Content-Type: multipart/mixed; boundary="-------------0105151125931" This is a multi-part message in MIME format. ---------------0105151125931 Content-Type: text/plain; name="01-178.comments" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="01-178.comments" We show in detail how to calculate the integral representing correlation funcitons in XXX spin 1/2 Heisenberg spin chain ---------------0105151125931 Content-Type: text/plain; name="01-178.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="01-178.keywords" Exactly Solvable Models, Correlation Functions, Quantum Spin Chains, Riemann Zeta Function, XXX Heisenberg Antiferromagnet, Bethe Ansatz ---------------0105151125931 Content-Type: application/x-tex; name="eval.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="eval.tex" \documentstyle[12pt]{article} \topmargin -.5cm \textheight 23cm \textwidth 175mm \hoffset -17mm \makeatletter \def\eqnarray{\stepcounter{equation}\let\@currentlabel=\theequation \global\@eqnswtrue \global\@eqcnt\z@\tabskip\@centering\let\\=\@eqncr $$\halign to \displaywidth\bgroup\@eqnsel\hskip\@centering $\displaystyle\tabskip\z@{##}$&\global\@eqcnt\@ne \hfil$\displaystyle{\hbox{}##\hbox{}}$\hfil &\global\@eqcnt\tw@ $\displaystyle\tabskip\z@ {##}$\hfil\tabskip\@centering&\llap{##}\tabskip\z@\cr} \def\@sect#1#2#3#4#5#6[#7]#8{\ifnum #2>\c@secnumdepth \def\@svsec{}\else \refstepcounter{#1}\edef\@svsec{\csname the#1\endcsname.\hskip 1em }\fi \@tempskipa #5\relax \ifdim \@tempskipa>\z@ \begingroup #6\relax \@hangfrom{\hskip #3\relax\@svsec}{\interlinepenalty \@M #8\par} \endgroup \csname #1mark\endcsname{#7}\addcontentsline {toc}{#1}{\ifnum #2>\c@secnumdepth \else \protect\numberline{\csname the#1\endcsname}\fi #7}\else \def\@svsechd{#6\hskip #3\@svsec #8\csname #1mark\endcsname {#7}\addcontentsline {toc}{#1}{\ifnum #2>\c@secnumdepth \else \protect\numberline{\csname the#1\endcsname}\fi #7}}\fi \@xsect{#5}} \def\label#1{\@bsphack\if@filesw {\let\thepage\relax \xdef\@gtempa{\write\@auxout{\string \newlabel{#1}{{\thesection.\@currentlabel}{\thepage}}}}}\@gtempa \if@nobreak \ifvmode\nobreak\fi\fi\fi\@esphack} \def\@eqnnum{(\thesection.\theequation)} \def\section{\setcounter{equation}{0} \@startsection {section}{1}{\z@} {-3.5ex plus -1ex minus -.2ex}{2.3ex plus .2ex}{\Large\bf}} \newcount\@minsofar \newcount\@min \newcount\@cite@temp \def\@citex[#1]#2{% \if@filesw \immediate \write \@auxout {\string \citation {#2}}\fi \@tempcntb\m@ne \let\@h@ld\relax \def\@citea{}% \@min\m@ne% \@cite{% \@for \@citeb:=#2\do {\@ifundefined {b@\@citeb}% {\@h@ld\@citea\@tempcntb\m@ne{\bf ?}% \@warning {Citation `\@citeb ' on page \thepage \space undefined}}% {\@minsofar\z@ \@for \@scan@cites:=#2\do {% \@ifundefined{b@\@scan@cites}% {\@cite@temp\m@ne} {\@cite@temp\number\csname b@\@scan@cites \endcsname \relax}% \ifnum\@cite@temp > \@min% select the next one to list \ifnum\@minsofar = \z@ \@minsofar\number\@cite@temp \edef\@scan@copy{\@scan@cites}\else \ifnum\@cite@temp < \@minsofar \@minsofar\number\@cite@temp \edef\@scan@copy{\@scan@cites}\fi\fi\fi}\@tempcnta\@min \ifnum\@minsofar > \z@ % some more \advance\@tempcnta\@ne \@min\@minsofar \ifnum\@tempcnta=\@minsofar % Number follows previous--hold on to it \ifx\@h@ld\relax \edef \@h@ld{\@citea\csname b@\@scan@copy\endcsname}% \else \edef\@h@ld{\ifmmode{-}\else--\fi\csname b@\@scan@copy\endcsname}% \fi \else \@h@ld\@citea\csname b@\@scan@copy\endcsname \let\@h@ld\relax \fi % no more \fi}% \def\@citea{,\penalty\@highpenalty\,}}\@h@ld}{#1}} \def\appendixname{Appendix} \def\appendix{\par \def\pre@section{\appendixname{}} \setcounter{section}{1} \@addtoreset{equation}{section} \def\thesection{\Alph{section}} \def\theequation{\arabic{equation}}} \def\appendix{\par \def\pre@section{\appendixname{}} \setcounter{section}{1} \@addtoreset{equation}{section} \def\thesection{\Alph{section}} \def\theequation{\arabic{equation}}} \makeatother \def\b{\beta} \def\d{\delta} \def\g{\gamma} \def\a{\alpha} \def\s{\sigma} \def\t{\tau} \def\th{\theta} \def\l{\lambda} \def\e{\epsilon} \def\r{\rho} \def\wid{\widehat} \def\ds{\displaystyle} \def\be{\begin{equation}} \def\ee{\end{equation}} \def\beq{\begin{eqnarray}} \def\eeq{\end{eqnarray}} \def\ov{\overline} \def\om{\omega} \begin{document} \begin{center} \bf {Evaluation of Integrals Representing Correlations in XXX Heisenberg Spin Chain} \end{center} \phantom{a} \vspace{1.5cm} \centerline{H.E. Boos\footnote{ E-mail: boos@mx.ihep.su}} \centerline{ Institute for High Energy Physics} \centerline{ Protvino, 142284, Russia} \phantom{a} \vspace{0.5cm} \phantom{a} \centerline{V.E. Korepin \footnote{E-mail: korepin@insti.physics.sunysb.edu}} \centerline{C.N.