knowledge of the leading and first subleading terms in the mutual information
long-distance expansion for spherical regions.

On a different front, it would be interesting to provide further evidence (or
disprove) the two conjectures we have put forward in this paper involving free
fields, namely: i) the fact that the free scalar $N$-partite information is
positive-definite in general dimensions and for arbitrary regions; ii) the
observation that, normalized by the disk EE universal coefficient, the mutual
information of the three-dimensional free scalar is greater than the free
fermion one for arbitrary regions. In relation with the latter, it would be
interesting to explore the possible existence of similar hierarchies between
theories beyond free fields and also beyond the mutual information case, i.e.,
for $N\geq 3$.

In the context of holographic theories, it would be interesting to extend the
arguments leading to the equality between boundary and bulk twist operators to
general shapes and arbitrary distances. This could help us to understand
further these mysterious objects, although, there is no guarantee that such
equality holds in those cases. The former case might require further input on
the bulk modular flow which might be possible in the bulk free field
approximation. On the other hand, moving away from the long-distance regime
for boundary spheres may be possible, given the universality of both, bulk and
boundary modular flows. This second exploration is of special interest, since
understanding the behaviour of the $N-$partite information at moderate
separation distances could elucidate the nature of the holographic phase
transition of the Ryu-Takayanagi prescription.

## Acknowledgements

We thank Horacio Casini for useful discussions. The work of AVL is supported
by the F.R.S.-FNRS Belgium through convention IISN 4.4514.08. The work of CA
is supported by the Simons Foundation through It from Qubit: Simons
Collaboration on Quantum Fields, Gravity, and Information.

##  Appendix A Coefficients for the $N$-partite information

Throughout the paper, we are interested in computing sums of combinations of
the following basic coefficient,

$\displaystyle
C_{ij}=\frac{1}{\sin^{2\Delta}\left[\frac{\pi(i-j)}{n}\right]}\,.$ (166)

Following [29, 28], where similar sums over $C_{ij}$ were computed, we will
make extensive use of the following representation for the sine function:

$\displaystyle\frac{1}{\sin^{\Delta}\left(\frac{\theta}{2}\right)}$
$\displaystyle=$
$\displaystyle\frac{2^{\Delta-2}}{\pi^{2}\Gamma\left(\Delta\right)}\int_{-\infty}^{\infty}dq\,e^{-\frac{q}{2}}e^{\frac{q\,\theta}{2\pi}}\Gamma\left(\frac{\Delta}{2}+i\frac{q}{2\pi}\right)\Gamma\left(\frac{\Delta}{2}-i\frac{q}{2\pi}\right)$
(167) $\displaystyle=$
$\displaystyle\frac{2^{\Delta-2}}{\pi^{2}\Gamma\left(\Delta\right)}\int_{-\infty}^{\infty}dq\,e^{-\frac{q}{2}}e^{\frac{q\,\theta}{2\pi}}\,\Big{|}\Gamma\left(\frac{\Delta}{2}+i\frac{q}{2\pi}\right)\Big{|}^{2}\,.$

This formula can be written in a more compact form in terms of the beta
function $B(x,y)$ due to its relation with the gamma function,

$\displaystyle
B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}=\int_{0}^{1}t^{x-1}(1-t)^{y-1}dt\,,$
(168)

where in the last equality we present its integral definition. Thus, we can
rewrite eq. (167) as

$\displaystyle\frac{1}{\sin^{\Delta}\left(\frac{\theta}{2}\right)}=\frac{2^{\Delta-2}}{\pi^{2}}\int_{-\infty}^{\infty}dq\,e^{-\frac{q}{2}}e^{\frac{q\,\theta}{2\pi}}B\left(\frac{\Delta}{2}+i\frac{q}{2\pi}\,,\,\frac{\Delta}{2}-i\frac{q}{2\pi}\right)\,.$
(169)

For the sake of simplicity, we introduce the following compact notation

$\displaystyle B_{q}(\Delta)\equiv
B\left(\Delta+i\frac{q}{2\pi},\Delta-i\frac{q}{2\pi}\right)=\frac{\Big{|}\Gamma\left(\Delta+i\frac{q}{2\pi}\right)\Big{|}^{2}}{\Gamma\left(2\Delta\right)}\,,$
(170)

to represent the beta function of our interest. In terms of this function we
have

$\displaystyle\frac{1}{\sin^{\Delta}\left(\frac{\theta}{2}\right)}=\frac{2^{\Delta-2}}{\pi^{2}}\int_{-\infty}^{\infty}dq\,e^{-\frac{q}{2}}e^{\frac{q\,\theta}{2\pi}}B_{q}\left(\frac{\Delta}{2}\right)\,,$
(171)

where the integral representation of the beta function (170) from eq. (168)
takes the simple form

$\displaystyle
B_{q}(\Delta)=\int_{0}^{1}dt\,t^{\Delta-1}(1-t)^{\Delta-1}e^{\frac{iq}{2\pi}\log\left(\frac{t}{1-t}\right)}\,.$
(172)

We will make use of this notation in what follows.

###  A.1 Computation of $I_{2}$ coefficients

We are interested in computing the sum

$\displaystyle c_{2:2}^{(2)}\equiv\lim_{n\to
1}\frac{1}{n-1}\sum_{i<j}C^{2}_{ij}=\lim_{n\to
1}\frac{n}{2(n-1)}\sum_{j=1}^{n-1}\frac{1}{\sin^{4\Delta}\left(\frac{\pi
j}{n}\right)}\,.$ (173)

For that purpose, we use the representation (171) with $\theta\to 2\pi j/n$ in
the above sum, which leads to

$\displaystyle c_{2:2}^{(2)}=\lim_{n\to
1}\frac{1}{n-1}\sum_{j=1}^{n-1}\frac{2^{4\Delta-3}}{\pi^{2}}\int_{-\infty}^{\infty}dq\,e^{-\frac{q}{2}}e^{\frac{qj}{n}}B_{q}\left(2\Delta\right)\,.$
(174)

We can carry out the sum over $j$ explicitly using

$\displaystyle\sum_{j=1}^{n-1}e^{\frac{qj}{n}}=\frac{e^{q/n}-e^{q}}{1-e^{q/n}}\approx(n-1)\frac{qe^{q}}{e^{q}-1}\,,$
(175)

which cancels out the $(n-1)$ in the $n\to 1$ limit and leads to

$\displaystyle
c_{2:2}^{(2)}=\frac{2^{4\Delta-3}}{\pi^{2}}\int_{0}^{1}dt\,t^{2\Delta-1}(1-t)^{2\Delta-1}\int_{-\infty}^{\infty}dq\,\frac{qe^{\frac{q}{2}}}{e^{q}-1}e^{\frac{iq}{2\pi}\log\left(\frac{t}{1-t}\right)}\,,$
(176)

after plugging in the integral representation (172). We can now perform the
integral over $q$, which is just the Fourier transform presented in eq. (263)
with $k\to[\log t-\log(1-t)]/2\pi$. This results in:

$\displaystyle\int_{-\infty}^{\infty}dq\,\frac{qe^{\frac{q}{2}}}{e^{q}-1}e^{\frac{iq}{2\pi}\log\left(\frac{t}{1-t}\right)}=4\pi^{2}t(1-t)\,.$
(177)

Plugging this result back into (176) leads to the well-known expression for
$c_{2:2}^{(2)}$,

$\displaystyle
c_{2:2}^{(2)}=2^{4\Delta-1}\int_{0}^{1}dt\,t^{2\Delta}(1-t)^{2\Delta}=2^{4\Delta-1}\frac{\Gamma\left(2\Delta+1\right)^{2}}{\Gamma\left(4\Delta+2\right)}\,.$
(178)

###  A.2 Computation of $I_{3}$ coefficients

We are interested in computing the coefficients $c^{(3)}_{3:2}$ and
$c^{(1,1,1)}_{3:3}$ which appear in the expression for the tripartite
information at long distances. We start with

$\displaystyle c^{(3)}_{3:2}$ $\displaystyle\equiv$ $\displaystyle\lim_{n\to
1}\frac{1}{n-1}\sum_{i<j}C^{3}_{ij}\,.$ (179)

This computation is identical to the one of the coefficient $c^{(2)}_{2:2}$
which determines $I_{2}$ at long distances with the simple replacement
$\Delta\to 3\Delta/2$, therefore, we conclude that

$\displaystyle
c^{(3)}_{3:2}=\frac{2^{6\Delta-1}\Gamma\left(3\Delta+1\right)^{2}}{\Gamma\left(6\Delta+2\right)}\,.$
(180)

Our second coefficient of interest is

$\displaystyle c^{(1,1,1)}_{3:3}$ $\displaystyle\equiv$
$\displaystyle\lim_{n\to
1}\frac{1}{n-1}\sum_{i<j<k}C_{ij}C_{jk}C_{ki}=\lim_{n\to
1}\frac{1}{3!(n-1)}\sum_{i\neq j\neq k}C_{ij}C_{jk}C_{ki}$ (181)
$\displaystyle=$ $\displaystyle\lim_{n\to
1}\frac{n}{3(n-1)}\sum_{0<j<k}C_{0j}C_{jk}C_{k0}=\lim_{n\to
1}\frac{n}{3(n-1)}\sum_{k=2}^{n-1}\sum_{j=1}^{k-1}C_{0j}C_{jk}C_{k0}\,.$

In the first line, we rewrite the ordered sum in a disordered form by
including the permutation factor that accounts for the multiple counting of
each configuration. In the second line, we fix one index to the particular
value of zero and multiply by $n$ the result (making use of the replica
symmetry). Furthermore, we reorder the resulting sum, which requires an extra
permutation factor of $2!$. Plugging in the explicit expression for the
coefficients, we get

$\displaystyle c^{(1,1,1)}_{3:3}=\lim_{n\to
1}\frac{n}{3(n-1)}\sum_{k=2}^{n-1}\frac{1}{\sin^{2\Delta}\left(\frac{\pi
k}{n}\right)}\sum_{j=1}^{k-1}\frac{1}{\sin^{2\Delta}\left(\frac{\pi(k-j)}{n}\right)}\frac{1}{\sin^{2\Delta}\left(\frac{\pi
j}{n}\right)}\,.$ (182)

Next, we use the integral representation (171) for each of the sine functions
above, leading to

$\displaystyle c^{(1,1,1)}_{3:3}=$ $\displaystyle\lim_{n\to
1}\frac{n}{3(n-1)}\left[\frac{2^{2\Delta-2}}{\pi^{2}}\right]^{3}$
$\displaystyle\times\sum_{k=2}^{n-1}\,\sum_{j=1}^{k-1}\int_{-\infty}^{\infty}dp\int_{-\infty}^{\infty}dq\int_{-\infty}^{\infty}ds\,e^{-\frac{p}{2}}\,e^{-\frac{q}{2}}e^{-\frac{s}{2}}e^{\frac{pk}{n}}e^{\frac{qj}{n}}e^{\frac{s(k-j)}{n}}B_{p}\left(\Delta\right)B_{q}\left(\Delta\right)B_{s}\left(\Delta\right)\,.$
(183)

In this form, we can explicitly carry out the sums over $j$ and $k$ and take
the $n\to 1$ limit of the resulting expression. This gives

