This question grew out of this one: Given an even integer $n\in 2\mathbb{N}$, compute the integral $$\int_0^1\mathrm{d} u_1 \cdots \int_0^1 \mathrm{d} u_n \frac{\delta(1-u_1-\cdots-u_n)}{(u_1+u_2)(u_2+u_3)\cdots(u_{n-1}+u_n)(u_n+u_1)}, $$ where $\delta$ is the Dirac distribution, i.e., this is a $n-1$ dimensional integral, integrating the cyclic integrand over one side of a $n$-dimensional simplex.

For $n=4$, Mathematica can do this, and \begin{multline} \int_0^1\mathrm{d} u_1 \int_0^1\mathrm{d} u_2 \int_0^1\mathrm{d} u_3 \int_0^1 \mathrm{d} u_4 \frac{\delta(1-u_1-\cdots-u_4)}{(u_1+u_2)(u_2+u_3)(u_3+u_4)(u_4+u_1)} \\ = \int_0^1 \int_0^{1-u_1}\int_0^{1-u_1-u_2} \frac{\mathrm{d} u_3\, \mathrm{d} u_2\, \mathrm{d} u_1}{(u_1+u_2)(u_2+u_3)(1-u_1-u_2)(1-u_2-u_3)} \\ = \frac{2}{3}\pi^2. \end{multline} For $n=6$, the value of the integral is numerically $\approx 51.95$, which may or may not be $\frac{8}{15}\pi^4$. I am interested in the general case, but lack the skills to find the answer.

Note: An even harder integral is the $n$ dimensional version of this Dirichlet-like integral.

  • $\begingroup$ I conjecture that for $n = 2m$ the value of the integral is $(2\pi)^{2m-2}\frac{((m-1)!)^2}{(2m-1)!}$. I have no idea how to prove this. $\endgroup$
    – heiner
    May 5, 2013 at 17:49
  • $\begingroup$ I'm sorry but in my opinion the question formulation is not quite correct. I do not really understand in which sense you use the delta-function? Is it a measure? If it is, how do you define it? I tried to consider it as a meausure but then your formula is incorrect, as the two-dimensional case gives us the measure of a diagonal of a square $\left[ 0, 1 \right]^2$, so it equals $\sqrt{2}$ not $1$, as it is according to your conjection. $\endgroup$
    – Vardashka
    Oct 31, 2013 at 19:48
  • $\begingroup$ The Dirac has to be understood in the following way: Let $\Omega\subseteq\mathbb{R}^n$ be open and $g:\Omega\to\mathbb{R}$ be continuously differentiable with $\mathrm{grad} g(x)\ne 0$ for all $x\in\Omega$. Let $M_r := \{x\in\Omega : g(x)=r\}$ be the level sets of $\Omega$ with respect to $g$. Then $\int_\Omega dx f(x) = \int_\mathbb{R} dr\int_{M_r} dS(\xi)f(\xi) \frac{1}{\lvert \mathrm{grad} g(\xi)\rvert}$ (a cruved version of Fubini; $S$ is the surface measure). Define $\int_\Omega d\xi \delta(r - g(\xi)) f(\xi) := \int_{M_r} \frac{dS(\xi)}{\lvert \mathrm{grad}g(\xi)\rvert}f(\xi).$ $\endgroup$
    – heiner
    Nov 20, 2013 at 10:42
  • 2
    $\begingroup$ A solution can be found at MathOverflow, or see below. mathoverflow.net/questions/129955/… $\endgroup$
    – dbkaplan
    Mar 7, 2014 at 18:46

2 Answers 2


This solution also appears on MathOverflow.

We can think of $I_{n}$ as being a classical partition function for $n$ beads on a circle which cannot pass through each other, with logarithmic interaction potential between each bead and its next-to-nearest neighbors on either side. For $I_{2n}$ the beads fall into two ``colors" which do not have logarithmic interactions with each other; while for $I_{2n+1}$ the beads do not fall into two independent groups.

