I asked the same question on math.se but got no answer there. Since it pertains to my current research, I decided to ask here:

Let $n\in 2\mathbb{N}$ be an even number. I want to evaluate $$I_n := \int_0^1\mathrm{d} u_1 \cdots \int_0^1 \mathrm{d} u_n \frac{\delta(1-u_1-\cdots-u_n)}{(u_1+u_2)(u_2+u_3)\cdots(u_{n-1}+u_n)(u_n+u_1)}. $$ For small $n$, this is computable by simply parameterizing the $\delta$ function, and $I_2 = 1$, $I_3 = \pi^2/4$, $I_4 = 2\pi^2/3$. The values of $I_5$ and $I_6$ are numerically $18.2642 \approx 3\pi^4/16$ and $51.9325\approx 8\pi^4/15$. I strongly suspect that $$ I_{2n+2} \stackrel{?}{=} (2\pi)^{2n} \frac{(n!)^2}{(2n+1)!} = \frac{(2\pi)^{2n}}{\binom{2n+1}{n}(n+1)} = (2\pi)^{2n}\mathrm{B}(n+1,n+1), $$ where $\mathrm{B}$ is the Beta function. Dividing by $(2\pi)^{2n}$, this is Sloane's A002457. For $I_6$, this conjecture is equivalent to $$\int_0^1\mathrm{d}x \Bigl(\mathrm{Li}_2(\frac{x-1}{x})\Bigr)^2 \stackrel{?}{=} \frac{17}{180}\pi^4$$ (with $\mathrm{Li}_2$ the dilogarithm), which seems to be true numerically, but I could neither prove it nor find it in the literature.

As a last remark, it is possible to get rid of the $\delta$ function by using the identity $$I_n = \int_{(0,\infty)^n}\mathrm{d}u \frac{f(\lvert u\rvert_1)}{(u_1+u_2)\cdots(u_n+u_1)} \Bigm/\int_0^\infty\mathrm{d}t \frac{f(t)}{t}$$ for any $f:(0,\infty)\to\mathbb{R}$ that makes both integrals finite. Using $f(t) = t 1_{[0,1]}(t)$ where $1_{[0,1]}$ is the characteristic function of the interval $[0,1]$, one can write $I_n$ as an integral over an $n$-dimensional simplex.


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

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 ${\cal G}$ in eq.(\ref{gdef}), 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.

This solution was found in collaboration with E. Mereghetti (we're physicists, so the language might look odd).


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.


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.