$\displaystyle
e^{-\frac{p}{2}}e^{-\frac{q}{2}}e^{-\frac{s}{2}}\sum_{k=2}^{n-1}\sum_{j=1}^{k-1}e^{\frac{pk}{n}}e^{\frac{qj}{n}}e^{\frac{s(k-j)}{n}}\approx(n-1)f(p,q,s)\,,$
(184)

where

$\displaystyle
f(p,q,s)\equiv-e^{\frac{p+q+s}{2}}\left[\frac{p}{(e^{-p}-e^{q})(1-e^{p+s})}-\frac{q}{(e^{s}-e^{q})(1-e^{p+q})}+\frac{s}{(e^{s}-e^{q})(1-e^{p+s})}\right]\,.$
(185)

This function is to be introduced inside eq. (A.2). We can therefore
manipulate it to get a more symmetric form by changing $p\to-p$, obtaining

$\displaystyle\tilde{f}(p,q,s)=e^{\frac{p+q+s}{2}}\left[\frac{p}{(e^{p}-e^{s})(e^{p}-e^{q})}+\frac{q}{(e^{q}-e^{s})(e^{q}-e^{p})}+\frac{s}{(e^{s}-e^{q})(e^{s}-e^{p})}\right]\,.$
(186)

The resulting function $\tilde{f}(p,q,s)$ is symmetric under permutations
among $\\{p,q,s\\}$. We could deduce from here that inside the integral each
term will contribute the same and thus it should be enough to focus on
evaluating any of the three terms and multiplying the answer by three.
However, this is not true because each term individually is not well defined
due to spurious divergences at coincident points of the $\\{p,q,s\\}$
variables. Such poles would cancel in the full expression, the following form
of the function makes it manifest

$\displaystyle\tilde{f}(p,q,s)=\frac{e^{\frac{p+q+s}{2}}}{(e^{p}-e^{q})}\left[\frac{p-s}{e^{p}-e^{s}}-\frac{q-s}{e^{q}-e^{s}}\right]\,.$
(187)

Thus, the integral of interest is

$\displaystyle c^{(1,1,1)}_{3:3}$ $\displaystyle=$
$\displaystyle\frac{2^{6(\Delta-1)}}{3\pi^{6}}\int_{-\infty}^{\infty}\,dp\,e^{\frac{p}{2}}B_{p}\left(\Delta\right)\int_{-\infty}^{\infty}\,dq\,e^{\frac{q}{2}}B_{q}\left(\Delta\right)\frac{1}{e^{p}-e^{q}}$
(188)
$\displaystyle\times\int_{-\infty}^{\infty}\,ds\,e^{\frac{s}{2}}\,B_{s}\left(\Delta\right)\left(\frac{p-s}{e^{p}-e^{s}}-\frac{q-s}{e^{q}-e^{s}}\right)\,.$

We start by doing the integral over $s$ of the $p$ dependent piece; the one
that depends on $q$ is identical. We have

$\displaystyle\int_{-\infty}^{\infty}ds\,e^{\frac{s}{2}}\,B_{s}\left(\Delta\right)\frac{p-s}{e^{p}-e^{s}}=e^{-\frac{p}{2}}\,\int_{-\infty}^{\infty}ds\,\frac{(s-p)\,e^{\frac{s-p}{2}}}{e^{s-p}-1}B_{s}\left(\Delta\right)\,=4\pi^{2}e^{-\frac{p}{2}}B_{p}\left(\Delta+1\right)\,,$
(189)

where in the last equality we used eq. (270). All in all, we obtain

$\displaystyle
c^{(1,1,1)}_{3:3}=\frac{2^{6\Delta}}{3\left(2\pi\right)^{4}}\int_{-\infty}^{\infty}\,dp\int_{-\infty}^{\infty}\,dq\,B_{p}\left(\Delta\right)B_{q}\left(\Delta\right)\left(\frac{e^{\frac{p}{2}}B_{q}\left(\Delta+1\right)-e^{\frac{q}{2}}B_{p}\left(\Delta+1\right)}{e^{q}-e^{p}}\right)\,.$
(190)

Once again, the resulting integral has a simple spurious pole if we split the
terms which does not appear in the full expression. Nevertheless, for
computational purposes we need to treat each term independently. We can do so
with eq. (271), obtaining

$\displaystyle
c^{(1,1,1)}_{3:3}=\frac{2^{6\Delta}}{3\left(2\pi\right)^{4}\Delta}\int_{-\infty}^{\infty}dq\,q\,e^{-\frac{q}{2}}\,B^{2}_{q}\left(\Delta\right)B_{q}\left(\Delta+1\right)\,.$
(191)

This final integral can also be carried out using eq. (273), which leads to
our final formula

$\displaystyle
c^{(1,1,1)}_{3:3}=-\frac{2^{6\Delta}\Gamma\left(\Delta+\frac{1}{2}\right)^{3}}{12\pi\Gamma\left(3\Delta+\frac{3}{2}\right)}\,.$
(192)

We have been very explicit in the manipulations leading to the final form of
this coefficient. We will be more sketchy from now on; the general strategy is
always the same: replacing the inverse powers of the sine by the integral
representation (171), summing and extracting the piece that contributes in the
$n\to 1$ limit, and then manipulating the resulting expression to be able to
integrate it.

###  A.3 Computation of $I_{4}$ coefficients

The first coefficient appearing in eq. (2.2.4) is straightforward to obtain,
as it is given by the single sum

$\displaystyle c_{4:2}^{(4)}=\lim_{n\to
1}\frac{1}{n-1}\sum_{j<k}C^{4}_{ij}\,.$ (193)

This can be written as a single sum using the symmetry under $i\leftrightarrow
j$ and the ciclicity of the replica geometry,

$\displaystyle c_{4:2}^{(4)}=\lim_{n\to
1}\frac{n}{2(n-1)}\sum_{j=1}^{n-1}C^{4}_{0j}\,.$ (194)

The computation is now analogous to the one of $c_{2:2}^{(2)}$ which produced
eq. (178), the only difference being the power of $C_{0j}$ in the sum. This is
accounted for by changing $\Delta\to 2\Delta$, producing

$\displaystyle
c_{4:2}^{(4)}=2^{8\Delta-1}\frac{\Gamma\left(4\Delta+1\right)^{2}}{\Gamma\left(8\Delta+2\right)}\,.$
(195)

The remaining coefficients in eq. (2.2.4) require more work, so before getting
into the details of the computations, let us write them in a convenient form.
Using arguments analogous to (181) based on replica symmetry and the
symmetries of the expression being summed, we can write them as

$\displaystyle c_{4:3}^{(1,1,2)}=\lim_{n\to
1}\frac{n}{3(n-1)}\sum_{k=2}^{n-1}\sum_{j=1}^{k-1}\left[C^{2}_{0j}C_{jk}C_{k0}+C_{0j}C^{2}_{jk}C_{k0}+C_{0j}C_{jk}C^{2}_{k0}\right]\,,$
(196) $\displaystyle c_{4:3}^{(2,2)}=\lim_{n\to
1}\frac{n}{3(n-1)}\sum_{k=2}^{n-1}\sum_{j=1}^{k-1}\left[C^{2}_{0j}C^{2}_{0k}+C^{2}_{0j}C^{2}_{jk}+C^{2}_{0k}C^{2}_{jk}\right]\,,$
(197) $\displaystyle c_{4:4}^{(1,1,1,1)}=\lim_{n\to
1}\frac{n}{4(n-1)}\sum_{l=3}^{n-1}\sum_{k=2}^{l-1}\sum_{j=1}^{k-1}\left[C_{0j}C_{jk}C_{kl}C_{l0}+C_{0j}C_{jl}C_{lk}C_{k0}+C_{0l}C_{lj}C_{jk}C_{k0}\right]\,,$
(198) $\displaystyle c_{4:4}^{(2,2)}=\lim_{n\to
1}\frac{n}{4(n-1)}\sum_{l=3}^{n-1}\sum_{k=2}^{l-1}\sum_{j=1}^{k-1}\left[C^{2}_{0l}C^{2}_{jk}+C^{2}_{0j}C^{2}_{kl}+C^{2}_{0k}C^{2}_{jl}\right]\,.$
(199)

####  A.3.1 Computation of $c_{4:3}^{(1,1,2)}$

Replace in eq. (196) the integral representation for each coefficien. Summing
and extracting the relevant piece in the $n\to 1$ limit, we get

$\displaystyle
e^{-\frac{p}{2}}e^{-\frac{q}{2}}e^{-\frac{s}{2}}\sum_{k=2}^{n-1}\sum_{j=1}^{k-1}\left(e^{\frac{pj}{n}}e^{\frac{q(k-j)}{n}}e^{\frac{sk}{n}}+e^{\frac{sj}{n}}e^{\frac{p(k-j)}{n}}e^{\frac{qk}{n}}+e^{\frac{s(k-j)}{n}}e^{\frac{qj}{n}}e^{\frac{pk}{n}}\right)\approx(n-1)f^{(1,1,2)}_{4:3}(p,q)\,,$
(200)

where

$\displaystyle f^{(1,1,2)}_{4:3}(p,q,s)=$
$\displaystyle\frac{e^{\frac{p+q+s}{2}}}{(e^{q}-e^{p})}\left[\frac{q+s}{e^{q+s}-1}-\frac{p+s}{e^{p+s}-1}\right]+\frac{e^{\frac{p+q+s}{2}}}{(e^{s}-e^{p})}\left[\frac{s+q}{e^{s+q}-1}-\frac{p+q}{e^{p+q}-1}\right]$
$\displaystyle+\frac{e^{\frac{p+q+s}{2}}}{(e^{s}-e^{q})}\left[\frac{s+p}{e^{s+p}-1}-\frac{q+p}{e^{q+p}-1}\right]\,.$
(201)

Therefore, we are left with the following triple integral to evaluate

$\displaystyle
c_{4:3}^{(1,1,2)}=\frac{2^{8\Delta-6}}{3\,\pi^{6}}\int_{-\infty}^{\infty}dp\,\int_{-\infty}^{\infty}dq\,\int_{-\infty}^{\infty}ds\,f^{(1,1,2)}_{4:3}(p,q,s)\,B_{p}(2\Delta)B_{q}(\Delta)B_{s}(\Delta)\,.$
(202)

The form of $f^{(1,1,2)}_{4:3}(p,q,s)$ allows to integrate each piece by means
of the same tricks than before. Integrals (270) and (271) reduce the
computation to the following single integral

$\displaystyle c_{4:3}^{(1,1,2)}=$
$\displaystyle\frac{2^{8\Delta-4}}{2\pi^{4}\Delta\,}\int_{-\infty}^{\infty}dq\,q\,e^{-\frac{q}{2}}\,B_{q}(2\Delta)B_{q}(\Delta)B_{q}(\Delta+1)$
$\displaystyle+\frac{2^{8\Delta-4}}{3\pi^{4}\Delta\,}\int_{-\infty}^{\infty}dq\,q\,e^{-\frac{q}{2}}\,B^{2}_{q}(\Delta)B_{q}(2\Delta+1)\,.$
(203)

We can now shift the argument of one of the beta functions of the first term
by means of basic gamma function identities to obtain:

$\displaystyle c_{4:3}^{(1,1,2)}$ $\displaystyle=$
$\displaystyle\frac{2^{8\Delta-3}(7\Delta+2)}{3\pi^{4}\Delta\,\left(2\Delta+1\right)}\int_{-\infty}^{\infty}dq\,q\,e^{-\frac{q}{2}}B^{2}_{q}\left(\Delta\right)B_{q}\left(2\Delta+1\right)$
(204) $\displaystyle-\frac{3\times
2^{8\Delta-6}}{\pi^{4}\left(2\Delta+1\right)}\int_{-\infty}^{\infty}dq\,q\,e^{-\frac{q}{2}}B^{2}_{q}\left(\Delta\right)B_{q}\left(2\Delta\right)\,.$