We make two changes of variable. First, we can label the coordinates of the $k^{th}$ bead as $y_k$, where $y_1=0$ is fixed (exploiting the translation invariance of the problem) and we define $y_{2n+k} = 1+y_k$ (because of the periodic nature of the circle): $$ u_i = y_{i+1}-y_i\ ,\qquad y_1=0\ ,\qquad y_{2n+i}\equiv 1+y_i \ . $$ Then the integral can be written as a path ordered expression without the delta function constraint as $$ I_{n}= \int_0^1 dy_{n} \int_0^{y_{n}} dy_{n-1}\cdots\int_0^{y_3} dy_2\, \prod_{k=1}^{2n}\frac{1}{y_{k+2}-y_k}\ . $$ The second change of variables to $\{y_2,\ldots,y_n\}\to \{s_2,\ldots,s_n\}$ in order change the integration domain to a unit hypercube: $$ y_{k} =\prod_{j=k}^{n} s_{j}\ , $$ with Jacobian $$ J_n = \prod_{j=3}^{n} s_j^{j-2}\ . $$ With this change of variables, $I_{2n}$ becomes (for $n\ge 2$) $$ I_{2n} = \int_0^1 d^{2n-1}{\bf s}\, \prod_{j=2}^{2n-1} \, \frac{1}{1-s_j s_{j+1}} \frac{1}{1-s_{2n}{\bf S}_{2n+1}}\equiv \int_0^1d^{2n-1}{\bf s}\, {\cal F}_{2n}({\bf s})\ $$ where $d^{2n-1}{\bf s}=ds_2\cdots ds_{2n}$, and the integral sign indicates that each of the $s$ variables is being integrated from zero to one, and we have defined $$ {\bf S}_{k}\equiv 1+(-1)^k\prod_{j=2}^{k-2} s_j\ ,\qquad {\cal F}_{2n}({\bf s}) = \prod_{j=2}^{2n-1} \, \frac{1}{1-s_j s_{j+1}} \frac{1}{1-s_{2n}{\bf S}_{2n+1}}\ , $$ with $$ S_3=0\ ,\qquad {\cal F}_{2}({\bf s}) =1\ . $$ Note that for odd $k$, ${\bf S}_k<1$, while for even $k$, ${\bf S}_k>1$. This object ${\bf S}_k$ has the property for any $k$ $$ {\bf S}_{k+1} -s_{k-1} = 1-s_{k-1} {\bf S}_k\ . $$

The strategy is to consider developing a recursion relation when integrating over $ds_{2n}$ and $ds_{2n-1}$, relating $I_{2n}$ to $I_{2n-2}$. To that end it is useful to define the following functions of $x$, $y$ in the domain $ 0<x<1,\ 0<y<1$: $$ {\cal P}_k(x,y) = \frac{1}{(2k)!} \prod_{i=1}^k \left(\pi^2 (2k-1)^2 + \ln^2\left[\frac{1-x}{x(1-y)}\right]\right)\ ,\qquad {\cal P}_0(x,y)\equiv 1\ , $$ and $$ {\cal G}(\alpha,x,y) = \sum_{n=0}^\infty \left(\frac{\alpha}{\pi}\right)^{2n} {\cal P}_n(x,y) = \frac{1}{2 \sqrt{1-\alpha ^2}}\left[\left(\frac{1-x}{x(1-y)}\right)^{c}+\left(\frac{1-x}{x(1-y) }\right)^{-c}\,\right]\ , $$ $$ c\equiv \frac{\sin ^{-1}(\alpha )}{\pi }\ . $$ We generalize the problem to considering the integral $$ {\cal I}_{2n}(\alpha) = \int_0^1d^{2n-1}{\bf s}\, {\cal F}_{2n}({\bf s})\,{\cal G}(\alpha,s_{2n},{\bf S}_{2n+1})\ . $$ We can perform the $s_{2n}$ and $s_{2n-1}$ integrals in ${\cal I}_{2n}$ using the results (using the properties of ${\bf S_k}$ above)

