I asked the [same question on math.se](https://math.stackexchange.com/questions/334962/int-01-mathrmd-u-1-cdots-int-01-mathrmd-u-n-frac-delta1-u-1-cdots) 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](http://oeis.org/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.



Evaluation of an $n$-dimensional integral
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).