Using the result (273) from Appendix B, we get

$\displaystyle
c_{4:3}^{(1,1,2)}=-\frac{9\sqrt{\pi}\Gamma\left(3\Delta\right)^{2}}{4\Gamma\left(\Delta\right)^{2}\Gamma\left(4\Delta+\frac{3}{2}\right)}=-\frac{9\times
2^{8\Delta}\,\Gamma\left(3\Delta\right)^{2}\Gamma\left(4\Delta+1\right)}{2\,\Gamma\left(\Delta\right)^{2}\Gamma\left(8\Delta+2\right)}\,.$
(205)

####  A.3.2 Computation of $c_{4:3}^{(2,2)}$

We start now from eq. (197) and use the integral representation of the
coefficients. We carry out the sum over the indices, obtaining

$\displaystyle
e^{-\frac{p}{2}}e^{-\frac{q}{2}}\sum_{k=2}^{n-1}\sum_{j=1}^{k-1}\left(e^{\frac{pl}{n}}e^{\frac{qk}{n}}+e^{\frac{pj}{n}}e^{\frac{q(k-j)}{n}}+e^{\frac{pk}{n}}e^{\frac{q(k-j)}{n}}\right)\approx(n-1)f^{(2,2)}_{4:3}(p,q)\,,$
(206)

where

$\displaystyle f^{(2,2)}_{4:3}(p,q)$ $\displaystyle=$
$\displaystyle\frac{e^{\frac{p+q}{2}}}{(e^{p}-1)}\left[\frac{q+p}{e^{q+p}-1}-\frac{q}{e^{q}-1}\right]+\,\frac{e^{\frac{p+q}{2}}}{(e^{q}-e^{p})}\left[\frac{q}{e^{q}-1}-\frac{p}{e^{p}-1}\right]$
(207)
$\displaystyle+\,\frac{e^{\frac{p+q}{2}}}{(e^{p+q}-1)}\left[\frac{q}{e^{q}-1}-\frac{p\,e^{p}}{e^{p}-1}\right]\,.$

We have an integral expression for our coefficient of the form

$\displaystyle
c_{4:3}^{(2,2)}=\frac{2^{8\Delta-4}}{3\pi^{4}}\int_{-\infty}^{\infty}dp\,\int_{-\infty}^{\infty}dq\,f^{(2,2)}_{4:3}(p,q)\,B_{p}(2\Delta)B_{q}(2\Delta)\,.$
(208)

Symmetrizing the integrands in $q$ we can obtain

$\displaystyle
c_{4:3}^{(2,2)}=-\frac{2^{8\Delta-4}}{6\pi^{4}\Delta}\int_{-\infty}^{\infty}dq\,q^{2}\,B^{2}_{q}(2\Delta)-\frac{2^{8\Delta-2}}{6\pi^{2}}\int_{-\infty}^{\infty}dq\,\,B_{q}(2\Delta)B_{q}(2\Delta+1)\,,$
(209)

and from here, by shifting the argument of the last beta function, we can use
eq. (273) to conclude:

$\displaystyle
c_{4:3}^{(2,2)}=-\frac{2^{8\Delta}\left[\Gamma\left(4\Delta+1\right)\right]^{2}}{2\,\Gamma\left(8\Delta+2\right)}\,.$
(210)

####  A.3.3 Computation of $c_{4:4}^{(1,1,1,1)}$

Doing the sum like in the previous cases, we obtain:

$\displaystyle
c_{4:4}^{(1,1,1,1)}=\frac{2^{8\Delta-8}}{4\,\pi^{8}}\int_{-\infty}^{\infty}dp\int_{-\infty}^{\infty}dq\int_{-\infty}^{\infty}dr\int_{-\infty}^{\infty}ds\,f^{(1,1,1,1)}_{4:4}(p,q,r,s)B_{p}(\Delta)B_{q}(\Delta)B_{r}(\Delta)B_{s}(\Delta)\,,$

where

$\displaystyle f_{4:4}^{(1,1,1,1)}(p,q,r,s)$ $\displaystyle=$ $\displaystyle
e^{\frac{p+q+r+s}{2}}\Bigg{[}\frac{(p+s)}{(e^{p+s}-1)(e^{p}-e^{r})(e^{p}-e^{q})}+\frac{(q+s)}{(e^{q+s}-1)(e^{q}-e^{r})(e^{q}-e^{p})}$
(212)
$\displaystyle\qquad\qquad\qquad+\frac{(r+s)}{(e^{r+s}-1)(e^{r}-e^{q})(e^{r}-e^{p})}\Bigg{]}$
$\displaystyle+$ $\displaystyle
e^{\frac{p+q+r+s}{2}}\Bigg{[}\frac{e^{q}(p+r-q-s)}{(e^{p+r}-e^{q+s})(e^{p+r}-1)(e^{p}-e^{q})}+\frac{e^{q+r}(q+s)}{(e^{q+s}-1)(e^{p+r}-1)(e^{q+r}-1)}$
$\displaystyle\qquad\qquad\qquad-\frac{(r-s)}{(e^{r}-e^{s})(e^{q+r}-1)(e^{p}-e^{q})}\Bigg{]}$
$\displaystyle+$ $\displaystyle
e^{\frac{p+q+r+s}{2}}\Bigg{[}\frac{e^{q}(p+s)}{(e^{p+s}-1)(e^{q+r}-1)(e^{q}-e^{p})}+\frac{e^{p}(q+s)}{(e^{q+s}-1)(e^{p+r}-1)(e^{p}-e^{q})}$
$\displaystyle\qquad\qquad\qquad+\frac{(r+s+p+q)}{(e^{r+s+p+q}-1)(e^{q+r}-1)(e^{p+r}-1)}\Bigg{]}\,.$

It is immediate to integrate over $s$ by means of eq. (270). After some
relabelling of the variables we arrive at the following expression

$\displaystyle c_{4:4}^{(1,1,1,1)}\\!\\!\\!$ $\displaystyle=$
$\displaystyle\\!\\!\\!\frac{2^{8\Delta-8}}{\pi^{6}}\int_{-\infty}^{\infty}dp\,\int_{-\infty}^{\infty}dq\,\int_{-\infty}^{\infty}dr\,B_{p}(\Delta)B_{q}(\Delta)B_{r}(\Delta)$
(213)
$\displaystyle\\!\\!\\!\\!\times\Bigg{[}\,\frac{3\,e^{\frac{q+r}{2}}B_{p}(\Delta+1)}{(e^{p}-e^{r})(e^{p}-e^{q})}+\frac{\left(e^{q+r}+1\right)\,e^{\frac{p+r}{2}}B_{q}(\Delta+1)}{(e^{q+r}-1)(e^{p+r}-1)}+\frac{e^{q}B_{q-p-r}(\Delta+1)}{(e^{p+r}-1)(e^{p}-e^{q})}$
$\displaystyle\\!\\!\\!\\!\qquad+\frac{B_{q+p+r}(\Delta+1)}{(e^{p+r}-1)(e^{q+r}-1)}+\frac{2\,e^{p+q}\,e^{\frac{p+r}{2}}B_{q}(\Delta+1)}{(e^{p+q}-1)(e^{p+r}-1)}\Bigg{]}\,.$

We can write the $B_{q-p-r}(\Delta+1)$ as $B_{p+q+r}(\Delta+1)$ by sending
$q\to-q$ and using the symmetry of the beta function
$B_{-s}(\Delta)=B_{s}(\Delta)$. That term then combines with the one in which
$B_{p+q+r}(\Delta+1)$ is already present, after some trivial relabelling.
Similarly, we can replace

$\displaystyle\frac{\left(e^{q+r}+1\right)\,e^{\frac{p+r}{2}}B_{q}(\Delta+1)}{(e^{q+r}-1)(e^{p+r}-1)}+\frac{2\,e^{p+q}\,e^{\frac{p+r}{2}}B_{q}(\Delta+1)}{(e^{p+q}-1)(e^{p+r}-1)}\to\frac{3\,e^{\frac{p+r}{2}}B_{q}(\Delta+1)}{(e^{p+r}-1)}+\frac{4\,e^{\frac{p+r}{2}}B_{q}(\Delta+1)}{(e^{p+r}-1)(e^{q+r}-1)}\,.$

The first term is odd under $(p,r)\to(-p,-r)$, while the prefactor in the
whole integral is even due to the symmetry of the beta functions. Therefore,
it will not contribute and only the second term is relevant. Let us recap and
split the integrals in a suitable way for further manipulation,

$c_{4:4}^{(1,1,1,1)}=J_{1}+J_{2}+J_{3}\,,$ (214)

where, explicitly indicating the $\Delta$ dependence

$\displaystyle J_{1}(\Delta)=$
$\displaystyle\frac{2^{8\Delta-8}}{\pi^{6}}\int_{-\infty}^{\infty}dp\,\int_{-\infty}^{\infty}dq\,\int_{-\infty}^{\infty}dr\,B_{p}(\Delta)B_{q}(\Delta)B_{r}(\Delta)\frac{2B_{p+q+r}\left(\Delta+1\right)}{(e^{p+r}-1)(e^{p+q}-1)}\,,$
(215) $\displaystyle J_{2}(\Delta)=$
$\displaystyle\frac{2^{8\Delta-8}}{\pi^{6}}\int_{-\infty}^{\infty}dp\,\int_{-\infty}^{\infty}dq\,\int_{-\infty}^{\infty}dr\,B_{p}(\Delta)B_{q}(\Delta)B_{r}(\Delta)\frac{3\,e^{\frac{q+r}{2}}B_{p}(\Delta+1)}{(e^{p}-e^{r})(e^{p}-e^{q})}\,,$
(216) $\displaystyle J_{3}(\Delta)=$
$\displaystyle\frac{2^{8\Delta-8}}{\pi^{6}}\int_{-\infty}^{\infty}dp\,\int_{-\infty}^{\infty}dq\,\int_{-\infty}^{\infty}dr\,B_{p}(\Delta)B_{q}(\Delta)B_{r}(\Delta)\frac{4\,e^{\frac{p+r}{2}}B_{q}(\Delta+1)}{(e^{p+r}-1)(e^{q+r}-1)}\,.$
(217)

Notice that the expression we have obtained for $c_{4:4}^{(1,1,1,1)}$, though
simpler than eq. (A.3.3), is ill-defined due to the appearance of spurious
poles. These are not present in the original expression, and we must remove
them in order to obtain a sensible answer. We do it by symmetrizing under the
variables of integration for each of the three pieces independently.