~Yang Institute for Theoretical Physics} \centerline{State University of New York at Stony Brook} \centerline{Stony Brook, NY 11794--3840, USA} \vspace{1.5cm} \vskip2em \begin{abstract} \noindent %Riemann zeta function is an important object of number theory. %It was also used for description of disordered systems in statistical %mechanics. We show that Riemann zeta function is also useful for the %description of %integrable models. We study XXX Heisenberg spin 1/2 anti-ferromagnet. We evaluate a probability of formation of a ferromagnetic string in the anti-ferromagnetic ground state in thermodynamics limit. We prove that for short strings the probability can be expressed in terms of Riemann zeta function with odd arguments. \end{abstract} \newpage \section{Introduction} Riemann zeta function for $Re (s) > 1 $ can be defined as follows: \be \zeta (s) = \sum_{n=1}^{\infty}\frac{1}{n^s} \label{zeta} \ee It also can be represented as a product with respect to all prime numbers $p$ \be \zeta (s) = \prod_{p} ( 1-p^{-s} ) ^{-1} \ee It can be analytically continued in the whole complex plane of $s$. It has only one pole, at $s=1$. Riemann zeta function satisfies a functional equation \be \zeta (s) = 2^s \pi^{s-1} \sin ( {s\pi /2}) \Gamma (1-s) \zeta (1-s) \ee It has 'trivial' zeros at $s=-2n$ ( $n>1$ is an integer). The famous Riemann hypothesis \cite{R} states that nontrivial zeros belong to the straight line $Re (s) = 1/2$. Recently Montgomery and Odlyzko conjectured that for large values of imaginary part of $s$ the distribution of zeros can be described by GUE of random matrices. Forrester and Odlyzko related the problem of distribution of zeros to Painleve differential equation and integrable integral operators \cite{fo}. Riemann zeta function is useful for study of distribution of prime numbers on the real axis \cite{TIT}. The values of Riemann zeta function at special points were studied in \cite{za} , \cite{bbbl} . At even values of its argument zeta function can be expressed in terms of powers of $\pi$ and Bernoulli's numbers \be \zeta (2n) = (-1)^{n+1} 2^{2n-1} \pi ^{2n} B_{2n} /(2n)! \label{zeta2n} \ee The values of Riemann zeta function at odd arguments provide infinitely many different irrational numbers \cite{tr} . % Mathematicians believe %that at odd values of its argument $\zeta (n)$ is an irrational %number [ this was proven for %$\zeta (3)$]. Riemann zeta function plays an important role, not only in pure mathematics %\cite{za} ,\cite{bbbl} but also theoretical physics. Some Feynman diagrams in quantum field theory can be expressed in terms of $\zeta (n)$, see, for example, \cite{KR} . %It appears also in string theory \cite{string}. In statistical mechanics Riemann zeta function was used for the description of chaotic systems \cite{kna}. %This is large field with many publications. %Important contributions to this field were made by Berry, Connes, %Julia, Kac, Katz, %Knauf, Pitkanen, Polya, Ruelle, Sarnak and Zagier. % This is definitely incomplete list. %We apologize to the authors, whose names we did not mention. One can find more information and citation on the following web-cite {http://www.maths.ex.ac.uk/~mwatkins/} . We argue that $\zeta (n)$ is also important for exactly solvable models. The most famous integrable models is the Heisenberg XXX spin chain. This model was first suggested by Heisenberg \cite{Heis} in 1928 and solved by Bethe \cite{B} in 1931 . Since that time it found multiple applications in solid state physics and statistical mechanics. %Recently the XXX spin chain was used for study of the entanglement in %quantum computations \cite{comp}. The Hamiltonian of the XXX spin chain can be written like this \be H = \sum_{i=1}^N \> (\s^x_i\s^x_{i+1}\; + \;\s^y_i\s^y_{i+1}\; +\; \s^z_i\s^z_{i+1}\;-1\;) \label{H} \ee Here $N$ is the length of the lattice and $\s^x_i,\s^y_i,\s^z_i$ are Pauli matrices. We consider thermodynamics limit , when $N$ goes to infinity. The sign in front of the Hamiltonian indicates that we are considering the anti-ferromagnetic