# A multiple integral that seems related to the $\zeta$ function at even integers

I came across this integral that seems related to the Riemann zeta function $$\zeta(2n)$$ evaluated at even integers $$2n \in 2\mathbb{Z}$$. Letting $$n$$ be an even integer, define the multiple integral over $$(2n+1)$$ variables $$u_1 \cdots u_{2n+1}$$

$$\begin{equation} \mathcal{I}_{2n} = \int_0^1 du_1 \cdots \int_0^1 du_{2n+1} \frac{1}{1+u_1} \frac{1}{u_1+u_2} \frac{1}{u_2+u_3} \cdots \frac{1}{u_{2n} + u_{2n+1}} \frac{1}{u_{2n+1}+1}. \end{equation}$$

For example, the case $$2n=4$$ has integrand $$\frac{1}{1+u_1} \frac{1}{u_1+u_2} \frac{1}{u_2+u_3} \frac{1}{u_{3} + u_{4}} \frac{1}{u_{4}+u_5} \frac{1}{u_{5}+1}$$.

Below are listed some exact and numerical results for the integrals for the first few values of $$2n$$ that could be numerically evaluated on my laptop. The $$2n=0$$ case is easy, and the $$2n=2$$ case could evaluated exactly by Mathematica in terms of a complicated expression of polylogarithms (although below, I show an alternate way to explicitly derive this).

• $$\mathcal{I}_0 = \frac{1}{2} = - \zeta(0)$$
• $$\mathcal{I}_2 = \frac{\pi^2}{6} = \zeta(2)$$
• $$\mathcal{I}_4 \approx 8 \frac{\pi^4}{90} = 8 \cdot \zeta(4)$$
• $$\mathcal{I}_6 \approx 54 \cdot \zeta(6)$$
• $$\mathcal{I}_8 \approx 384 \cdot \zeta(8)$$
• $$\mathcal{I}_{10} \approx 2880 \cdot \zeta(10)$$

These integrals are almost exactly integer multiples of the zeta function at even integers, up to the small errors given by Mathematica. The series of integers $$1,8,54,384,2880$$ don't appear in the OEIS, although the first terms $$1,8,54,384$$ go as $$k^2 \cdot k!$$, and $$2880$$ isn't far off from $$5^2 \cdot 5! = 3000$$.

This integral seems closely related to the one considered in this question. However, their method for their integral seemed quite magical and I was not able to generalize it. While I'm specifically interested in this integral, it would be nice to know if there's a general method to deal with these integrals.

Below are alternate formulations of the integral that may help, as well as showing the result for $$2n=2$$. First, one can change variables to $$u_i = \frac{1-v_i}{1+v_i}$$ to rewrite it as $$\begin{equation} \mathcal{I}_{2n} = \frac{1}{2} \int_0^1 dv_1 \cdots \int_0^1 dv_{2n+1} \frac{1}{1 - v_1 v_2} \frac{1}{1 - v_2 v_3} \cdots \frac{1}{1 - v_{2n} v_{2n+1}}, \end{equation}$$ which is similar to a form found in the other MathOverflow question linked above. As such, the same substitutions used there done backwards yield other expressions $$\begin{equation} \begin{split} \mathcal{I}_{2n} &= \frac{1}{2} \int_0^1 dy_{2n+2} \int_0^{y_{2n+2}} dy_{2n+1} \cdots \int_{0}^{y_3} dy_{2} \frac{1}{y_3} \frac{1}{y_4-y_2} \cdots \frac{1}{y_{2n+2}-y_{2n+1}} \frac{1}{1-y_{2n+1}} \\ &= \frac{1}{2} \int_0^1 d \tilde{u}_{1} \cdots \int_0^1 d \tilde{u}_{2n+2} \frac{\delta(1-\tilde{u}_1 - \cdots - \tilde{u}_{2n+2})}{(\tilde{u}_1 + \tilde{u}_2)(\tilde{u}_2 + \tilde{u}_3)\cdots(\tilde{u}_{2n+1} + \tilde{u}_{2n+2})} \end{split} \end{equation}$$ which almost match the integral considered in linked question with $$2n+2$$ variables, except missing the last factor $$\frac{1}{\tilde{u}_{2n+2} + \tilde{u}_{1}}$$

For $$2n=2$$, one can use Feynman Parameterization to write $$\begin{equation} \begin{split} \mathcal{I}_{2} &= \frac{1}{2} \int_0^1 dv_1 \int_0^1 dv_2 \int_0^1 dv_3 \frac{1}{1 - v_1 v_2} \frac{1}{1 - v_2 v_3} \\ &= \frac{1}{2} \int_0^1 du \int_0^1 dv_1 \int_0^1 dv_2 \int_0^1 dv_3 \frac{1}{\left[(1 - v_1 v_2)u + (1 - v_2 v_3)(1-u)\right]^2} \\ &= \frac{1}{2} \int_0^1 du \left( \frac{\log(1-u)}{u} + \frac{\log(u)}{1-u} \right) = \int_0^1 du \frac{\log(1-u)}{u} = \sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}, \end{split} \end{equation}$$ where the second-to-last equality follows from term-by-term integration of the series expansion.

I haven't found a way to generalize the method for any of the other cases and still don't have analytical proof for the next equalities.