For $J_{1}$, after sending $p\to-p$, replace inside the integral,

$\displaystyle\frac{2e^{2p}B_{q+r-p}\left(\Delta+1\right)}{(e^{p}-e^{r})(e^{p}-e^{q})}\to\frac{2}{3}\Bigg{[}\frac{e^{2p}\,B_{q+r-p}\left(\Delta+1\right)}{(e^{p}-e^{r})(e^{p}-e^{q})}+\frac{e^{2r}\,B_{q+p-r}\left(\Delta+1\right)}{(e^{r}-e^{p})(e^{r}-e^{q})}+\frac{e^{2q}\,B_{p+r-q}\left(\Delta+1\right)}{(e^{q}-e^{p})(e^{q}-e^{r})}\Bigg{]}\,.$

If we now use in the last term the following identity

$\displaystyle\frac{1}{(e^{p}-e^{r})(e^{p}-e^{q})}+\frac{1}{(e^{q}-e^{p})(e^{q}-e^{r})}+\frac{1}{(e^{r}-e^{q})(e^{r}-e^{p})}=0\,,$
(218)

we get:

$\displaystyle J_{1}(\Delta)=$
$\displaystyle\frac{2}{3}\frac{2^{8\Delta-8}}{\pi^{6}}\int_{-\infty}^{\infty}dp\int_{-\infty}^{\infty}dq\int_{-\infty}^{\infty}dr\,\frac{B_{p}(\Delta)B_{q}(\Delta)B_{r}(\Delta)}{e^{p}-e^{r}}\times$
(219)
$\displaystyle\times\left[\frac{e^{2p}\,B_{q+r-p}\left(\Delta+1\right)-e^{2q}\,B_{p+r-q}\left(\Delta+1\right)}{e^{p}-e^{q}}-\frac{e^{2r}\,B_{q+p-r}\left(\Delta+1\right)-e^{2q}\,B_{p+r-q}\left(\Delta+1\right)}{e^{r}-e^{q}}\right]\,.$

This formula is pole free and consequently it is well defined. Unfortunately,
we were not able to integrate it in a closed form analytically; not even by
means of the manipulations we will describe for $J_{2}$ and $J_{3}$. The most
we managed to do is obtaining numerical results for particular (integer or
semi-integer) values of $\Delta$. We present them after the results for
$J_{2}$ and $J_{3}$.

The same manipulations in $J_{2}$ —symmetrizing and using (218)— transform the
integrand into the following pole-free form

$\displaystyle J_{2}(\Delta)=$
$\displaystyle\frac{2^{8\Delta-8}}{\pi^{6}}\int_{-\infty}^{\infty}dp\int_{-\infty}^{\infty}dq\int_{-\infty}^{\infty}dr\,B_{p}(\Delta)B_{q}(\Delta)B_{r}(\Delta)\frac{e^{\frac{p+q+r}{2}}}{e^{p}-e^{q}}\times$
(220)
$\displaystyle\times\left[\frac{e^{-\frac{p}{2}}B_{p}(\Delta+1)-e^{-\frac{r}{2}}B_{r}(\Delta+1)}{e^{p}-e^{r}}-\frac{e^{-\frac{q}{2}}B_{q}(\Delta+1)-e^{-\frac{r}{2}}B_{r}(\Delta+1)}{e^{q}-e^{r}}\right]\,,$

which, after integrating over $q$ in the first term and $p$ in the second one,
we get

$J_{2}(\Delta)=\frac{2^{8\Delta-8}}{\pi^{6}\Delta}\int_{-\infty}^{\infty}dp\int_{-\infty}^{\infty}dr\,B^{2}_{p}(\Delta)B_{r}(\Delta)\,p\,e^{\frac{r}{2}}\left(\frac{e^{-\frac{p}{2}}B_{p}(\Delta+1)-e^{-\frac{r}{2}}B_{r}(\Delta+1)}{e^{p}-e^{r}}\right)\,.$
(221)

This integral can be done for integer or semi-integer $\Delta$ through the
following procedure. We will present the details for $\Delta=(2k+1)/2$ with
$k\in\mathbb{N}$; analogous manipulations provide the results for $\Delta=k$.
First of all, write the difference in the denominator as $\sinh((p-r)/2)$ and
use the symmetries of the integrand to obtain

$\displaystyle
J_{2}\left(\frac{2k+1}{2}\right)=\frac{2^{8k-4}}{(2k+1)\pi^{6}}\int_{-\infty}^{\infty}dp$
$\displaystyle\int_{-\infty}^{\infty}dr\,\frac{p\,B_{p}\left(\frac{2k+1}{2}\right)^{2}B_{r}\left(\frac{2k+1}{2}\right)}{\sinh\left(\frac{p-r}{2}\right)}$
(222)
$\displaystyle\times\left[\cosh(p)B_{p}\left(\frac{2k+3}{2}\right)-\cosh\left(\frac{p+r}{2}\right)B_{r}\left(\frac{2k+3}{2}\right)\right]\,.$

Now, use the shift formula for the beta function (269) to write all arguments
as $(2k+1)/2$. The final step is the most involved one; we will illustrate it
for the first term within brackets. The basic idea is that we can decouple the
$p$ and $r$ integrals by introducing an extra integration variable $u$ with a
$\delta$-function at $u=(p-r)/2$; so that we can write the denominator as
$\sinh(u/2)$. Writing then the $\delta$-function by means of its integral
Fourier representation, we get, dropping the global prefactor in the previous
integral

$\displaystyle\int_{-\infty}^{\infty}\frac{d\xi}{2\pi}$
$\displaystyle\left[\int_{-\infty}^{\infty}dp\,p\cosh(p)\frac{p^{2}+\pi^{2}(2k+1)^{2}}{4\pi^{2}(2k+1)(2k+2)}B_{p}\left(\frac{2k+1}{2}\right)^{3}e^{-i\xi
p}\right]$
$\displaystyle\times\left[\int_{-\infty}^{\infty}dr\,B_{r}\left(\frac{2k+1}{2}\right)e^{i\xi
r}\right]\left[\int_{-\infty}^{\infty}du\,\frac{e^{i\xi
u}}{\sinh(u/2)}\right]\,.$ (223)

The integral over $r$ can be done directly by using the hyperbolic
representation of the beta function in (268), while the one over $u$ is just
eq. (262). The only really challenging one is the integral over $p$. We can
use the expansion (277) to write the integrand as some polynomial in $p$ times
$\cosh(p)/\cosh(p/2)$. The powers of $p$ can be traded for derivatives
$i\partial_{\xi}$ and taken outside of the integral, while using
$\cosh(p)=2\cosh^{2}(p/2)-1$ we can write the remaining integral as a
combination of eq. (261) and eq. (264). All in all, the previous expression
becomes

$-\frac{\pi^{2(2-3k)}}{2^{8k+1}(2k)!^{3}(2k+1)(2k+2)}\int_{-\infty}^{\infty}d\xi\,F_{2}^{(1)}(\xi,k)\frac{\tanh(\pi\xi)}{\cosh^{2k+1}(\pi\xi)}\,,$
(224)

where $F_{2}^{(1)}(\xi,k)$ is the result of extracting powers of $p$ as we
described

$F_{2}^{(1)}(\xi,k)=\left[\prod_{l=0}^{k-1}\mathcal{D}_{2l+1}^{3}\right]\mathcal{D}_{2k+1}\partial_{\xi}\left(\frac{3-4\xi^{2}}{\cosh(\pi\xi)}\right)\,,$
(225)

and $\mathcal{D}_{n}$ is the following differential operator:

$\mathcal{D}_{n}\equiv-\partial_{\xi}^{2}+n^{2}\pi^{2}\,.$ (226)

This expression can be evaluated for any non-negative integer $k$
algorithmically and, in particular, it is not difficult to introduce it in a
standard symbolic algebra software. The remaining integral over $\xi$ can be
done by means of the general formulas presented in Appendix B, because it is
just an inverse power of $\cosh(\pi\xi)$ times some complicated function of
$\xi$ and hyperbolic functions $\cosh(\alpha\xi)$ or
$\sinh(\alpha\xi)$.151515For this one might need to express powers of
hyperbolic functions in the numerator as hyperbolic functions with different
argument, _e.g._ , $4\cosh^{3}(x)=3\cosh(x)+\cosh(3x)$. Alternatively, for low
enough $k$, Mathematica is able to do the resulting integral over $\xi$
directly. One can show, as an example, that for $k=0$ ($\Delta=1/2$) the
previous integral evaluates to $4\pi^{4}+16\pi^{6}/15$.

The second term in (222) can be treated analogously (after separating the $p$
and $r$ parts of the $\cosh$ by means of the identity for the $\cosh$ of a
sum), although in this case both the $p$ and the $r$ integrals have to be done
by means of the expansion (277). The remaining manipulations are identical, so
we will just write the final result,

$\displaystyle
J_{2}\left(\frac{2k+1}{2}\right)=-\frac{1}{16\pi^{6k+2}(2k)!^{3}(2k+1)^{2}(2k+2)}$
$\displaystyle\int_{-\infty}^{\infty}d\xi\,\tanh(\pi\xi)$
$\displaystyle\times\left[\frac{F_{2}^{(1)}(\xi,k)}{2\cosh^{2k+1}(\pi\xi)}-\frac{F_{2}^{(2)}(\xi,k)}{(2k)!\pi^{2k}}\right]\,,$
(227)

where $F_{2}^{(1)}$ was presented in (225) and $F_{2}^{(2)}$ is

$\displaystyle F_{2}^{(2)}(\xi,k)=$
$\displaystyle\left[\left(\prod_{l=0}^{k-1}\mathcal{D}_{2l+1}^{2}\right)\partial_{\xi}\left(\frac{1}{\cosh(\pi\xi)}\right)\right]\left[\left(\prod_{l=0}^{k-1}\mathcal{D}_{2l+1}^{2}\right)\mathcal{D}_{2k+1}\left(\frac{1}{\cosh(\pi\xi)}\right)\right]$
$\displaystyle+4\left[\left(\prod_{l=0}^{k-1}\mathcal{D}_{2l+1}^{2}\right)\partial_{\xi}\left(\frac{\xi}{\cosh(\pi\xi)}\right)\right]\left[\left(\prod_{l=0}^{k-1}\mathcal{D}_{2l+1}^{2}\right)\mathcal{D}_{2k+1}\left(\frac{\xi}{\cosh(\pi\xi)}\right)\right]\,.$
(228)

The case $\Delta=k$ is completely analogous, obtaining as a final result,

$\displaystyle
J_{2}\left(k\right)=\frac{1}{16\pi^{6k-1}(2k-1)!^{3}(2k)^{2}(2k+1)}$
$\displaystyle\int_{-\infty}^{\infty}d\xi\,\tanh(\pi\xi)$
$\displaystyle\times\left[\frac{G_{2}^{(1)}(\xi,k)}{2\cosh^{2k}(\pi\xi)}+\frac{G_{2}^{(2)}(\xi,k)}{(2k-1)!\pi^{2k-1}}\right]\,,$
(229)

where

$\displaystyle G_{2}^{(1)}(\xi,k)=$
$\displaystyle\left[\prod_{l=1}^{k-1}\mathcal{D}_{2l}^{3}\right]\mathcal{D}_{2k}\partial^{4}_{\xi}\left((3-4\xi^{2})\tanh(\pi\xi)\right)\,,$
(230) $\displaystyle G_{2}^{(2)}(\xi,k)=$
$\displaystyle\left[\left(\prod_{l=1}^{k-1}\mathcal{D}_{2l}^{2}\right)\partial^{3}_{\xi}\left(\tanh(\pi\xi)\right)\right]\left[\left(\prod_{l=1}^{k-1}\mathcal{D}_{2l}^{2}\right)\mathcal{D}_{2k}\partial_{\xi}^{2}\left(\tanh(\pi\xi)\right)\right]$
$\displaystyle+4\left[\left(\prod_{l=1}^{k-1}\mathcal{D}_{2l}^{2}\right)\partial^{3}_{\xi}\left(\xi\tanh(\pi\xi)\right)\right]\left[\left(\prod_{l=1}^{k-1}\mathcal{D}_{2l}^{2}\right)\mathcal{D}_{2k}\partial_{\xi}^{2}\left(\xi\tanh(\pi\xi)\right)\right]\,.$
(231)

We come finally to the last piece, $J_{3}$. We first manipulate a bit the
integrand: changing $(p,q)\to(-p,-q)$ and symmetrizing afterwards over $p$ and
$r$, the final part of the integral becomes

