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    $\begingroup$ Welcome to MathOverflow! As you may be able to tell, you have a well-written first-time question here that is appreciated by the community. I wish you success in your studies, and look forward to more quality questions from you. Gerhard "Also Try Answers And Comments" Paseman, 2015.09.17 $\endgroup$ – Gerhard Paseman Sep 17 '15 at 20:28

As indicated in my comment, some of these integrals are essentially known, and involve the hyperbolic "Beukers-Kolk-Calabi" change of variables.

In particular, in this paper, Z. Silagadze shows in (27) that \begin{equation*} \zeta(n) = \frac{2^n}{2^n-1}\int_0^1\cdots\int_0^1 \frac{dx_1\cdots dx_n}{1-x_1^2\cdots x_n^2}, \end{equation*} which clearly yields (7) above.

The hyperbolic change of variables is used later in that paper to obtain a related formula over the $2n$-simplex $\Delta_{2n}$, namely \begin{equation*} \zeta(2n+1) = -\frac{1}{2n}\frac{2^{2n+1}}{2^{2n+1}-1}\int_{\Delta_{2n}} \log(\tan x_1 \tan x_2\ldots \tan x_{2n})dx_1\cdots dx_{2n}. \end{equation*} This integral is amenable to further interesting modifications.


  1. The arXiv paper cited above has been published in the journal "Resonance"
  2. Another paper by the same author (Z. Silagadze) comments on these integrals.
  3. The "Beukers-Kolk-Calabi" change of variables was also considered in this paper.
  4. This paper (cited below by Benjamin Dickman) is also relevant here.
  5. The relation of these representations to Amoebas is quite interesting, and worthy of further exploration.
  • $\begingroup$ Is there a way to notify prof Silagadze of this question? Will writing @ZurabSilagadze work? mathoverflow.net/users/32389/zurab-silagadze $\endgroup$ – Vít Tuček Sep 17 '15 at 21:03
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    $\begingroup$ Around the same time as the former paper there is also ζ(n) via hyperbolic functions (D'Avanzo & Krylov, 2010) which cites the paper of Silagadze. I point this out because it was (perhaps) a bit easier to find given the OP's presentation using sinh, cosh, tanh: google ζ(n) hyperbolic and it ought to be the first result. [Edit: Separately, why on earth is Suvrit's answer voted down?!] $\endgroup$ – Benjamin Dickman Sep 17 '15 at 21:28
  • $\begingroup$ @BenjaminDickman thanks for the very nice reference! $\endgroup$ – Suvrit Sep 17 '15 at 23:19
  • $\begingroup$ @VítTuček I tried to do that (as a comment to Gjergji's post, but I think I used the wrong syntax; your choice seems more like it). $\endgroup$ – Suvrit Sep 17 '15 at 23:19
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    $\begingroup$ I noticed this MO question only now and I have not much to add to already given answers. My paper mentioned in Suvrit's answer is now published: ias.ac.in/resonance/php/cissue.php hyperbolic version of the "Beukers-Kolk-Calabi" change of variables was considered also in this paper (I cite it in my work): arxiv.org/abs/1003.2170 Relation to amoeba's is quite interesting and worth to further exploration, in my opinion. $\endgroup$ – Zurab Silagadze Sep 23 '15 at 8:24

I recommend this recent preprint "Sums of generalized harmonic series for kids from five to fifteen" by Z.K. Silagadze. The author explains identities for $\zeta(n)$ that come from your equation (6) by using hyperbolic function substitutions. It is pointed out that the Beukers-Kolk-Calabi substitution and its hyperbolic analogue can be explained by relating them to the amoeba's of a certain Laurent polynomial (the regions you have been computing in mathematica). Unfortunately I can't retell that part of the story beyond $\zeta(2)$, but this is somewhat discussed in the last section.

  • $\begingroup$ This paper is the one linked in my comment to the OP! $\endgroup$ – Suvrit Sep 17 '15 at 20:52
  • $\begingroup$ I just noticed that :) $\endgroup$ – Gjergji Zaimi Sep 17 '15 at 20:53
  • $\begingroup$ Maybe @zurab-silagadze reads this thread and himself comments! $\endgroup$ – Suvrit Sep 17 '15 at 20:59

Yes, it is known by using Bernoulli polynomials:

$$\zeta (2n+1)=(-1)^{n+1}\frac {(2\pi)^{2n+1}}{2 (2n+1)!}\int_0^1B_{2n+1}(x)\cot (\pi x)\,dx,$$

where $B_n (x) $ is the Bernoulli polynomial, which is very important in numerical problems.

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    $\begingroup$ I'm having a hard time seeing how this integral in the case $\zeta(3)$ can be transformed into the integral in Equation (4), or vice-versa. I understand the expressions are equal; I do not understand how to show this without appealing to the fact that both expressions equal $\zeta(3)$. Could you elaborate on this, or provide a reference that does? $\endgroup$ – Andrew Knapp Sep 17 '15 at 9:45
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    $\begingroup$ I think it will not be too hard , because if you take $n=1$ for the case $\zeta (3) $ and by using $B_3(x) =x^3-\frac{3}{2}x^2+\frac{1}{2}x$, we can see the right hand sides are equal $\endgroup$ – user21574 Sep 17 '15 at 20:37
  • $\begingroup$ Maybe, it answers another question: is there known representation? $\endgroup$ – Fedor Petrov Sep 17 '15 at 20:38
  • $\begingroup$ I think it will be an simple exercise to see, right hand sides are equal. $\endgroup$ – user21574 Sep 17 '15 at 20:45
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    $\begingroup$ @HassanJolany : the fact that you can easily show it for $n=1$ using an explicit formula for the Bernoulli polynomial does not really indicate anything on how difficult it is in general. $\endgroup$ – Vladimir Dotsenko Sep 17 '15 at 21:20

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