# Integrate $\int_0^1\mathrm{d} u_1 \cdots \int_0^1\mathrm{d} u_n \frac{\delta(1-u_1-\cdots-u_n)}{(u_1+u_2)(u_2+u_3)\cdots(u_{n-1}+u_n)(u_n+u_1)}$

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.

• 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. May 5, 2013 at 17:49
• 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. Oct 31, 2013 at 19:48
• 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 $\int_\Omega d\xi \delta(r - g(\xi)) f(\xi) := \int_{M_r} \frac{dS(\xi)}{\lvert \mathrm{grad}g(\xi)\rvert}f(\xi).$ Nov 20, 2013 at 10:42
• A solution can be found at MathOverflow, or see below. mathoverflow.net/questions/129955/… Mar 7, 2014 at 18:46

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.

• 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. Mar 10, 2014 at 14:46
• 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. 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: $$I_n := \int_{[0,1]^n}\mathrm{d}u\,\delta(1-\lvert u\rvert_1) \frac{1}{\prod_{j=1}^n (u_j + u_{j+1})} = (2\pi)^{n-2} \frac{[\Gamma(\frac{n}{2})]^2}{\Gamma(n)}.$$

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 $$J_n = \int_{(0,\infty)^n}\mathrm{d}u\, \frac{f(\lvert u\rvert_1)}{\prod_{j=1}^{n-1}(u_j + u_{j+1})} \Bigm/\! \int_0^\infty\mathrm{d}t\, f(t).$$ In particular, $$J_n = \int_{(0,\infty)^n}\mathrm{d}u\, \frac{e^{-\lvert u\rvert_1}}{\prod_{j=1}^{n-1}(u_j + u_{j+1})},$$ 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 $$(Tf)(x) := \int_0^\infty \mathrm{d}y\, \frac{e^{-(x+y)/2}}{x+y} f(y) \quad (x\in(0,\infty)).$$ 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 $H:\ell_2(\mathbb{N})\to\ell_2(\mathbb{N})$, $$(H x)_j = \sum_{k=1}^\infty \frac{x_k}{j+k-1} \quad(j\in\mathbb{N}, x\in\ell_2(\mathbb{N}))$$ 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 $$(Uf)(k) = \pi^{-1}\sqrt{k\sinh 2\pi k} \, \lvert \Gamma(1/2 - ik)\rvert \int_0^\infty\mathrm{d}x\, x^{-1} W_{0,ik}(x)f(x)$$ for $f\in L_2(0,\infty)$ and $k\in(0,\infty)$, where the Whittaker functions are given by $$W_{0,\nu}(x) = \sqrt{x/\pi} K_\nu(x/2) \quad (\nu, x\in(0,\infty)),$$ 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: $$(UTf)(k) = \frac{\pi}{\cosh(k\pi)}(Uf)(k) \quad (k\in(0,\infty), f\in L_2(0,\infty).$$

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 $$J_n = \langle f_0, T^{n-1}f_0\rangle$$ with $f_0(x) := e^{-x/2}$. From this and the identity of $UT$ above, we obtain $$J_n = \langle Uf_0, UT^{n-1}f_0\rangle = \int_0^\infty\mathrm{d}k\, \lvert \hat{f}_0(k)\rvert^2 \Bigl(\frac{\pi}{\cosh(k\pi)}\Bigr)^{n-1},$$ where $\hat{f}_0 := Uf_0$. In order to compute $\hat{f}_0$, we employ the classical formula $$\lvert\Gamma(1/2 - ik)\rvert^2 = \frac{\pi}{\cosh(k\pi)} \quad (k\in\mathbb{R}),$$ which is a consequence of the reflection formula for the Gamma function, and $$\int_0^\infty\mathrm{d}x\, x^{-1} W_{0,ik}(x)e^{-x/2} = \frac{\pi}{\cosh(k\pi)} \quad(k > 0),$$ 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 $$\lvert\hat{f}_0(k)\rvert^2 = 2\pi k\frac{\sinh(k\pi)}{\cosh(k\pi)^2} \quad (k > 0).$$ This yields $$J_n = 2\pi^{n-2}\int_0^\infty\mathrm{d}k\, k \frac{\sinh(k)}{\cosh(k)^{n+1}} = \frac{2\pi^{n-2}}{n}\int_0^\infty\mathrm{d}k\,\frac{1}{\cosh(k)^n}$$ 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.