$\frac{4\,e^{\frac{p+r}{2}}B_{q}(\Delta+1)}{(e^{p+r}-1)(e^{q+r}-1)}\to
2\,e^{\frac{p+q+r}{2}}e^{q/2}B_{q}(\Delta+1)\left[\frac{1}{(e^{r}-e^{p})(e^{r}-e^{q})}+\frac{1}{(e^{p}-e^{r})(e^{p}-e^{q})}\right]\,.$
(232)

Applying now (218), symmetrizing the result over the three variables and then
using (218) again to write the integrand in a pole free form like in the
previous $J_{1}$ and $J_{2}$, we get

$\displaystyle J_{3}(\Delta)=$
$\displaystyle-\frac{2}{3}\frac{2^{8\Delta-8}}{\pi^{6}}\int_{-\infty}^{\infty}dp\int_{-\infty}^{\infty}dq\int_{-\infty}^{\infty}dr\,B_{p}(\Delta)B_{q}(\Delta)B_{r}(\Delta)\frac{e^{\frac{p+q+r}{2}}}{e^{p}-e^{q}}\times$
(233)
$\displaystyle\times\left[\frac{e^{\frac{p}{2}}B_{p}(\Delta+1)-e^{\frac{r}{2}}B_{r}(\Delta+1)}{e^{p}-e^{r}}-\frac{e^{\frac{q}{2}}B_{q}(\Delta+1)-e^{\frac{r}{2}}B_{r}(\Delta+1)}{e^{q}-e^{r}}\right]\,.$

Integration over $q$ in the first term and over $p$ in the second one produces

$J_{3}(\Delta)=-\frac{2}{3}\frac{2^{8\Delta-8}}{\pi^{6}\Delta}\int_{-\infty}^{\infty}dp\int_{-\infty}^{\infty}dr\,B^{2}_{p}(\Delta)B_{r}(\Delta)\,p\,e^{\frac{r}{2}}\left(\frac{e^{\frac{p}{2}}B_{p}(\Delta+1)-e^{\frac{r}{2}}B_{r}(\Delta+1)}{e^{p}-e^{r}}\right)\,.$
(234)

Already from the form of this integral we can see that it is very similar to
$J_{2}(\Delta)$. The technique we use is exactly the same one, so we will not
give details about the intermediate steps, but instead we will present the
final result directly. We need to separate semi-integer and integer $\Delta$.
In the first case, we have

$\displaystyle J_{3}\left(\frac{2k+1}{2}\right)=$
$\displaystyle\frac{1}{6\pi^{6k+2}(2k)!^{3}(2k+1)^{2}(2k+2)}$
$\displaystyle\times\int_{-\infty}^{\infty}d\xi\,\left[\frac{F_{3}^{(1)}(\xi,k)}{8}\frac{\tanh(\pi\xi)}{\cosh^{2k+1}(\pi\xi)}-\frac{F_{3}^{(2)}(\xi,k)}{(2k)!\pi^{2k}}\coth(\pi\xi)\right]\,,$
(235)

where

$\displaystyle F_{3}^{(1)}(\xi,k)=$
$\displaystyle\left[\prod_{l=0}^{k-1}\mathcal{D}_{2l+1}^{3}\right]\mathcal{D}_{2k+1}\partial_{\xi}\left(\frac{1+4\xi^{2}}{\cosh(\pi\xi)}\right)\,,$
(236) $\displaystyle F_{3}^{(2)}(\xi,k)=$
$\displaystyle\left[\left(\prod_{l=0}^{k-1}\mathcal{D}_{2l+1}^{2}\right)\partial_{\xi}\left(\frac{\xi}{\sinh(\pi\xi)}\right)\right]\left[\left(\prod_{l=0}^{k-1}\mathcal{D}_{2l+1}^{2}\right)\mathcal{D}_{2k+1}\left(\frac{\xi}{\sinh(\pi\xi)}\right)\right]\,.$
(237)

While for integer $\Delta=k$ we get

$\displaystyle J_{3}\left(k\right)=$
$\displaystyle\frac{1}{6\pi^{6k-1}(2k-1)!^{3}(2k)^{2}(2k+1)}$
$\displaystyle\times\int_{-\infty}^{\infty}d\xi\,\left[\frac{G_{3}^{(1)}(\xi,k)}{8}\frac{\tanh(\pi\xi)}{\cosh^{2k}(\pi\xi)}-\frac{G_{3}^{(2)}(\xi,k)}{(2k-1)!\pi^{2k-1}}\coth(\pi\xi)\right]\,,$
(238)

where

$\displaystyle G_{3}^{(1)}(\xi,k)=$
$\displaystyle\left[\prod_{l=1}^{k-1}\mathcal{D}_{2l}^{3}\right]\mathcal{D}_{2k}\partial^{4}_{\xi}\left((1+4\xi^{2})\tanh(\pi\xi)\right)\,,$
(239) $\displaystyle G_{3}^{(2)}(\xi,k)=$
$\displaystyle\left[\left(\prod_{l=1}^{k-1}\mathcal{D}_{2l}^{2}\right)\partial^{3}_{\xi}\left(\xi\coth(\pi\xi)\right)\right]\left[\left(\prod_{l=1}^{k-1}\mathcal{D}_{2l}^{2}\right)\mathcal{D}_{2k}\partial_{\xi}^{2}\left(\xi\coth(\pi\xi)\right)\right]\,.$
(240)

We conclude with Table 1, where we show numerical results for $J_{1}$
—obtained from eq. (219)— as well as exact results for $J_{2}$ and $J_{3}$,
for the smallest semi-integer and integer values of $\Delta$. We also compute
the total value of the coefficient $c_{4:4}^{(1,1,1,1)}$ for those cases.

$\Delta$ | $J_{1}$ | $J_{2}$ | $J_{3}$ | $c_{4:4}^{(1,1,1,1)}$
---|---|---|---|---
$1/2$ | $\frac{8}{135}+\frac{1}{9\pi^{2}}$ | $\frac{2}{45}+\frac{1}{3\pi^{2}}$ | $-\frac{2}{135}+\frac{2}{9\pi^{2}}$ | $\frac{4}{45}+\frac{2}{3\pi^{2}}\simeq 0.156436$
$1$ | $\frac{64}{945}-\frac{4}{27\pi^{2}}$ | $\frac{32}{315}+\frac{4}{9\pi^{2}}$ | $\frac{32}{945}-\frac{8}{27\pi^{2}}$ | $\frac{64}{315}\simeq 0.203175$
$3/2$ | $0.04382$ | $\frac{10496}{45045}+\frac{32}{27\pi^{2}}$ | $-\frac{10496}{135135}+\frac{64}{81\pi^{2}}$ | $0.399298$
$2$ | $0.03827$ | $\frac{1482752}{2297295}+\frac{256}{81\pi^{2}}$ | $\frac{1482752}{6891885}-\frac{512}{243\pi^{2}}$ | $1.005593$
$5/2$ | $0.03440$ | $\frac{1507328}{793611}+\frac{26624}{2835\pi^{2}}$ | $-\frac{1507328}{2380833}+\frac{53248}{8505\pi^{2}}$ | $2.886498$
$3$ | $0.03151$ | $\frac{1097859072}{185910725}+\frac{32768}{1125\pi^{2}}$ | $\frac{365953024}{185910725}-\frac{65536}{3375\pi^{2}}$ | $8.888975$
Table 1: Values of the different terms contributing to $c_{4:4}^{(1,1,1,1)}$.
$J_{1}$ is obtained by numerical integration of (219), while $J_{2}$ and
$J_{3}$ are obtained from the expressions (A.3.3), (A.3.3), (A.3.3), and
(A.3.3).

For $\Delta=1/2$, there is a fortunate numerical coincidence, easy to
discover, which allows us to write an analytic value of the coefficient. The
total value of $c_{4:4}^{(1,1,1,1)}(\Delta=1/2)$ is twice that of $J_{2}(1/2)$
to an extremely good level of precision, and this last one we know
analytically. Therefore, we conclude161616This coefficient can alternatively
be read from [81] —more precisely, from the last two terms on the right hand
side of their equation (4.19). Similarly, one can also find that expression in
eqs (79) and (80) of [29]. In order to identify the exact result, it is useful
to notice the following relations
$C_{ij}=2\langle\phi_{i}(1)\phi_{j}(1)\rangle$ and $C_{ij}=1/s_{ij}$ between
our coefficients and the correlators in [81] and [29] respectively. In their
context, the computation appears in the subleading contribution to the mutual
information of two disjoint disks for a free scalar in three space-time
dimensions.

$c_{4:4}^{(1,1,1,1)}(\Delta=1/2)=\frac{4}{45}+\frac{2}{3\pi^{2}}~{}.$ (241)

Unfortunately, it is not obvious that a similar relation holds for other
values of $\Delta$. Still, for $\Delta=1$ we are reasonably confident that the
numerical value obtained corresponds to

$J_{1}(\Delta=1)\overset{?}{=}\frac{64}{945}-\frac{4}{27\pi^{2}}\quad\Rightarrow\quad
c_{4:4}^{(1,1,1,1)}(\Delta=1)\overset{?}{=}\frac{64}{315}\,.$ (242)

More generally, we do believe $J_{1}(\Delta)$ might have a reasonably simple
closed form, much like the others. This is a technically challenging problem
which we leave open for further investigation.171717It is also conceivable
that all the integrals, or at least the coefficient as a whole, can be done
for general $\Delta$ in a closed form. This is what happens for the other
coefficients $c_{4:2}^{(4)}$, $c_{4:3}^{(1,1,2)}$, etc.

####  A.3.4 Computation of $c_{4:4}^{(2,2)}$

Finally, we come to the coefficient (199). With the integral representation
for the coefficients, we must perform the following sum

$\displaystyle
e^{-\frac{p}{2}}e^{-\frac{q}{2}}\sum_{l=3}^{n-1}\sum_{k=2}^{l-1}\sum_{j=1}^{k-1}\left(e^{\frac{pl}{n}}e^{\frac{q(k-j)}{n}}+e^{\frac{pj}{n}}e^{\frac{q(l-k)}{n}}+e^{\frac{pk}{n}}e^{\frac{q(l-j)}{n}}\right)\approx(n-1)f^{(2,2)}_{4:4}(p,q)\,,$
(243)

where

$\displaystyle
f^{(2,2)}_{4:4}(p,q)=\frac{e^{\frac{p+q}{2}}}{e^{p}-1}\left[\frac{e^{p}\left(p+q\right)}{e^{p+q}-1}-\frac{p-q}{e^{p}-e^{q}}\right]\,.$
(244)

The full integral form of the coefficient is

$\displaystyle
c_{4:4}^{(2,2)}=\frac{1}{4}\left[\frac{2^{8\Delta-4}}{\pi^{4}}\right]\int_{-\infty}^{\infty}dp\,\int_{-\infty}^{\infty}dq\,f^{(2,2)}_{4:4}(p,q)\,B_{p}(2\Delta)B_{q}(2\Delta)\,,$
(245)

and the integral over $q$ is trivial to do by means of the same tricks as in
the previous cases. The final integral to be done is

$\displaystyle
c_{4:4}^{(2,2)}=\frac{2^{8\Delta-4}}{\pi^{2}}\int_{-\infty}^{\infty}dp\,B_{p}(2\Delta)B_{p}(2\Delta+1)\,,$
(246)

but this is of the form (272), so

$\displaystyle
c_{4:4}^{(2,2)}=\frac{2^{8\Delta}\left[\Gamma\left(4\Delta+1\right)\right]^{2}}{4\,\Gamma\left(8\Delta+2\right)}\,.$
(247)

