19
$\begingroup$

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 $n$.

$\endgroup$
9
  • 1
    $\begingroup$ Are you missing a factor $\frac{1}{u_1+u_2}$ in the integrand? $\endgroup$ Sep 1 '21 at 19:56
  • $\begingroup$ yes, editing it now $\endgroup$
    – Joe
    Sep 1 '21 at 20:01
  • 1
    $\begingroup$ 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. $\endgroup$
    – Joe
    Sep 2 '21 at 14:46
  • 2
    $\begingroup$ I am reminded of a hypercube integral from arXiv:1608.03174 that equals a zeta function for any (odd or even) integer: $$\zeta(k)=\int_0^1 \frac{dx_1}{x_1} \int_0^{x_1}\frac{dx_2}{x_2} \cdots \int_0^{x_{k-2}}\frac{dx_{k-1}}{x_{k-1}} \int_0^{x_{k-1}} \frac{dx_k}{1-x_k}$$ $\endgroup$ Sep 2 '21 at 20:52
  • 5
    $\begingroup$ 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. $\endgroup$
    – Terry Tao
    Sep 4 '21 at 3:54
11
+200
$\begingroup$

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.)

$\endgroup$
6
  • 1
    $\begingroup$ 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 $0$.) $\endgroup$
    – Terry Tao
    Sep 4 '21 at 18:31
  • $\begingroup$ Wow this is beautiful! One can note that these fit quite neatly into the generating functional $\sum_{n=0}^{\infty} I_{2n} (\frac{\alpha}{\pi})^{2n+2} = \frac{1}{2 \pi^2}\mathrm{ArcSin}({\alpha})^2$. $\endgroup$
    – Joe
    Sep 4 '21 at 20:12
  • 1
    $\begingroup$ Certainly beautiful, but can someone explain to non experts what the first few lines of the proof mean ? In addition, I do not have access to the given papers. What is Mehler-Fock, etc... $\endgroup$ Sep 4 '21 at 20:40
  • $\begingroup$ 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 $\endgroup$ Sep 4 '21 at 20:53
  • 2
    $\begingroup$ 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). $\endgroup$ Sep 5 '21 at 9:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.