case. We consider periodic boundary conditions. Notice that this Hamiltonian annihilates the ferromagnetic state [ all spins up]. The construction of the anti-ferromagnetic ground state wave function $|AFM>$ can be credited to Hulth\'{e}n \cite{H}. An important correlation function was defined in \cite{KIEU}. It was called the emptiness formation probability $$ P(n) = $$ %$P_j=\frac{(\s^z_j+1)}{2}$ where $P_j= (1+\s^z_j)/2$ is a projector on the state with spin up in $j$th lattice site. Averaging is over the anti-ferromagnetic ground state. It describes the probability of formation of a ferromagnetic string of the length $n$ in the anti-ferromagnetic background $|AFM>$ . %Our main results is $P(4)$. In this paper we shall first study short strings ( $n$ is small), in the end we shall discuss long distance asymptotics ( at finite temperature). The four first values of the emptiness-formation probability look as follows: \beq &{\ds P(1)\;=\;{1\over 2}}\;=\;0.5,&\label{P1}\\ &{\ds P(2)\;=\;{1\over 3} (1\; -\; \ln{2})}\;=\;0.102284273,& \label{P2}\\ &{\ds P(3)\;=\;{1\over 4}\; - \;\ln{2}\;+\;{3\over 8}\;\zeta(3)}\;= \;0.007624158,& \label{P3}\\ &{\ds P(4)\;=\;{1\over 5}\; -\; 2\ln{2} \; + \; {173\over 60} \,\zeta(3) \; -\; {11\over 6} \,\zeta(3)\, \ln{2}\;-\; {51\over 80} \, \zeta^2(3) \; }& \nonumber\\ &{\ds -\; {55\over 24}\,\zeta(5) \; + \; {85\over 24}\,\zeta(5)\, \ln{2}\;=\;0.000206270}& %\nonumber\\ %&\quad& \label{P4} \eeq Let us comment. The value of $P(1)$ is evident from the symmetry, $P(2)$ can be extracted from the explicit expression of the ground state energy \cite{H}. $P(3)$ can be extracted from the results of M.Takahashi \cite{T1} on the calculation of the nearest neighbor correlation. It was confirmed in paper \cite{DI}. One should also mention independent calculation of $P(3)$ in \cite{BGSS}. One can express $P(3)$ in terms of next to the nearest neighbor correlation %Namely, %as was shown in \cite{BGSS} one can express $P(3)$ in terms of %next to the nearest neighbor correlation via the formula \be <\;S^z_{i}S^z_{i+2}\;>\;=\;2\,P(3)\;-\;2\,P(2)\;+\;{1\over 2}\,P(1) \label{G2a} \ee The calculation of $P(3)$ and $P(4)$ is discussed in this paper. {\bf The expression above for $P(4)$ is our main result here.} We briefly announced our results in \cite{BK}, here we provide the detailed derivation. The plan of the paper is as follows. In the next section we discuss a general procedure of the calculation of $P(n)$. We also show how this scheme works for $P(2)$. %The thermodynamics of $P(n)$ for the non-zero temperature %is briefly discussed in section 3. In the Appendices A and B we describe in detail the calculation of $P(3)$ and $P(4)$ respectively by means of the technique elaborated in the Section 2. The main results are summarized in the conclusion. \section{General discussion of the calculation of $P(n)$} There are several different approaches to investigate $P(n)$: \begin{itemize} \item {\bf Representation of correlation functions as determinants of Fredholm integral operators} .This approach is based on following steps: i. Quantum correlation function should be represented as a determinant of a Fredholm integral operators of a special type. We call these operators {\it integrable} integral operators. ii. The determinant can be described by completely integrable equation of Painleve type. iii. Asymptotic of correlation function [and the determinant] can be described by Riemann-Hilbert problem. This approach was discovered in \cite{iiks}, it is described in detail in the book \cite{KIB} . It is interesting to mention that this approach was successfully applied also to matrix models \cite{tw} \item {\bf Vertex operator approach} was developed in Kyoto by Foda, Jimbo, Miki, Miwa and Nakayashiki. This approach is based on study of representations of infinite dimensional quantum group $U_{q}\widehat {SL(2)}$, see \cite{JMMN} . \end{itemize} We shall use the integral representation obtained in \cite{KIEU} \be P(n)=\int_C {d\lambda_1\over 2\pi i\lambda_1} \int_C {d\lambda_2\over 2\pi i\lambda_2}\ldots \int_C {d\lambda_n\over 2\pi i\lambda_n} \prod_{a=1}^n (1+{i\over\lambda_a})^{n-a} ({\pi\lambda_a\over\sinh{\pi\lambda_a}})^n \prod_{1\le j {\prod_{1\leq k < j\leq n}\sinh{\pi(\lambda_j-\lambda_k)} \over \prod_{j=1}^n\sinh^n{\pi\lambda_j} } \label{U_n} \ee and \be