##  Appendix B Some useful integrals

We present here some very general Fourier transforms that are needed at
different points of our computations. Defining the function

$f_{n}(z)=\frac{e^{(a+ib)z}}{\cosh^{n}(z)}\,,$ (248)

we can compute the integral over the real line for $|a|<n$ by means of a
rectangular contour, with basis on the real line and height $i\pi$. The only
pole enclosed is that at $z=i\pi/2$, and the periodicity properties of the
$\cosh$ allow to relate the horizontal integrals to obtain

$\int_{-\infty}^{\infty}dx\,f_{n}(x)=\begin{cases}\displaystyle 2\pi
iR_{n}(a,b)\frac{\sinh\left((b+ia)\pi/2\right)}{\cosh(b\pi)-\cos(a\pi)}\,,\quad\text{if
}n\text{ is even}.\vspace{0.2cm}\\\ \displaystyle 2\pi
iR_{n}(a,b)\frac{\cosh\left((b+ia)\pi/2\right)}{\cosh(b\pi)+\cos(a\pi)}\,,\quad\text{if
}n\text{ is odd}.\end{cases}\,$ (249)

Here $R_{n}(a,b)$ is defined as

$R_{n}(a,b)\equiv e^{-i(a+ib)\pi/2}\,{\rm
Res}\left[f_{n}(z),z=i\pi/2\right]~{}.$ (250)

This is a polynomial in $(a,b)$ which can be easily computed for any $n$ with
the aid of a symbolic algebra software. Separating real and imaginary parts,
and rescaling the variables of integration and the parameters we arrive at the
general Fourier transforms for $k=1,2,\dots$,

$\displaystyle\int_{-\infty}^{\infty}dx\,\frac{\cosh(\alpha
x)}{\cosh^{2k-1}(\gamma x)}e^{i\xi
x}=\frac{2\pi/\gamma}{\cosh(\frac{\pi\xi}{\gamma})+\cos(\frac{\pi\alpha}{\gamma})}{\rm
Re}\left[iR_{2k-1}\left(\frac{\alpha}{\gamma},\frac{\xi}{\gamma}\right)\cosh\left((\xi+i\alpha)\frac{\pi}{2\gamma}\right)\right]\,,$
(251) $\displaystyle\int_{-\infty}^{\infty}dx\,\frac{\cosh(\alpha
x)}{\cosh^{2k}(\gamma x)}e^{i\xi
x}=\frac{2\pi/\gamma}{\cosh(\frac{\pi\xi}{\gamma})-\cos(\frac{\pi\alpha}{\gamma})}{\rm
Re}\left[iR_{2k}\left(\frac{\alpha}{\gamma},\frac{\xi}{\gamma}\right)\sinh\left((\xi+i\alpha)\frac{\pi}{2\gamma}\right)\right]\,,$
(252) $\displaystyle\int_{-\infty}^{\infty}dx\,\frac{\sinh(\alpha
x)}{\cosh^{2k-1}(\gamma x)}e^{i\xi x}=\frac{2\pi
i/\gamma}{\cosh(\frac{\pi\xi}{\gamma})+\cos(\frac{\pi\alpha}{\gamma})}{\rm
Im}\left[iR_{2k-1}\left(\frac{\alpha}{\gamma},\frac{\xi}{\gamma}\right)\cosh\left((\xi+i\alpha)\frac{\pi}{2\gamma}\right)\right]\,,$
(253) $\displaystyle\int_{-\infty}^{\infty}dx\,\frac{\sinh(\alpha
x)}{\cosh^{2k}(\gamma x)}e^{i\xi
x}=\frac{2\pi/\gamma}{\cosh(\frac{\pi\xi}{\gamma})-\cos(\frac{\pi\alpha}{\gamma})}{\rm
Im}\left[iR_{2k}\left(\frac{\alpha}{\gamma},\frac{\xi}{\gamma}\right)\sinh\left((\xi+i\alpha)\frac{\pi}{2\gamma}\right)\right]\,.$
(254)

We can play the same game with

$\tilde{f}_{n}(z)=\frac{e^{(a+ib)z}}{\sinh^{n}(z)}\,.$ (255)

In this case, there are poles lying on the contour at $z=0,i\pi$. We
circumvent them in the usual way, thus computing the principal value of the
integrals, and obtain

$\displaystyle\int_{-\infty}^{\infty}dx\,\frac{\cosh(\alpha
x)}{\sinh^{2k-1}(\gamma x)}e^{i\xi x}=\frac{i\pi}{\gamma}{\rm
Im}\left[i\tilde{R}_{2k-1}\left(\frac{\alpha}{\gamma},\frac{\xi}{\gamma}\right)\tanh\left((\xi-i\alpha)\frac{\pi}{2\gamma}\right)\right]\,,$
(256) $\displaystyle\int_{-\infty}^{\infty}dx\,\frac{\cosh(\alpha
x)}{\sinh^{2k}(\gamma x)}e^{i\xi x}=\frac{\pi}{\gamma}{\rm
Re}\left[i\tilde{R}_{2k}\left(\frac{\alpha}{\gamma},\frac{\xi}{\gamma}\right)\coth\left((\xi-i\alpha)\frac{\pi}{2\gamma}\right)\right]\,,$
(257) $\displaystyle\int_{-\infty}^{\infty}dx\,\frac{\sinh(\alpha
x)}{\sinh^{2k-1}(\gamma x)}e^{i\xi x}=\frac{\pi}{\gamma}{\rm
Re}\left[i\tilde{R}_{2k-1}\left(\frac{\alpha}{\gamma},\frac{\xi}{\gamma}\right)\tanh\left((\xi-i\alpha)\frac{\pi}{2\gamma}\right)\right]\,,$
(258) $\displaystyle\int_{-\infty}^{\infty}dx\,\frac{\sinh(\alpha
x)}{\sinh^{2k}(\gamma x)}e^{i\xi x}=\frac{i\pi}{\gamma}{\rm
Im}\left[i\tilde{R}_{2k}\left(\frac{\alpha}{\gamma},\frac{\xi}{\gamma}\right)\coth\left((\xi-i\alpha)\frac{\pi}{2\gamma}\right)\right]\,,$
(259)

where, in this case,

$\tilde{R}_{n}(a,b)\equiv{\rm Res}\left[\tilde{f}_{n}(z),z=0\right]~{}.$ (260)

We will be interested in particular cases of the previous Fourier transforms,
trivially obtained from the previous general expressions. Let us list them
here. The simplest ones are

$\displaystyle\int_{-\infty}^{\infty}dx\,\frac{e^{i\xi
x}}{\cosh(x/2)}=2\int_{-\infty}^{\infty}dx\,\frac{e^{x/2}}{e^{x}+1}e^{i\xi
x}=\frac{2\pi}{\cosh(\pi\xi)}~{},$ (261)
$\displaystyle\int_{-\infty}^{\infty}dx\,\frac{e^{i\xi
x}}{\sinh(x/2)}=2\int_{-\infty}^{\infty}dx\,\frac{e^{x/2}}{e^{x}-1}e^{i\xi
x}=2\pi i\tanh(\pi\xi)~{}.$ (262)

Derivatives of these functions allow to include polynomials in $x$. As an
example, we will need the integral

$\int_{-\infty}^{\infty}dx\,\frac{xe^{\frac{x}{2}}}{e^{x}-1}e^{i\xi
x}=\frac{\pi^{2}}{\cosh^{2}(\pi\xi)}~{}.$ (263)

We list some other relevant integrals for the computations in Appendix A,

$\displaystyle\int_{-\infty}^{\infty}dx\,\frac{e^{i\xi
x}}{\cosh^{2}(x/2)}=\frac{4\pi\xi}{\sinh(\pi\xi)}~{},$
$\displaystyle\int_{-\infty}^{\infty}dx\,\frac{e^{i\xi
x}}{\cosh^{3}(x/2)}=(1+4\xi^{2})\frac{\pi}{\cosh(\pi\xi)}\,,$ (264)
$\displaystyle\int_{-\infty}^{\infty}dx\,\frac{e^{i\xi
x}}{\sinh^{2}(x/2)}=-\frac{4\pi\xi}{\tanh(\pi\xi)}~{},$
$\displaystyle\int_{-\infty}^{\infty}dx\,\frac{e^{i\xi
x}}{\sinh^{3}(x/2)}=-i\pi(1+4\xi^{2})\tanh(\pi\xi)\,,$ (265)
$\displaystyle\int_{-\infty}^{\infty}dx\,\frac{\cosh(x/2)}{\sinh(x/2)}e^{i\xi
x}=\frac{2\pi i}{\tanh(\pi\xi)}~{},$
$\displaystyle\int_{-\infty}^{\infty}dx\,\frac{\cosh(x/2)}{\sinh^{2}(x/2)}e^{i\xi
x}=-4\pi\xi\tanh(\pi\xi)\,,$ (266)
$\displaystyle\int_{-\infty}^{\infty}dx\,\frac{\sinh(x/2)}{\cosh^{2}(x/2)}e^{i\xi
x}=\frac{4\pi i\xi}{\cosh(\pi\xi)}\,.$ (267)

In Appendix A, we follow essentially two strategies when dealing with the
integrals. Whenever we find it computationally feasible, we work with the beta
functions for general argument. When doing this, it is convenient to remember
the integral representations of the beta function defined in eq. (170),

$B_{q}(\Delta)=\int_{0}^{1}dt\,t^{\Delta-1}(1-t)^{\Delta-1}e^{\frac{iq}{2\pi}\log\left(\frac{t}{1-t}\right)}=\frac{1}{4^{\Delta}}\int_{-\infty}^{\infty}dx\frac{e^{i\frac{q}{2\pi}x}}{\cosh^{2\Delta}(x/2)}\,.$
(268)

We will refer to the first one as the _logarithmic_ representation, while the
second one will be the _hyperbolic_ one. From properties of the gamma function
it is trivial to show that

$B_{q}(\Delta+1)=\frac{1}{4\pi^{2}(2\Delta)(2\Delta+1)}\left(q^{2}+4\pi^{2}\Delta^{2}\right)B_{q}(\Delta)\,.$
(269)

From the logarithmic representation (or from the hyperbolic one, for what
matters), and using the previous Fourier transforms, it is possible to prove a
number of useful identities. The first one is

$\int_{-\infty}^{\infty}dp\,\frac{e^{\frac{p\pm q}{2}}\left(p\pm
q\right)}{e^{p\pm q}-1}\,B_{p}(\Delta)=4\pi^{2}B_{q}(\Delta+1)\,,$ (270)

which follows directly from eq. (263) after introducing the beta function
integral representation. From eq. (262) we similarly obtain

$\int_{-\infty}^{\infty}dp\,\frac{e^{\frac{p\pm q}{2}}}{e^{p\pm
q}-1}\,B_{p}(\Delta)=\pm\frac{q}{2\Delta}B_{q}(\Delta)\,.$ (271)

It is also straightforward to obtain the identity

$\int_{-\infty}^{\infty}dq\,B_{q+Q}(\Delta)B_{q+\bar{Q}}(\bar{\Delta})=4\pi^{2}B_{Q-\bar{Q}}(\Delta+\bar{\Delta})\,,$
(272)

Finally, a bit more involved is the following result

$\int_{-\infty}^{\infty}dq\,q\,e^{-\frac{q}{2}}B^{2}_{q}(\Delta_{1})B_{q}(\Delta_{2})=-\frac{2^{4-2\Delta_{1}-2\Delta_{2}}\pi^{4}\Gamma\left(\Delta_{1}+\frac{1}{2}\right)\left(\Gamma\left(\Delta_{1}+\Delta_{2}\right)\right)^{2}}{\Gamma\left(\Delta_{1}\right)\Gamma\left(\Delta_{2}\right)\Gamma\left(2\Delta_{1}+\Delta_{2}+\frac{1}{2}\right)}\,.$
(273)

It can be proved to be valid by considering the integral

$I(\alpha)=\int_{-\infty}^{\infty}dq\,e^{i\alpha
q}B^{2}_{q}(\Delta_{1})B_{q}(\Delta_{2})\,,$ (274)

which can be integrated over $q$ and one of the variables appearing in the
representation of the beta function straightforwardly for ${\rm
Im}(\alpha)\in(-1/2,1/2)$. Using the trick presented in Appendix B of [82],
taking $-i\partial_{\alpha}$ to recover the $q$ term in the integrand of eq.
(273), and approaching $\alpha\to i/2$ from below, it is possible to reproduce
the value of the integral in terms of gamma functions.