  1. For $0<s_{2n-1}<1$ and $0<{\bf S}_{2n+1}<1$: $$ \frac{1}{2} \int_0^1 ds_{2n} \frac{1}{(1-s_{2n-1} s_{2n})(1-s_{2n}{\bf S}_{2n+1})}\left[ \left(\frac{1-s_{2n}}{s_{2n}(1-{\bf S}_{2n+1})}\right)^c+ \left(\frac{1-s_{2n}}{s_{2n}(1-{\bf S}_{2n+1})}\right)^{-c}\right] = \frac{\pi \csc (\pi c) \left(\left(\frac{1-s_{2n-1}}{1-{\bf S}_{2n+1}}\right)^{-c}-\left(\frac{1-s_{2n-1}}{1-{\bf S}_{2n+1}}\right)^c\right)}{2(s_{2n-1}-{\bf S}_{2n+1})} =\frac{\pi \csc (\pi c)\left(\left(\frac{1-s_{2n-1}}{s_{2n-1}({\bf S}_{2n}-1)}\right)^{-c}-\left(\frac{1-s_{2n-1}}{s_{2n-1}({\bf S}_{2n}-1)}\right)^c\right)}{2(1-s_{2n-1}{\bf S}_{2n})}\ $$

  2. For $0<s_{2n-2}<1$ and $1<{\bf S}_{2n}$: $$\frac{1}{2} \int_0^1 ds_{2n-1} \frac{1}{(1-s_{2n-2}s_{2n-1})(1-{\bf S}_{2n}s_{2n-1})} \qquad\times \left[ \left(\frac{1-s_{2n-1}}{s_{2n-1}({\bf S}_{2n}-1)}\right)^c- \left(\frac{1-s_{2n-1}}{s_{2n-1}({\bf S}_{2n}-1)}\right)^{-c}\right] = -\frac{\pi \csc (\pi c) \left(\left(\frac{1-s_{2n-2}}{{\bf S}_{2n}-1}\right)^{-c}+\left(\frac{1-s_{2n-2}}{{\bf S}_{2n}-1}\right)^c-2 \cos (\pi c)\right)}{2 (s_{2n-2}-{\bf S}_{2n})} = -\frac{\pi \csc (\pi c) \left(\left(\frac{1-s_{2n-2}}{1-s_{2n-2}{\bf S}_{2n-1}}\right)^{-c}+\left(\frac{1-s_{2n-2}}{1-s_{2n-2}{\bf S}_{2n-1}}\right)^c-2 \cos (\pi c)\right)}{2 (1-s_{2n-2}{\bf S}_{2n-1})} $$

With these integrals we can perform the integrations over $s_{2n}$ and $s_{2n-1}$ in our generalized integral ${\cal I}_{2n}(\alpha)$, obtaining
$$ {\cal I}_{2n}(\alpha) =\int d^{2n-3}{\bf s} \, {\cal F}_{2n-2}({\bf s})\, \left[\pi^2\csc^2(c\pi)\left({\cal G}(\alpha,s_{2n-2},{\bf S}_{2n-1}) -\frac{ \cos c\pi}{ \sqrt{1-\alpha^2}}\right) \right] \, =\int d^{2n-3}{\bf s}{\cal F}_{2n-2}({\bf s})\, \left[\frac{\pi^2}{\alpha^2}\left({\cal G}(\alpha,s_{2n-2},{\bf S}_{2n-1})-1\right) \right] $$ where to get the second line we just plugged in $\pi c=\sin^{-1}\alpha$. Referring to the definition of ${\cal G}$, we can equate powers of $\alpha$ on both sides of the above equation with the result that for every $k\ge 0$, $$ \int_0^1d^{2n-1}{\bf s}\, {\cal F}_{2n}({\bf s})\, {\cal P}_k(s_{2n},{\bf S}_{2n+1}) = \int_0^1d^{2n-3}{\bf s}\, {\cal F}_{2n-2}({\bf s})\, {\cal P}_{k+1}(s_{2n-2},{\bf S}_{2n-1}) $$ which is a pretty result.