T_n(\l_1,\ldots,\l_n)\;=\; {\prod_{j=1}^n\l_j^{j-1}(\lambda_j+i)^{n-j}\over \prod_{1\leq k < j\leq n}(\lambda_j-\lambda_k-i) } \label{T_n} \ee First of all, let us note that in principle the contour $C$ can be chosen between $0$ and $-i$ arbitrary. Let us denote $C_{\a}$ the contour that goes from $i\a - \infty$ to $i\a +\infty$. In what follows it will be convenient to choose $\a=-1/2$ i.e. to integrate over the contour $C_{-1/2}$. As appeared we can make a lot of simplifications without taking integrals but using some simple observations and properties of the function in the r.h.s. of (\ref{intPna}) which has to be integrated. Let us define a "weak" equality $\sim$. Namely, let us say that two functions $F_n(\l_1,\ldots,\l_n)$ and $G_n(\l_1,\ldots,\l_n)$ are "weakly" equivalent \be F_n(\l_1,\ldots,\l_n)\;\sim\;G_n(\l_1,\ldots,\l_n) \label{sim} \ee if \be \prod_{j=1}^n\int_{C_{-1/2}}{{d\l_j}\over{2\pi i}}\> U_n(\l_1,\ldots,\l_n)\; (F_n(\l_1,\ldots,\l_n)\;-\;G_n(\l_1,\ldots,\l_n))\>=\>0. \ee Let us also introduce a "canonical" form of the function by the following formula \be T_n^c(\l_1,\ldots,\l_n)\;=\; \sum_{j=0}^{[{n\over 2}]}\; P_j^{(n)} \prod_{k=1}^j{1\over{\l_{2k}-\l_{2k-1}}} \label{T_n^c} \ee where $P_j^{(n)}$ are some polynomials of the form $$ P_j^{(n)}\equiv P_j^{(n)}(\l_1,\l_3,\ldots,\l_{2j-1}|\l_{2j+1},\l_{2j+2},\ldots,\l_n) \;=\; $$ \be =\;\sum_ {\tiny {\begin{array}{lcr} \phantom{a} & 0 \le i_1, i_3, \ldots, i_{2j-1} \le n-2 & \phantom{a}\\ \phantom{a} & 0 \le i_{2j+1}< i_{2j+2}< \ldots < i_n\le n-1 &\phantom{a} \end{array} }} C_{i_1,\quad i_3,\;\ldots,\; i_{2j-1}}^ { i_{2j+1}, i_{2j+2},\ldots, i_n}\quad \l_1^{i_1}\>\l_3^{i_3}\>\ldots\>\l_{2j-1}^{i_{2j-1}}\> \l_{2j+1}^{i_{2j+1}}\>\l_{2j+2}^{i_{2j+2}}\>\ldots\> \l_n^{i_n} \label{P_j} \ee where $$ C_{i_1,\quad i_3,\;\ldots,\; i_{2j-1}}^ { i_{2j+1}, i_{2j+2},\ldots, i_n}= i^{\b}\hat C_{i_1,\quad i_3,\;\ldots,\; i_{2j-1}}^ { i_{2j+1}, i_{2j+2},\ldots, i_n} $$ with $\b=0$ or $1$ in accordance with the equality $$ \b+i_1+i_3+\ldots+i_{2j-2}+i_{2j+1}+i_{2j+2}+\ldots+i_n \equiv j+n\quad\mbox{mod}\quad 2 $$ and some rational numbers $\hat C_{i_1,\quad i_3,\;\ldots,\; i_{2j-1}}^ { i_{2j+1}, i_{2j+2},\ldots, i_n}$. This form has some arbitrariness because if we substitute $\l_j=x_j-i/2$ where all $x_j$ are real then it is easy to see that the function $\tilde U_n(x_1,\ldots,x_n)= U_n(x_1-i/2,\ldots,x_n-i/2)$ transforms when $\{x_1,\ldots,x_n\}\rightarrow\{-x_1,\ldots,-x_n\}$ as follows \be \tilde U_n(-x_1,\ldots,-x_n)\;=\;(-1)^n\>\tilde U_n(x_1,\ldots,x_n). \label{tildeU} \ee Therefore any function $\tilde F_n(x_1,\ldots,x_n)$ that satisfies \be \tilde F_n(-x_1,\ldots,-x_n)=(-1)^{n+1}\;\tilde F_n(x_1,\ldots,x_n) \label{tildeF} \ee being integrated makes zero contribution $$ \prod_{j=1}^n\int_{-\infty}^{\infty}{{dx_j}\over{2\pi i}} \tilde U_n(x_1,\ldots,x_n)\;\tilde F_n(x_1,\ldots,x_n)\;=\;0. $$ It means that in order to get a nonzero result one should have the function $\tilde F_n(x_1,\ldots,x_n)$ of the same parity as of the function $\tilde U_n(x_1,\ldots,x_n)$. Then if we re-expand the form (\ref{T_n^c}) in terms of variables $x_j$ instead of $\l_j$ we can fix the arbitrariness by imposing some additional constraints, namely, $$ \tilde P_j^{(n)}(x_1,x_3,\ldots,x_{2j-1}|x_{2j+1},x_{2j+2},\ldots,x_n) \;=\; $$ $$ =\;P_j^{(n)}(x_1-i/2,x_3-i/2,\ldots,x_{2j-1}-i/2| x_{2j+1}-i/2,x_{2j+2}-i/2,\ldots,x_n-i/2)\;=\; $$ \be \sum_ {\tiny {\begin{array}{lcr} \phantom{a} & 0 \le i_1, i_3, \ldots, i_{2j-1} \le n-2 & \phantom{a}\\ \phantom{a} & 0 \le i_{2j+1}< i_{2j+2}< \ldots < i_n\le n-1 & \phantom{a}\\ \phantom{a} & i_1+i_3+\ldots+i_{2j-2}+i_{2j+1}+i_{2j+2}+\ldots+i_n \equiv j+n \>\mbox{mod}\>2 \end{array} }} \tilde C_{i_1,\quad i_3,\;\ldots,\; i_{2j-1}}^ { i_{2j+1}, i_{2j+2},\ldots, i_n} \quad x_1^{i_1}\>x_3^{i_3}\>\ldots\>x_{2j-1}^{i_{2j-1}}\> x_{2j+1}^{i_{2j+1}}\>x_{2j+2}^{i_{2j+2}}\>\ldots\> x_n^{i_n} \label{P_ja}. \ee In comparison with the coefficients $C_{i_1,\quad i_3,\;\ldots,\; i_{2j-1}}^ { i_{2j+1}, i_{2j+2},\ldots, i_n}$ which can be pure imaginary all the coefficients $\tilde C_{i_1,\quad i_3,\;\ldots,\; i_{2j-1}}^ { i_{2j+1}, i_{2j+2},\ldots, i_n}$ are real and rational numbers. So we can expect that the function \be \tilde T_n^c(x_1,\ldots,x_n)\equiv T_n^c(x_1-i/2,\ldots,x_n-i/2) \label{tildeT_n^c} \ee should satisfy the following