This last computation shows very clearly that things become arbitrarily
involved when one starts adding beta functions and/or more complicated
prefactors. At some point in the computation of the $c_{4:4}^{(1,1,1,1)}$
coefficient in Appendix A, we were not able to carry out the integrals in full
generality, for any value of $\Delta$. In those cases, we will need to restore
to another approach, still able to generate results for integer or semi-
integer $\Delta$, but in an algorithmic way. This approach is therefore well-
suited for implementation in a standard symbolic algebra software, and it is
based on the following observation. For $a=0$, the $R_{n}(a,b)$ notably
simplify,

$\displaystyle
R_{2k+1}(0,p/\pi)=-\frac{i}{(2k)!\,\pi^{2k}}\prod_{l=0}^{k-1}\left(p^{2}+(2l+1)^{2}\pi^{2}\right)\,,$
(275) $\displaystyle
R_{2k}(0,p/\pi)=-\frac{ip}{(2k-1)!\,\pi^{2k-1}}\prod_{l=1}^{k-1}\left(p^{2}+4l^{2}\pi^{2}\right)\,.$
(276)

From the hyperbolic representation of the beta function, it is then immediate
to obtain

$\displaystyle
B_{p}\left(\frac{2k+1}{2}\right)^{n}=\frac{\pi^{(1-2k)n}}{2^{2kn}(2k)!^{n}}\left[\prod_{l=0}^{k-1}\left(p^{2}+(2l+1)^{2}\pi^{2}\right)\right]\frac{1}{\cosh^{n}(p/2)}\,,$
(277) $\displaystyle
B_{p}\left(k\right)^{n}=\frac{\pi^{2(1-k)n}}{2^{(2k-1)n}(2k-1)!^{n}}\left[\prod_{l=1}^{k-1}\left(p^{2}+4l^{2}\pi^{2}\right)\right]\frac{p^{n}}{\sinh^{n}(p/2)}\,.$
(278)

It is also possible to obtain these results by repeated application of (269),
knowing the seed values

$B_{p}(1/2)=\frac{\pi}{\cosh(p/2)}\,,\qquad
B_{p}(1)=\frac{1}{2}\frac{p}{\sinh(p/2)}~{}.$ (279)

##  Appendix C Identifying the leading scaling in the lattice

In this appendix we present a table with the coefficient of determination
$\mathcal{R}^{2}$ obtained from different linear fits to the continuum lattice
results for $I_{N}$ in the case of the free scalar in the long-distance
regime. We do this for the scaling predicted from our general discussion as
well as for the immediately greater and the immediately smaller power. In all
cases, we observe that the scaling predicted by eq. (124) is strongly
favoured.

$N=2$ | Blocks | $\mathcal{R}^{2}[1/x]$ | $\mathcal{R}^{2}[1/x^{2}]$ | $\mathcal{R}^{2}[1/x^{3}]$ | $N=3$ | $\mathcal{R}^{2}[1/x^{2}]$ | $\mathcal{R}^{2}[1/x^{3}]$ | $\mathcal{R}^{2}[1/x^{4}]$
---|---|---|---|---|---|---|---|---
| $L\times L$ | $0.99759$ | $\mathbf{1}.$ | $0.99772$ |  | $0.99387$ | $\mathbf{0.99994}$ | $0.99229$
| Circles | $0.99761$ | $\mathbf{1}.$ | $0.99770$ |  | $0.99378$ | $\mathbf{0.99995}$ | $0.99239$
| $L\times 2L$ | $0.99763$ | $\mathbf{1}.$ | $0.99769$ |  | $0.99366$ | $\mathbf{0.99996}$ | $0.99252$
| $L\times 4L$ | $0.99764$ | $\mathbf{1}.$ | $0.99768$ |  | $0.99350$ | $\mathbf{0.99997}$ | $0.99269$
| $L\times 6L$ | $0.99764$ | $\mathbf{1}.$ | $0.99767$ |  | $0.99345$ | $\mathbf{0.99997}$ | $0.99275$

$N=4$ | Blocks | $\mathcal{R}^{2}[1/x^{3}]$ | $\mathcal{R}^{2}[1/x^{4}]$ | $\mathcal{R}^{2}[1/x^{5}]$ | $N=5$ | $\mathcal{R}^{2}[1/x^{4}]$ | $\mathcal{R}^{2}[1/x^{5}]$ | $\mathcal{R}^{2}[1/x^{6}]$
---|---|---|---|---|---|---|---|---
| $L\times L$ | $0.99630$ | $\mathbf{0.99974}$ | $0.99479$ |  | $0.99726$ | $\mathbf{0.99884}$ | $0.98852$
| Circles | $0.99613$ | $\mathbf{0.99975}$ | $0.99498$ |  | $0.99655$ | $\mathbf{0.99920}$ | $0.98984$
| $L\times 2L$ | $0.99589$ | $\mathbf{0.99977}$ | $0.99524$ |  | $0.99599$ | $\mathbf{0.99960}$ | $0.99103$
| $L\times 4L$ | $0.99558$ | $\mathbf{0.99978}$ | $0.99555$ |  | $0.99474$ | $\mathbf{0.99987}$ | $0.99268$
| $L\times 6L$ | $0.99546$ | $\mathbf{0.99978}$ | $0.99567$ |  | $0.99422$ | $\mathbf{0.99993}$ | $0.99326$

$N=6$ | Blocks | $\mathcal{R}^{2}[1/x^{5}]$ | $\mathcal{R}^{2}[1/x^{6}]$ | $\mathcal{R}^{2}[1/x^{7}]$
---|---|---|---|---
| $L\times L$ | $0.99903$ | $\mathbf{0.99968}$ | $0.99665$
| Circles | $0.99873$ | $\mathbf{0.99979}$ | $0.99716$
| $L\times 2L$ | $0.99944$ | $\mathbf{0.99955}$ | $0.99598$
| $L\times 4L$ | $0.99812$ | $\mathbf{0.99987}$ | $0.99791$
| $L\times 6L$ | $0.99903$ | $\mathbf{0.99981}$ | $0.99686$

Table 2: We show the values of the coefficient of determination,
$\mathcal{R}^{2}$, of different fits to the long-distance results obtained for
$I_{N}$, with $N=2,3,4,5,6$, for a free scalar field in the lattice. In each
case, we consider sets of $N$ identical entangling regions (squares, disks and
increasingly thinner rectangles) placed on a regular $N$-gon arrangement.
$\mathcal{R}^{2}$ is computed for fits of the form $I_{N}\sim(1/x)^{\alpha}$
with $\alpha=N-1,N,N+1$.

##  Appendix D Analytic mutual-information toy models

In this appendix we present two toy models which resemble the mutual
information results corresponding to free scalars and fermions, respectively,
while allowing for analytic calculations. For both models, we present analytic
formulas for the mutual information corresponding to the different shapes
considered in Section 4.3.

###  D.1 EMI model

The mutual information is a positive semi-definite quantity for general
regions and theories, $I_{2}\geq 0$. On the other hand, as we mentioned in the
introduction, the tripartite information does not have a definite sign and, in
fact, there exist theories for which $I_{3}(A_{1},A_{2},A_{3})\geq 0$ and
others with $I_{3}(A_{1},A_{2},A_{3})\leq 0$, corresponding to “subextensive”
and “superextensive” mutual informations, respectively. On the other hand, a
theory with a vanishing tripartite information for arbitrary regions,
$I_{3}(A_{1},A_{2},A_{3})=0$ for all $A_{1},A_{2},A_{3}$, would have a mutual
information which is extensive in its arguments, since in that case
$I_{2}(A_{1},A_{2}\cup A_{3})=I_{2}(A_{1},A_{2})+I_{2}(A_{1},A_{3})$.

Interestingly, imposing $I_{3}(A_{1},A_{2},A_{3})=0$ for general regions along
with the set of known axioms satisfied by mutual information in a relativistic
QFT, strongly restricts the form of the mutual information in general
dimensions. The result defines the so-called “Extensive Mutual Information”
(EMI) model. Further imposing conformal invariance, the expression for its
mutual information is fully determined up to a global multiplicative constant
and reads [39]181818One can also arrive at the EMI formula by assuming the
Rényi twist operators to be exponentials of free fields [83]. The EMI model
formula can also be interpreted naturally as a “partial entanglement entropy”
[84].

$I_{\rm\scriptscriptstyle EMI}(A_{1},A_{2})=2\kappa_{(d)}\,\int_{\partial
A_{1}}d\sigma_{1}\,\int_{\partial A_{2}}d\sigma_{2}\,\frac{(n_{1}\cdot
n_{2})(\bar{n}_{1}\cdot\bar{n}_{2})-(n_{1}\cdot\bar{n}_{2})(\bar{n}_{1}\cdot
n_{2})}{|x_{1}-x_{2}|^{2(d-2)}}\,,$ (280)

where $\kappa_{(d)}$ is such (positive) constant and $\bar{n}_{1}$ and $n_{1}$
are unit vectors orthogonal to the surface and to each other,
$n_{1}\cdot\bar{n}_{1}=0$. For fixed time slices, one can choose
$n_{1}=n_{2}=\hat{t}$ and the integrand reduces to
$-(\bar{n}_{1}\cdot\bar{n}_{2})/|x_{1}-x_{2}|^{2(d-2)}$.

In spite of its simplicity, the EMI model produces physically reasonable
results in general dimensions [52, 13, 85, 57, 86, 58]. Such results are
rather similar to the free fermion ones and, in fact, the EMI model exactly
coincides with a free massless fermion in $d=2$ [87].191919Besides, as shown
in [24], the result for the mutual information of two arbitrarily boosted
spheres in the EMI model exactly matches the contribution from the free
fermion leading conformal block in general dimensions. However, it does not
contain all the rest of conformal blocks present in the free fermion theory.
On the other hand, using the long-distance expansion of the mutual information
to bootstrap its putative operatorial content, it has been shown that the EMI
model cannot describe the mutual information of any actual CFT in higher
dimensions [24].202020This implies that the set of known axioms satisfied by
EE in QFT is incomplete. Still, as we have said, the EMI model produces
reasonable results which tend to resemble the free fermion ones in most
situations, so we find it useful to include its predictions in the present
cases. In particular, in general dimensions the long-distance expression for
the EMI model for two regions in the same time slice takes the universal form

$I_{\rm\scriptscriptstyle EMI}(A_{1},A_{2})\overset{r\gg
R_{A_{1},A_{2}}}{=}4\kappa_{(d)}(d-1)(d-2)\frac{{\rm vol}(A_{1})\cdot{\rm
vol}(A_{2})}{r^{2(d-1)}}+\dots$ (281)

which is the same scaling as the one corresponding to a free fermion.

