Asymptotics of $\int_1^L\int_1^L\int_1^L\int_1^L \frac{\mathrm dx_1~\mathrm dx_2 ~ \mathrm dx_3 ~ \mathrm dx_4}{(x_1+x_2)(x_2+x_3)(x_3+x_4)(x_4+x_1)}$ Let $L>1$. I am looking for the value, or the leading asymptotics for $L\to\infty$, of
$$\int_1^L\int_1^L\int_1^L\int_1^L \dfrac{\mathrm dx_1~\mathrm dx_2 ~ \mathrm  dx_3 ~ \mathrm dx_4}{(x_1+x_2)(x_2+x_3)(x_3+x_4)(x_4+x_1)}$$
More generally, I'd like to know the leading asymptotics of an expression like this with $2n$ terms, where the above has $2n=4$.
 A: The general asymptotics as $L \to\infty$ of
$$
I_m(L) = \int_1^L \cdots \int_1^L \frac {dx_1 \cdots dx_{m}} {(x_1+x_2)(x_2+x_3) \cdots (x_{m} +x_1)}
$$
for $m\in\mathbb N$ can be computed in the following way:

*

*Show that
$$
I_m(L) - \int_1^L \int_1^\infty \cdots \int_1^\infty \frac {dx_1 \cdots dx_{m}} {(x_1+x_2)(x_2+x_3) \cdots (x_{m} +x_1)} = O(1)
$$
as $L\to\infty$, i.e. you perform all integrals up to $\infty$ except one integral.


*You write the integral as the trace of an operator
$$
\int_1^L \int_1^\infty \cdots \int_1^\infty \frac {dx_1 \cdots dx_{m}} {(x_1+x_2)(x_2+x_3) \cdots (x_{m} +x_1)}
=
\operatorname{tr} 1_{[1,L]} K^m 1_{[1,L]}
$$
where $K$ is the Hankel operator
$$
K:L^2([0,\infty)) \to L^2([0,\infty))
$$
with kernel $K(x,y) = \frac 1 {x+y}$, i.e. $(Kf)(x) = \int_0^\infty \frac1{x+y}f(y)$ where $\operatorname{tr}$ is the trace and $1_{[1,L]}$ is the orthogonal projection onto $L^2([1,L])$. The operator $K$ is also called Carleman operator.


*The important point why this helps is that $K$ can be explicitly diagonalized using the Mellin transform, i.e.
$$
K \psi_k = f(k) \psi_k
$$
with $(\psi_k)(x) = \frac 1 {\sqrt{2\pi}}x^{-1/2 + ik}$ and $f(k) = \frac \pi{\cosh(\pi k)}$ and $k\in\mathbb R$. Noting that the $\psi_k$ define a unitary transform (Mellin transform) one can write
$$
K = \int_{\mathbb R} f(k) |\psi_k\rangle\langle\psi_k| dk
$$
and $\langle\psi_k,\psi_m\rangle =\delta_{km}$.


*Inserting this in $\operatorname{tr} 1_{[1,L]} K^m 1_{[1,L]}$ we obtain
$$
\operatorname{tr} 1_{[1,L]} K^m 1_{[1,L]}
= \int_1^L d x \int_{\mathbb R} d k f(k)^m |\psi_k(x)|^2
$$
Now $|\psi_k(x)|^2 = \frac 1 {2\pi x}$ independently of $k$ and integrating over $x$ gives
$$
\int_1^L d x \int_{\mathbb R} d k f(k)^m |\psi_k(x)|^2 = \frac{\log L}{2\pi}  \int_{\mathbb R} d k \big(\frac \pi{\cosh(\pi x)}\big)^m.
$$
This implies that
$$
I_m(L) \sim \frac{\log L}{2\pi}  \int_{\mathbb R} d k \big(\frac \pi{\cosh(\pi x)}\big)^m.
$$
For $m=4$ computing the integral gives
$$
I_4(L) \sim \log L \frac 2 3 \pi^2
$$
which is consistent with the previous answer.
