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.

- $f$ is surjective, with $\lim\limits_{x\to 0} f(x) = 0$ and $\lim\limits_{x \to \infty} f(x) = \infty$
- $f$ is invertible
- $f$ 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:

- Are Equations (4), (5), and (7) known results?
- 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?

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