Disk regions. The result for the mutual information of two identical disks of
diameters $R$ and separated a distance $r$ in the three-dimensional EMI model
can be extracted from the result presented in [24], where it was computed in
the more general case of boosted spheres in general dimensions. Defining $r$
as the distance between disks such that $r\rightarrow 0$ when the disks touch
at a point and $x\equiv r/R$, one finds

$I_{\rm\scriptscriptstyle
EMI}(A_{1},A_{2})=2\pi^{2}\kappa_{(3)}\frac{\left[\sqrt{x(2+x)}-2x(2+x)(1+x-\sqrt{x(2+x)})\right]}{x(1+x)(2+x)}\,.$
(282)

In the almost-touching regime, as well as in the long-distance one, one finds

$\displaystyle I_{\rm\scriptscriptstyle EMI}$
$\displaystyle\overset{x\rightarrow
0}{=}\frac{\sqrt{2}\pi^{2}\kappa_{(3)}}{\sqrt{x}}-4\pi^{2}\kappa_{(3)}+\mathcal{O}(\sqrt{x})\,,$
(283) $\displaystyle I_{\rm\scriptscriptstyle EMI}$
$\displaystyle\overset{x\rightarrow\infty}{=}\frac{\pi^{2}\kappa_{(3)}}{2x^{4}}-\frac{2\pi^{2}\kappa_{(3)}}{x^{5}}+\mathcal{O}(1/x^{6})\,.$
(284)

On the other hand, the EE disk coefficient $F_{0}$, which we use to normalize
our curves in Section 4.3, is given for the EMI model by
$F_{0}^{\rm\scriptscriptstyle EMI}=2\pi^{2}\kappa_{(3)}$.

Rectangle regions. Let us now consider the other case of interest for us,
which corresponds to two identical rectangles of sides $\xi R\times R$
separated along the $X$ direction a distance $r$ —measured between the
rightmost edge of the left rectangle and the leftmost edge of the right one.
In the $r\rightarrow 0$ limit, the two rectangles would touch each other. The
boundary of the leftmost rectangle is defined as $\partial
A=\\{[X,0],X\in[0,\xi R];[X,R],X\in[0,\xi R];[0,Y],Y\in[0,R];[\xi
R,Y],Y\in[0,R]\\}$. The boundary of $B$ is obtained by shifting the resulting
figure along the $X$ axis a distance $\xi R+r$, $\partial B=\\{[X,0],X\in[\xi
R+r,2\xi R+r];[X,R],X\in[\xi R+r,2\xi R+r];[\xi R+r,Y],Y\in[0,R];[2\xi
R+r,Y],Y\in[0,R]\\}$.

In principle there would be 16 contributions to the mutual information coming
from the $4\times 4$ pairings of segments from the two rectangles. However,
observe that 8 of them vanish, corresponding to segments perpendicular to each
other. It is not difficult to see that the remaining contributions can be
written as

$I_{\rm\scriptscriptstyle
EMI}(A_{1},A_{2})=-2\kappa_{(3)}\left[2\Delta_{1}+\Delta_{2}+\Delta_{3}+2\Delta_{4}+2\Delta_{5}\right]\,,$
(285)

where

$\displaystyle\Delta_{1}$ $\displaystyle=\Delta[R,0,R,r+\xi
R]\,,\quad\Delta_{2}=-\Delta[R,0,R,r+2\xi
R]\,,\quad\Delta_{3}=-\Delta[R,0,R,r]\,,$ (286) $\displaystyle\Delta_{4}$
$\displaystyle=\Delta[\xi R,\xi R+r,2\xi R+r,0]\,,\quad\Delta_{5}=-\Delta[\xi
R,\xi R+r,2\xi R+r,R]\,,$ (287)

and we defined

$\Delta[\alpha,\beta,\gamma,\chi]\equiv\int_{0}^{\alpha}\int_{\beta}^{\gamma}\frac{dx_{1}dx_{2}}{\chi^{2}+(x_{1}-x_{2})^{2}}\,.$
(288)

Performing the integrals and simplifying a bit, one finds the final result

$\displaystyle I_{\rm\scriptscriptstyle
EMI}=4\kappa_{(3)}\left[\frac{\operatorname{arccot}(x)}{x}-\frac{2\operatorname{arccot}(x+\xi)}{x+\xi}+\frac{\operatorname{arccot}(x+2\xi)}{x+2\xi}-2(x+\xi)\arctan(x+\xi)\right.$
$\displaystyle\left.+x\arctan(x)+(x+2\xi)\arctan(x+2\xi)-\log\left[\frac{(x+\xi)^{4}(1+x^{2})(1+(x+2\xi)^{2})}{x^{2}(x+2\xi)^{2}(1+(x+\xi)^{2})^{2}}\right]\right]\,,$
(289)

where we defined $x\equiv r/R$. For almost-touching rectangles and in the
long-distance regime one finds, respectively,

$\displaystyle I_{\rm\scriptscriptstyle EMI}$
$\displaystyle\overset{x\rightarrow 0}{=}\frac{k_{\rm\scriptscriptstyle
EMI}^{(3)}}{x}+8\kappa_{(3)}\log x+g(\xi)+\mathcal{O}(x)\,,$ (290)
$\displaystyle I_{\rm\scriptscriptstyle EMI}$
$\displaystyle\overset{x\rightarrow\infty}{=}\frac{8\kappa_{(3)}\xi^{2}}{x^{4}}-\frac{32\kappa_{(3)}\xi^{3}}{x^{5}}+\mathcal{O}(x^{-6})\,,$
(291)

where $k_{\rm\scriptscriptstyle EMI}^{(3)}=2\pi\kappa_{(3)}$ is the
coefficient appearing in the entanglement entropy of a strip region, $g(\xi)$
is a complicated function of $\xi$ and $\xi^{2}/x^{4}$ in the second
expression is the product of the areas of the two rectangles $A,B$ divided by
their separation to the fourth power, $\xi^{2}/x^{4}={\rm area}(A){\rm
area}(B)/r^{4}$, as expected from the general formula, eq. (281).

###  D.2 MEMI model

A simple modified version of the EMI model can be constructed such that the
long-distance scaling reproduces the one corresponding to a free scalar field.
This “Modified” EMI (MEMI) model is defined, in the case of a fixed time
slice, by a mutual information of the form

$I_{\rm\scriptscriptstyle MEMI}(A_{1},A_{2})=2\mu_{(d)}\,\int_{\partial
A_{1}}d\sigma_{1}\,\int_{\partial
A_{2}}d\sigma_{2}\,\frac{1}{|x_{1}-x_{2}|^{2(d-2)}}\,,$ (292)

where $\mu_{(d)}$ is a positive constant. It is easy to see that, for long
separations, this behaves as

$I_{\rm\scriptscriptstyle MEMI}(A_{1},A_{2})\overset{r\gg
R_{A_{1},A_{2}}}{=}2\mu_{(d)}\,\frac{{\rm area}(\partial A_{1})\cdot{\rm
area}(\partial A_{2})}{r^{2(d-2)}}\,,$ (293)

which is indeed the same scaling as for the free scalar. Note that this
formula seems problematic in the case of regions $A_{1}$, $A_{2}$ which have
very small volumes but are bounded by finite area surfaces. In that situation,
one would expect the mutual information to vanish, a behavior which is not
reproduced by eq. (293), since $\mu_{(d)}$ is a fixed constant. In other
words, the MEMI model does not obey the monotonicity property of the mutual
information and thus, it can not describe an actual mutual information for
general geometric regions.

On the other hand, interestingly, we have verified that the MEMI model for the
case of disjoint spheres is proportional to the conformal block of an
intermediate scalar operator with scaling dimension $d-2$. In those cases the
MEMI model can thus be interpreted as the resumed contribution of the replica
primary operator $\phi^{i}\phi^{j}$ and its descendants to the mutual
information of a theory with a free scalar $\phi$ in its spectrum. A similar
statement holds for the EMI model, where the associated replica operator is
the conserved current $\bar{\psi}_{i}\gamma_{\mu}\psi_{j}$ associated to a
free fermion [24]. It would be interesting to explore the extent to which
similar statements might exist for higher spin generalizations or interacting
fields. Another unusual feature of the MEMI model is that the universal term
in the EE of a disk region turns out to vanish for this model, i.e.,
$F_{0}^{\rm\scriptscriptstyle MEMI}=0$. Hence, when presenting the results for
this model we will normalize the mutual information in a way such that at
short distances the result tends to the free scalar one (in the main text, we
denote the model with such normalization by MEMI⋆).

Disk regions. In the case of two identical disks of diameters $R$ separated a
distance $r$ defined as in the EMI case above, we have for the MEMI model

$I_{\rm\scriptscriptstyle
MEMI}=\frac{2\pi^{2}\mu_{(3)}}{(x+1)\sqrt{x(x+2)}}\,.$ (294)

For small and large separations one finds, respectively,

$\displaystyle I_{\rm\scriptscriptstyle MEMI}$
$\displaystyle\overset{x\rightarrow
0}{=}\frac{\sqrt{2}\pi^{2}\mu_{(3)}}{\sqrt{x}}-\frac{5\pi^{2}\mu_{(3)}\sqrt{x}}{2\sqrt{2}}+\mathcal{O}(x^{3/2})\,,$
(295) $\displaystyle I_{\rm\scriptscriptstyle MEMI}$
$\displaystyle\overset{x\rightarrow\infty}{=}\frac{2\pi^{2}\mu_{(3)}}{x^{2}}-\frac{4\pi^{2}\mu_{(3)}}{x^{3}}+\mathcal{O}(1/x^{4})\,.$
(296)

Rectangle regions. Following analogous steps to the EMI model ones, one can
also obtain the mutual information corresponding to two rectangle regions of
sides $\xi R\times R$ in the MEMI model. The final result reads

$\displaystyle I_{\rm\scriptscriptstyle MEMI}=4\mu_{(3)}\Big{[}$
$\displaystyle+\frac{\operatorname{arccot}(x)}{x}+\frac{2\operatorname{arccot}(x+\xi)}{x+\xi}+\frac{\operatorname{arccot}(x+2\xi)}{x+2\xi}-2(x+\xi)\arctan(x+\xi)$
$\displaystyle+x\arctan(x)+(x+2\xi)\arctan(x+2\xi)+\log\left[\frac{(x+\xi)^{4}}{(1+x^{2})(1+(x+2\xi)^{2})}\right]$
(297)
$\displaystyle+2\int_{0}^{\infty}dt\,\,te^{t}\left[\frac{x}{e^{2t}x^{2}+1}-\frac{2\xi+x}{e^{2t}(x+2\xi)^{2}+1}\right]\Big{]}\,.$

This behaves respectively, in the almost-touching and long-distance regimes as

$\displaystyle I_{\rm\scriptscriptstyle MEMI}$
$\displaystyle\overset{x\rightarrow 0}{=}\frac{k_{\rm\scriptscriptstyle
MEMI}^{(3)}}{x}+\frac{2\pi\mu_{(3)}}{\xi}\log
x+\tilde{g}(\xi)+\mathcal{O}(x)\,,$ (298) $\displaystyle
I_{\rm\scriptscriptstyle MEMI}$
$\displaystyle\overset{x\rightarrow\infty}{=}\frac{8\mu_{(3)}(1+\xi)^{2}}{x^{2}}-\frac{16\mu_{(3)}\xi(1+\xi)^{2}}{x^{3}}+\mathcal{O}(x^{-4})\,,$
(299)

where $\tilde{g}(\xi)$ is some complicated function of $\xi$.

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