property \be \tilde T_n^c(-x_1,\ldots,-x_n)=(-1)^{n}\;\tilde T_n^c(x_1,\ldots,x_n) \label{tildeT_n^ca} \ee Below the property (\ref{tildeT_n^ca}) will be implied when we will speak about the "canonical" form (\ref{T_n^c}-\ref{P_j}). Besides, one can note that the function $\tilde T_n^c(x_1,\ldots,x_n)$ should be real for real variables $x_j$. Our hypothesis is that for any $n$ one can reduce the function $T_n$ defined by (\ref{T_n}) to the canonical form i.e. there exist polynomials $P_j$ in (\ref{T_n^c}) such that \be T_n(\l_1,\ldots,\l_n)\;\sim\;T_n^c(\l_1,\ldots,\l_n). \label{T_nT_n^c} \ee Unfortunately, for the moment we do not have a proof of this statement for any $n$ but we will demonstrate below how it works for $n=2,3,4$. In fact, the problem of the calculation of $P(n)$ given by the integral (\ref{intPna}) can be reduced to the two steps. The first step corresponds to the obtaining of the "canonical" form for $T_n$. The second step is the calculation of the integral by means of this "canonical" form. To do this one needs the following simple facts: I. $\quad$ Since the function $U_n(\l_1,\ldots,\l_n)$ is antisymmetric with respect to transposition of any pair of integration variables, say, $\l_j$ and $\l_k$ the following integral \be \prod_{j=1}^n\int_{C}{d\lambda_j\;\over 2\pi i }\; U(\l_1,\ldots,\l_n)\;S(\l_1,\ldots,\l_n)\;=\;0 \label{int} \ee if the function $S$ is symmetric for at least one pair of $\l$-s. Therefore for an arbitrary function $F_n(\l_1,\ldots,\l_n)$ one can transpose any pair of $\l$-s taking into consideration appearance of additional sign because of the antisymmetry of $U_n(\l_1,\ldots,\l_n)$. For example, if one transposes $\l_j$ with $\l_k$ one gets \be F_n(\ldots,\l_j,\ldots,\l_k,\ldots)\;\sim\; -F_n(\ldots,\l_k,\ldots,\l_j,\ldots). \label{transp} \ee II. $\quad$ The reduction of the power of denominator for $T_n$ is based on two relations which can be checked directly $$ {1\over \l_k-\l_l-i}\;{1\over \l_j-\l_l-i}\;{1\over \l_j-\l_k-i}\;= $$ \be i\;{1\over \l_j-\l_l-i}\;{1\over \l_j-\l_k-i}\;+\; i\;{1\over \l_k-\l_l-i}\;{1\over \l_j-\l_l-i}\;-\; i\;{1\over \l_k-\l_l-i}\;{1\over \l_j-\l_k-i} \label{re1} \ee \be \prod_{k=1}^{j-1}{1\over \l_j-\l_k-i}= \sum_{k=1}^{j-1}{1\over \l_j-\l_k-i}\prod_{ \tiny \begin{array}{lcr} \phantom{a} & l=1 &\phantom{a}\\ \phantom{a} & l\ne k &\phantom{a} \end{array} }^{j-1} {1\over \l_k-\l_l} \label{re2}. \ee In Appendices A and B we will show how the reduction can be performed for $n=3$ and $n=4$. Unfortunately, so far we have not succeeded in finding a result for general $n$. III. $\quad$ The ratio \be {{T_{n+1}(\l_1,\ldots,\l_{n+1})}\over {T_{n}(\l_1,\ldots,\l_{n})}}\;=\; {\prod_{j=1}^n(\l_j+i)\>\l_{n+1}^n\over \prod_{j=1}^n(\l_{n+1}-\l_j-i)} \label{ratio} \ee is symmetric with respect to any permutation of $\l_1,\ldots,\l_n$. Therefore the relation (\ref{ratio}) allows us to use the result $T_n$ also for derivation of $T_{n+1}$ if this result was obtained by applying the relations (\ref{transp}-\ref{re2}) from I and II. IV. {\bf Proposition 1}\\ Let the function $f(\l_1,\ldots,\l_n)$ have only poles of the form $1/(\l_j-\l_k+i a)$ with $a$ an integer i.e. the product $U_n(\l_1,\ldots,\l_n)f(\l_1,\ldots,\l_n)$ does not have poles of that kind. Then \be \l_j^m\,f(\ldots,\l_j,\ldots)\;\sim\;-(\l_j+i)^m\, f(\ldots,\l_j+i,\ldots) \label{l^m} \ee where $m$ is an integer and $m\ge n$. $\quad$\\ {\it Proof}\\ Let us suppose that all variables $\l_1,\ldots,\l_{j-1},\l_{j+1},\ldots, \l_n$ are fixed. Extracting from $U_n(\l_1,\ldots,\l_n)$ the function which depends on $\l_j$ one gets $$ \int_{C_{-1/2}}{d\l_j\over 2\pi i} {\prod_{k\ne j}\sinh{\pi(\l_j-\l_k)}\over \sinh^n{\pi\l_j}} \l_j^m\;w(\l_j)\;=\; -\int_{C_{-3/2}}{d\l_j\over 2\pi i} {\prod_{k\ne j}\sinh{\pi(\l_j-\l_k)}\over \sinh^n{\pi\l_j}} (\l_j+i)^m\;w(\l_j+i)\;= $$ $$ =\;-(\int_{C_{-3/2}}-\int_{C_{-1/2}}+\int_{C_{-1/2}}) {d\l_j\over 2\pi i} {\prod_{k\ne j}\sinh{\pi(\l_j-\l_k)}\over \sinh^n{\pi\l_j}} (\l_j+i)^m\;w(\l_j+i) $$ where $w(\l_j)=f(\ldots,\l_j,\ldots)$. The first step here was to shift integration variable $\l_j\rightarrow\l_j+i$ and to use the fact that $\sinh{\pi(x+i)}=-\sinh{\pi x}$. The two first integrals in the last expression are equal to a contour integral around the point $\l_j=-i$ in a complex plane of the variable $\l_j$. Since, $m\ge n$ the term $(\l_j+i)^m$ which is in the numerator and corresponds to a zero of order $m$ compensates the pole from the term $\sinh^n{\pi\l_j}$ in the denominator. Therefore the contribution of those two integrals is zero and we immediately come to the statement (\ref{l^m}). \vspace{0.3cm} One can get two useful corollaries from the proposition 1. $\quad$\\ {\it Corollary 1} \be \l_j^m\;g(\l_1,\ldots,\hat\l_j,\ldots,\l_n)\;\sim\; {(-i)\over 2}\sum_{k=0}^{m-1}\l_j^k(\l_j+i)^{m-1-k}\; g(\l_1,\ldots,\hat\l_j,\ldots,\l_n) \label{cor1} \ee where the function $g(\l_1,\ldots,\hat\l_j,\ldots,\l_n)$ does not depend on $\l_j$ and as above it is implied that $m\ge n$. $\quad$\\ {\it Proof}\\ The relation (\ref{cor1}) is easy to derive using the relation $(\l_j^m+(\l_j+i)^m)g(\l_1,\ldots,\hat\l_j,\ldots,\l_n)\sim 0$ or equivalently $\l_j^m g(\l_1,\ldots,\hat\l_j,\ldots,\l_n)\sim 1/2(\l_j^m-(\l_j+i)^m) g(\l_1,\ldots,\hat\l_j,\ldots,\l_n)$. $\quad$\\ {\it Corollary 2} $$ {\l_j^{m-1}\over \l_k-\l_j}\; g(\l_1,\ldots,\hat\l_k,\ldots,\hat\l_j,\ldots,\l_n)\;\sim\; $$ \be \sim\;{i\over m}\Bigl(\sum_{l=2}^m {\Bigl( \begin{array}{c} m\\ l \end{array} \Bigr)\; i^l\l_j^{m-l}\over \l_k-l_j} \>+\> \sum_{l=0}^{m-1}\l_k^l(\l_j+i)^{m-1-l}\Bigr) g(\l_1,\ldots,\hat\l_k,\ldots,\hat\l_j,\ldots,\l_n) \label{cor2} \ee where $$ \Bigl( \begin{array}{c} m\\ l \end{array} \Bigr)={m!\over l!\,(m-l)!} $$ is binomial coefficient and the function $g(\l_1,\ldots,\hat\l_k,\ldots,\hat\l_j,\ldots,\l_n)$ does not depend on $\l_k$ and $\l_j$ and $m\ge n$. $\quad$\\ {\it Proof}\\ Using the proposition 1 we get $$ {\l_j^{m}\over \l_k-\l_j}\> g(\l_1,\ldots,\hat\l_k,\ldots,\hat\l_j,\ldots,\l_n)\;\sim\; -{(\l_j+i)^{m}\over \l_k-\l_j-i}\> g(\l_1,\ldots,\hat\l_k,\ldots,\hat\l_j,\ldots,\l_n)\;=\; $$ $$ =\;\Bigl(-{\l_k^{m}\over \l_k-\l_j-i}+ {\l_k^{m}-(\l_j+i)^{m}\over \l_k-\l_j-i}\Bigr)\> g(\l_1,\ldots,\hat\l_k,\ldots,\hat\l_j,\ldots,\l_n)\;\sim\; $$ $$ \sim\;\Bigl({(\l_k+i)^{m}\over \l_k-\l_j}+ {\l_k^{m}-(\l_j+i)^{m}\over \l_k-\l_j-i}\Bigr)\> g(\l_1,\ldots,\hat\l_k,\ldots,\hat\l_j,\ldots,\l_n)\;\sim\; $$ $$ \sim\;\Bigl({(\l_j+i)^{m}\over \l_k-\l_j}+ {\l_k^{m}-(\l_j+i)^{m}\over \l_k-\l_j-i}\Bigr)\> g(\l_1,\ldots,\hat\l_k,\ldots,\hat\l_j,\ldots,\l_n)\;\sim\; $$ or $$ \Bigl({(\l_j+i)^{m}-\l_j^m\over \l_k-\l_j}+ {\l_k^{m}-(\l_j+i)^{m}\over \l_k-\l_j-i}\Bigr)\> g(\l_1,\ldots,\hat\l_k,\ldots,\hat\l_j,\ldots,\l_n)\;\sim\;0. $$ Then expanding both numerators according to the formulae $$ (\l_j+i)^{m}-\l_j^m=\sum_{l=1}^m \Bigl( \begin{array}{c} m\\ l \end{array} \Bigr)\;i^l\l_j^{m-l} $$ $$ \l_k^m-(\l_j+i)^m=(\l_k-\l_j-i)\sum_{l=0}^{m-1}\l_k^l(\l_j+i)^{m-1-l} $$ we arrive at the formula (\ref{cor2}). With the help of the corollaries 1 and 2 one can effectively reduce the power of the numerator in $T_n$. V. $\quad$ For the calculation of integrals we need the following formula\\ {\bf Proposition 2} \be \int_{C_{-1/2}}{d\l\over 2\pi i}\>{f(\l)\over \sinh^n{\pi\l}}\;=\; {d^{(n)}(\e)}_{\e\rightarrow 0}\sum_{l=0}^{\infty}(-1)^{ln}f(il+\e)\;=\; -{d^{(n)}(\e)}_{\e\rightarrow 0}\sum_{l=1}^{\infty}(-1)^{ln}f(-il+\e) \label{int1} \ee where $f(\l)$ is an arbitrary analytic function and a differential operator $d^{(n)}(\e)$ looks as follows \be d^{(n)}(\e)\;=\;{1\over \pi^n(n-1)!} \sum_{l=0}^{n-1} \Bigl( \begin{array}{c} n-1\\ l \end{array} \Bigr) {\Bigl({\partial\over\partial\a}\Bigr)}^{n-1-l}_{\a\rightarrow 0} {\Bigl(\sum_{0\le 2kF_n(\l_1,\ldots,\l_n) \;=\;D^{(n)}\;\tilde F_n(\e_1,\ldots,\e_n) \label{IS} \ee where the product $U_n(\l_1,\ldots,\l_n)\>F_n(\l_1,\ldots,\l_n)$ does not have any other poles besides the poles of the denominator $\prod_{j=1}^n\sinh^n{\pi \l_j}$ of the function $U_n(\l_1,\ldots,\l_n)$ and $D^{(n)}$ is the differential operator \be D^{(n)}=\pi^{n(n+1)\over 2} \prod_{j=1}^n{d^{(n)}(\e_j)}_{\e_j\rightarrow 0} \prod_{0\le k0$. For example, if $l_j+l_k>0$ then we can use (\ref{ints}) to write down the following expansion $$ {1\over l_j+l_k+i(\e_j-\e_k)}\;=\;\int_{0}^1{ds\over s} s^{l_j+l_k+i(\e_j-\e_k)}\;=\; $$ \be =\;\int_{0}^1{ds\over s}s^{l_j+l_k}(1\;+\; \ln{s}\,\bigl(i(\e_j-\e_k)\bigr)\;+\; {1\over 2}\ln^2{s}\,\bigl(i(\e_j-\e_k)\bigr)^2\;+\;\ldots) \label{expand} \ee Then after using the formulae like (\ref{expand}) and taking all summations in (\ref{F}) one can apply differential operator (\ref{Da}) to the series in power of $\e$-s. As the