EDIT: See below for a beautiful answer! It turns out that it's not related to the $$\zeta$$ function except for the first few values of $$n$$.

• Are you missing a factor $\frac{1}{u_1+u_2}$ in the integrand? Sep 1, 2021 at 19:56
• yes, editing it now
– Joe
Sep 1, 2021 at 20:01
• My strategy to compute these was that one can do various partial integrations to turn the expressions into integrals over fewer variables. I can reduce $\mathcal{I}_4$ to a one-dimensional integral, so I can indeed check it up to >50 decimal places. For the rest, I can reduce them to two- or three-dimensional integrals and can check that the LHS/RHS ratio is 1 up to about 7 or 8 decimal places. For the higher values of $2n$, I can only reduce them to integrals over at least 4 variables and a badly behaved integrand, so I don't know how to get numerically stable results.
– Joe
Sep 2, 2021 at 14:46
• I am reminded of a hypercube integral from arXiv:1608.03174 that equals a zeta function for any (odd or even) integer: $$\zeta(k)=\int_0^1 \frac{dx_1}{x_1} \int_0^{x_1}\frac{dx_2}{x_2} \cdots \int_0^{x_{k-2}}\frac{dx_{k-1}}{x_{k-1}} \int_0^{x_{k-1}} \frac{dx_k}{1-x_k}$$ Sep 2, 2021 at 20:52
• The key seems to be to compute the eigenvalues and eigenvectors of the integral operator $T$ on $L^2([0,1])$ with kernel $\frac{1}{x+y}$, since $I_{2n}$ is the kernel of $T^{2n+2}$ evaluated at $(1,1)$. In mathscinet.ams.org/mathscinet-getitem?mr=656439 it is observed that this operator commutes with the second order operator $\frac{d}{dx} x^2(1-x^2) \frac{d}{dx} - 2x^2$. Presumably the eigenfunctions here are essentially some classical special functions and this should give a zeta-like formula for $I_{2n}$, but I didn't carry out the computations further. Sep 4, 2021 at 3:54

We have $$I_{2n}=\frac{(2n)!!}{(2n+1)!!}\cdot \frac1{2n+2}\cdot \pi^{2n}.$$ To see this, we follow the suggestion by Terry Tao in the comments and apply the diagonalization of the integral operator with the kernel $$1/(x+y)$$ on $$[0,1]$$. Change the variable to $$1/x\in [1,\infty)$$ and use (1.18) here (this is Mehler integral operator, as I understand) to diagonalize. Substituting the value of Legendre functions at 1, we get $$I_{2n}=\pi^{2n}\int_0^\infty x\tanh x/\cosh^{2n+2} x dx.$$ (The above part is suggested by Vladimir Petrov, I understand nothing about all this special functions and integral transforms stuff. But the answer for small $$n$$ coincides, so I guess everything is ok:) Well, you are free to ask for more details if necessary.)
For evaluating these integrals, integrate by parts noting that $$\tanh x/\cosh^{2n+2} x dx=\tanh x\cosh^{-2n} xd\tanh x=\frac12(1-\tanh^2 x)^nd\tanh^2 x\\=\frac{-1}{2(n+1)}(1-\tanh^2x)^{n+1},$$ thus $$I_{2n}=\pi^{2n}\cdot \frac1{2(n+1)}\int_0^\infty (1-\tanh^2x)^{n+1}dx=\pi^{2n}\cdot \frac1{2(n+1)}\int_0^1 (1-t^2)^{n}dt\\=\frac{\pi^{2n}}{2(n+1)}\cdot \frac {(2n)!!}{(2n+1)!!}$$ (here $$t=\tanh^2 x$$, the last integral is well known and may be calculated by induction or using Beta function or how do you prefer.)
• Glad to see the calculation completed! One minor thing: I think (1.16) and (1.17) from that paper of V. Petrov are slightly more relevant than (1.18) (it uses the Mehler-Fock transform to diagonalise the integral operator $\tilde T f(x) := \int_1^\infty \frac{f(y)}{x+y}\ dy$ as a multiplier with symbol $\pi / \cosh \pi \xi$). Was a little surprised to see the operator having continuous spectrum instead of discrete, but I was misled by the compact nature of $[0,1]$. (The operator fails to be compact due to a divergence at $0$.) Sep 4, 2021 at 18:31
• Wow this is beautiful! One can note that these fit quite neatly into the generating functional $\sum_{n=0}^{\infty} I_{2n} (\frac{\alpha}{\pi})^{2n+2} = \frac{1}{2 \pi^2}\mathrm{ArcSin}({\alpha})^2$.
• Say, convolution operator $A$ (with kernel $a(x-y)$ on the real line) is diagonalized by Fourier transform (that means that $\mathcal{F}^{-1}A\mathcal{F}$ is an operator of multiplying by $a(x)$). Mehler--Fock is an integral tranform which diagonalizes the operator with kernel $1/(x+y)$ on $[1,\infty)$. en.wikipedia.org/wiki/Mehler–Fock_transform Sep 4, 2021 at 20:53
• By the way, $I_{2n+1}$ also has a similar formula, since $\int (1-t^2)^kdt$ is pretty calculable for half-integer $k$ aswell (all previous part is literally the same). Sep 5, 2021 at 9:50