The above result allows us to write for the desired $2n$-dimensional integrals as one-dimensional integrals $$ I_{2n}=\int_0^1d^{2n-1}{\bf s}\, {\cal F}_{2n}({\bf s})\, {\cal P}_0(s_{2n},{\bf S}_{2n+1}) =\int_0^1ds_2 {\cal P}_{n-1}(s_{2},0)\ . $$ The above results then imply that $$ \sum_{n=0}^\infty \left(\frac{\alpha}{\pi}\right)^{2n} I_{2n+2} = \int_0^1dx\,\sum_{n=0}^\infty \left(\frac{\alpha}{\pi}\right)^{2n} {\cal P}_n(x,0) = \int_0^1dx\, {\cal G}(\alpha,x,0) =\int_0^1dx\frac{1}{2 \sqrt{1-\alpha ^2}}\left[\left(\frac{1-x}{x}\right)^{c}+\left(\frac{1-x}{x }\right)^{-c}\right] = \frac{\sin^{-1}\alpha}{\alpha\sqrt{1-\alpha^2}} = \sum_{n=0}^\infty (2\alpha)^{2n} B(n+1,n+1)\ . $$ Equating powers of $\alpha$ between the first and last expressions answers the posted question: $$ I_{2n+2} = (2\pi)^{2n}\mathrm{B}(n+1,n+1){=} (2\pi)^{2n} \frac{(n!)^2}{(2n+1)!}\ . $$

This solution was found in collaboration with E. Mereghetti.

  • $\begingroup$ Thanks a lot, very impressive! A colleague recently showed my another solution that uses the spectral representation of the Hilbert matrix and its integral kernel. Yours is in some sense more straightforward, though. $\endgroup$
    – heiner
    Mar 10, 2014 at 14:46
  • 1
    $\begingroup$ I'd like to see that other solution...I was attracted to this problem in part because it looked similar to, but simpler than a problem I encountered in my physics, but I was disappointed to find that the solution of this problem didn't help with the other one. $\endgroup$
    – dbkaplan
    Mar 10, 2014 at 18:54

Here is another proof, the main part of which was communicated to me by Dr. Peter Otte of Bochum University: \begin{equation} I_n := \int_{[0,1]^n}\mathrm{d}u\,\delta(1-\lvert u\rvert_1) \frac{1}{\prod_{j=1}^n (u_j + u_{j+1})} = (2\pi)^{n-2} \frac{[\Gamma(\frac{n}{2})]^2}{\Gamma(n)}. \end{equation}

First, define $$J_n(t) := \int_{[0,1]^n}\mathrm{d}u\,\delta(t-\lvert u\rvert_1) \frac{1}{\prod_{j=1}^{n-1}(u_j + u_{j+1})}.$$ for $t>0$. By scaling, $J_n(t) = J_n(1) =: J_n$ for all $t > 0$. Also, \begin{align} I_n & = \frac{1}{2}\int_{[0,1]^n}\mathrm{d}u\, \delta(1-\lvert u\rvert_1) \frac{2\lvert u\rvert_1}{\prod_{j=1}^n (u_j + u_{j+1})} \notag\\ & = \frac{1}{2}\sum_{k=1}^n \int_{[0,1]^n}\mathrm{d}u\, \delta(1-\lvert u\rvert_1) \frac{u_k+u_{k+1}}{\prod_{j=1}^n (u_j + u_{j+1})} \notag\\ & = \frac{n}{2} \int_{[0,1]^n}\mathrm{d}u\, \frac{\delta(1-\lvert u\rvert_1)}{(u_1+u_2)\dotsm(u_{n-1}+u_n)} = \frac{n}{2} J_n. \end{align} Next, let $f\in L_1(0,\infty)$. Then \begin{equation} J_n = \int_{(0,\infty)^n}\mathrm{d}u\, \frac{f(\lvert u\rvert_1)}{\prod_{j=1}^{n-1}(u_j + u_{j+1})} \Bigm/\! \int_0^\infty\mathrm{d}t\, f(t). \end{equation} In particular, \begin{equation} J_n = \int_{(0,\infty)^n}\mathrm{d}u\, \frac{e^{-\lvert u\rvert_1}}{\prod_{j=1}^{n-1}(u_j + u_{j+1})}, \end{equation} We will need the Rosenblum-Rovnyak integral operator $T: L_2(0,\infty)\to L_2(0,\infty)$, see Rosenblum (1958) and Rovnyak (1970), defined via \begin{equation} (Tf)(x) := \int_0^\infty \mathrm{d}y\, \frac{e^{-(x+y)/2}}{x+y} f(y) \quad (x\in(0,\infty)). \end{equation} for $f\in L_2(0,\infty)$. This is the special case $T = \mathcal{H}_0$ in Rosenblum (1958), Formula (2.3). The operator $T$ is unitary equivalent to the Hilbert matrix $H:\ell_2(\mathbb{N})\to\ell_2(\mathbb{N})$, \begin{equation} (H x)_j = \sum_{k=1}^\infty \frac{x_k}{j+k-1} \quad(j\in\mathbb{N}, x\in\ell_2(\mathbb{N})) \end{equation} and can be explicitly diagonalized: Following Yafaev (2010), Sec. 4.2, we define the unitary operator $U: L_2(0,\infty)\to L_2(0,\infty)$ via \begin{equation} (Uf)(k) = \pi^{-1}\sqrt{k\sinh 2\pi k} \, \lvert \Gamma(1/2 - ik)\rvert \int_0^\infty\mathrm{d}x\, x^{-1} W_{0,ik}(x)f(x) \end{equation} for $f\in L_2(0,\infty)$ and $k\in(0,\infty)$, where the Whittaker functions are given by \begin{equation} W_{0,\nu}(x) = \sqrt{x/\pi} K_\nu(x/2) \quad (\nu, x\in(0,\infty)), \end{equation} with $K_\nu$ as the modified Bessel function of the second kind, see DLMF.

In order to compute $J_n$, we will employ the following result due to Rosenblum, see Yafaev, Prop. 4.1: \begin{equation} (UTf)(k) = \frac{\pi}{\cosh(k\pi)}(Uf)(k) \quad (k\in(0,\infty), f\in L_2(0,\infty). \end{equation}

Proof of $I_n = (2\pi)^{n-2} \frac{[\Gamma(\frac{n}{2})]^2}{\Gamma(n)}$. Let $n\in\mathbb{N}_{\ge 2}$. From the definition of $T$ and the identity of $J_n$ above, we see that \begin{equation} J_n = \langle f_0, T^{n-1}f_0\rangle \end{equation} with $f_0(x) := e^{-x/2}$. From this and the identity of $UT$ above, we obtain \begin{equation} J_n = \langle Uf_0, UT^{n-1}f_0\rangle = \int_0^\infty\mathrm{d}k\, \lvert \hat{f}_0(k)\rvert^2 \Bigl(\frac{\pi}{\cosh(k\pi)}\Bigr)^{n-1}, \end{equation} where $\hat{f}_0 := Uf_0$. In order to compute $\hat{f}_0$, we employ the classical formula \begin{equation} \lvert\Gamma(1/2 - ik)\rvert^2 = \frac{\pi}{\cosh(k\pi)} \quad (k\in\mathbb{R}), \end{equation} which is a consequence of the reflection formula for the Gamma function, and \begin{equation} \int_0^\infty\mathrm{d}x\, x^{-1} W_{0,ik}(x)e^{-x/2} = \frac{\pi}{\cosh(k\pi)} \quad(k > 0), \end{equation} which follows from the special case $z=1/2$ and $\nu = \kappa = 0$ in DLMF. From the definition of $U$ above and the last two equations, we deduce \begin{equation} \lvert\hat{f}_0(k)\rvert^2 = 2\pi k\frac{\sinh(k\pi)}{\cosh(k\pi)^2} \quad (k > 0). \end{equation} This yields \begin{equation} J_n = 2\pi^{n-2}\int_0^\infty\mathrm{d}k\, k \frac{\sinh(k)}{\cosh(k)^{n+1}} = \frac{2\pi^{n-2}}{n}\int_0^\infty\mathrm{d}k\,\frac{1}{\cosh(k)^n} \end{equation} where we applied the substitution $\tilde{k} = k\pi$ and integrated by parts. This integral can be evaluated using the substitutions $y = \cosh(k)^{-1}$ and $x = y^2$, one after the other: \begin{align} J_n = \frac{2\pi^{n-2}}{n} \int_0^1\mathrm{d}y\, \frac{y^{n-1}}{\sqrt{1-y^2}} & = \frac{\pi^{n-2}}{n} \int_0^1\mathrm{d}x\, x^{n/2-1}(1-x)^{-1/2} \\ & = \frac{\pi^{n-2}}{n} \mathrm{B}(n/2, 1/2), \end{align} since $k'(y) = - y^{-1}(1-y^2)^{-1/2}$. The claim then follows by expressing the Beta function via the Gamma function and then applying the classical duplication formula.


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