last step one should take integrals on auxiliary variables like variable $s$ in (\ref{expand}). If the function $F_n$ has the form (\ref{Fmon}) then after applying the summation formula (\ref{F}) one faces the series that is formally divergent. Let us consider more attentively how to handle this problem. When extracting the integral over $\l_1$ in the l.h.s. of (\ref{IS}) one gets \be \int_{C_{-1/2}}{d\l_1\over 2\pi i} {\sinh{\pi(\l_2-\l_1)}\ldots\sinh{\pi(\l_n-\l_1)} \over \sinh^n{\pi\l_1}}\>\l_1^a\;=\; \label{intl^a} \ee $$ =\;\lim_{\d\rightarrow 0^+} \int_{C_{-1/2}}{d\l_1\over 2\pi i} {\sinh{\pi(\l_2-\l_1)}\ldots\sinh{\pi(\l_n-\l_1)} \over \sinh^n{\pi\l_1}}\>\l_1^a\;e^{i\l_1\d}\;=\; $$ $$ =\;\lim_{\d\rightarrow 0^+} {d^{(n)}(\e_1)}_{\e_1\rightarrow 0} \sinh{\pi(\l_2-\e_1)}\ldots\sinh{\pi(\l_n-\e_1)} \sum_{l_1=0}^{\infty}(-1)^{l_1}(il_1+\e_1)^a\;e^{-l_1\d+i\e_1\d}\;= $$ $$ =\;\lim_{\d\rightarrow 0^+} \bigr[{d^{(n)}(\e_1)}_{\e_1\rightarrow 0} \sinh{\pi(\l_2-\e_1)}\ldots\sinh{\pi(\l_n-\e_1)} \sum_{l_1=0}^{\infty}(-e^{-\d})^{l_1}(il_1+\e_1)^a\;+\;\mbox{O}(\d) \bigr]\;=\; $$ \be =\;{d^{(n)}(\e_1)}_{\e_1\rightarrow 0} \sinh{\pi(\l_2-\e_1)}\ldots\sinh{\pi(\l_n-\e_1)} \lim_{\d\rightarrow 0^+} \sum_{l_1=0}^{\infty}(-e^{-\d})^{l_1}(il_1+\e_1)^a \label{intmon} \ee So the integral (\ref{intl^a}) reduces to calculation of the sum \be \r(b)\;=\;\lim_{\d\rightarrow 0^+}\sum_{l_1=0}^{\infty} (-e^{-\d})^{l_1}l_1^b \label{ro} \ee with an integer $b\ge 0$. Let us adduce a number of first values of $\r(b)$ \be \r(0)={1\over 2},\quad \r(1)=-{1\over 4},\quad \r(2)=0,\quad \r(3)={1\over 8},\quad \r(4)=0,\quad \r(5)=-{1\over 4} \label{ro5} \ee Below for integrals like (\ref{intl^a}) we shall take $\d$ to be zero at once implying the limiting procedure $\lim_{\d\rightarrow 0^+}$ described above. Now let us illustrate how the whole procedure works for a simple case $P(2)$ \be P(2)\;=\;\pi^3\int_{C_{-1/2}}{d\l_1\over 2\pi i} \int_{C_{-1/2}}{d\l_2\over 2\pi i}\> {\sinh{\pi(\l_2-\l_1)}\over \sinh^2{\pi\l_1} \sinh^2{\pi\l_2}}\> T_2(\l_1,\l_2) \label{PP2} \ee In this case it is very simple to perform the first step, namely, to get the "canonical" form (\ref{T_n^c}) described in the beginning of the Section because we do not need to reduce a power of denominator in this case. Indeed, \be T_2(\l_1,\l_2)\;=\;{(\l_1+i)\l_2\over\l_2-\l_1-i}\;=\; \l_1+i\;+\;{(\l_1+i)^2\over\l_2-\l_1-i}\;\sim\; \l_1\;-\;{\l_1^2\over\l_2-\l_1} \label{T_2} \ee where we have used the property I and the formula (\ref{l^m}) from the proposition 1 of the item IV. Then using the formula (\ref{cor1}) of the corollary 1 for $m=2,3$ one gets $$ -\;{\l_1^2\over\l_2-\l_1}\;\sim\;-{i\over 3}\; \biggl({3i^2\l_1+i^3\over\l_2-\l_1}\;+\;\l_2^2\;+\l_2(\l_1+i)\;+\; (\l_1+i)^2\biggr)\;\sim\; $$ $$ \sim\;{i\l_1\over\l_2-\l_1}\; -\;{1\over 3}\,{1\over\l_2-\l_1}\;-\;{i\over 3}(i\l_1)\;\sim\; {1\over 2}\,{1\over\l_2-\l_1}\;-\;{1\over 3}\,{1\over \l_2-\l_1}\;+\; {1\over 3} \l_1\;=\;{1\over 3}\l_1\;+\;{1\over 6}\,{1\over \l_2-\l_1} $$ Substituting it to the formula (\ref{T_2}) we get \be T_2(\l_1,\l_2)\;\sim\;T_2^c(\l_1,\l_2) \label{T_2^c} \ee where \be T_2^c(\l_1,\l_2)\;=\;{4\over 3}\,\l_1\;+\;{1\over 6}\,{1\over \l_2-\l_1} \label{T_2^cdef} \ee and this is the "canonical" form for $T_2$ i.e. the polynomials $P_0^{(2)}$ and $P_1^{(2)}$ from (\ref{T_n^c}-\ref{P_j}) are equal to $4/3\l_1$ and $1/6$ respectively. Let us take the integral from the first term using the formula (\ref{F}), our comments about the limiting procedure for the integral (\ref{intl^a}) and the result (\ref{intmon}) $$ J_0^{(2)}\;=\;\pi^3\int_{C_{-1/2}}{d\l_1\over 2\pi i} \int_{C_{-1/2}}{d\l_2\over 2\pi i}\> {\sinh{\pi(\l_2-\l_1)}\over \sinh^2{\pi\l_1} \sinh^2{\pi\l_2}}\> {4\over 3}\,\l_1\;=\;D^{(2)}{4\over 3}\sum_{l_1=0}^{\infty} (-1)^{l_1}(il_1+\e_1)\sum_{l_1=0}^{\infty}(-1)^{l_2}\;=\; $$ \be =\;D^{(2)}{4\over 3}(i\r(1)+\e_1\r(0))\r(0)\;=\; D^{(2)}{4\over 3}(-{i\over 4}+{\e_1\over 2}){1\over 2}\;=\; {({\partial\over\partial\e_1}-{\partial\over\partial\e_2})}_ {\e_1,\e_2\rightarrow 0} {2\over 3}(-{i\over 4}+{\e_1\over 2})\;=\;{1\over 3} \label{J02} \ee where we have used the formulae (\ref{ro5}) for $\r(b)$ given by (\ref{ro}) implying the limiting procedure as it was explained above. The second integral is treated as it was described in item VI with the help of the integral representation (\ref{ints}) $$ J_1^{(2)}\;=\;\pi^3\int_{C_{-1/2}}{d\l_1\over 2\pi i} \int_{C_{-1/2}}{d\l_2\over 2\pi i}\> {\sinh{\pi(\l_2-\l_1)}\over \sinh^2{\pi\l_1} \sinh^2{\pi\l_2}}\> {1\over 6}\;{1\over \l_2-\l_1}\,=\, $$ $$ =\;D^{(2)}\>{(-1)\over 6}\sum_{l_1=1}^{\infty}\sum_{l_2=0}^{\infty} \>{1\over il_2+il_1+\e_2-\e_1}\;=\; D^{(2)}\>{i\over 6}\sum_{l_1=1}^{\infty}\sum_{l_2=0}^{\infty} \>{1\over l_1+l_2+i(\e_1-\e_2)}\;=\; $$ $$ =\;D^{(2)}\>{i\over 6}\int_0^1{ds\over s} \sum_{l_1=1}^{\infty}\sum_{l_2=0}^{\infty}\> s^{l_1+l_2+i(\e_1-\e_2)}\;=\; D^{(2)}\>{(-i)\over 6}\int_0^1 ds\; {s^{i(\e_1-\e_2)}\over (1+s)^2}\;=\; $$ \be =\;{({\partial\over\partial\e_1}-{\partial\over\partial\e_2})}_ {\e_1,\e_2\rightarrow 0} {(-i)\over 6}\int_0^1 ds\; {s^{i(\e_1-\e_2)}\over (1+s)^2}\;=\; {1\over 3}\int_0^1 ds {\ln{s}\over (1+s)^2}\;=\;-{\ln{2}\over 3} \label{J12} \ee Summing up two answers (\ref{J02}) and (\ref{J12}) we get the result \be P(2)\;=\;J_0^{(2)}\;+\;J_1^{(2)}\;=\;{1\over 3}\;-\;{\ln(2)\over 3}. \label{P2res} \ee which coincides with the formula (\ref{P2}). In the Appendices A and B we shall derive the formulae (\ref{P3}) and (\ref{P4}) for $P(3)$ and $P(4)$ respectively. In the end of this Section let us note that both results (\ref{P3}) and (\ref{P4}) are expressed in terms of the logarithmic function and the Riemann zeta function of odd arguments and do not depend on polylogarithms like, for example, $\mbox{Li}_4(1/2)$. All coefficients before those functions in (\ref{P1}-\ref{P4}) are rational. Also they do not contain any powers of $\pi$ which could be considered as Riemann zeta functions of even arguments, see the formula (\ref{zeta2n}) from the Introduction. {\bf Our conjecture is that the final answer for any $P(n)$ will also be expressed in terms of logarithm $\ln{2}$ and Riemann zeta functions $\zeta(k)$ with odd integers $k$ and with rational coefficients.} In fact, this conjecture is intimately connected with our hypothesis from the beginning of this Section that the function $T_n$ (\ref{T_n}) can be reduced to the "canonical" form. Looking at the "canonical" form (\ref{T_n^c}) one can conclude that only Riemann zeta functions and their products can enter into the final answer because all the denominators in the r.h.s. of (\ref{T_n^c}) are split out. It means that after applying the formula (\ref{F}) the multiple summation can be performed by pairs, say, $\sum_{l_{2k-1}}$ and $\sum_{l_{2k}}$. Each pair of these summations results in some combination of zeta functions. \section{Conclusion} We want to emphasize an interesting connection between integrable and disordered models. In order to describe correlations in integrable models one can use integrable integral operators \cite{iiks}. On the other hand Tracy and Widom showed that in matrix models the distribution of eigenvalues and level spacing can be described by the integral operators, belonging to the same integrable class \cite{tw}. Our current work supports this link between integrable models and chaotic models. Riemann zeta function appears in the description of both kind of models. Let us repeat that the main result of this paper is the calculation of $P(3)$ and $P(4)$ (\ref{P3}-\ref{P4}) by means of the multi-integral representation (\ref{intPn}). The fact that only the logarithm $\ln{2}$ and Riemann zeta function with odd arguments participate in the answers for $P(1),\ldots, P(4)$ and with rational coefficients before these functions allows us to suppose that this is the general property of $P(n)$. One could compare the calculation of $P(n)$ with the many-loop calculation of the self-energy diagrams in the renormalizable quantum field theory which can also be expressed in terms of $\zeta$ functions of odd arguments \cite{KR} . Unfortunately, so far we have not got even a conjecture for $P(n)$ but we believe that it is not an unsolvable problem. May be already after calculation of $P(5)$ one could guess the right formula for a generic case $P(n)$. It would give an answer to the question discussed in the previous section, namely, the question about the law of decay of $P(n)$ when $n$ tends to infinity. Also it would be interesting to generalize above results to the XXZ spin chain. Some interesting conjectures were recently invented by Razumov and Stroganov \cite{RS} for the special case of the XXZ model with $\Delta=-1/2$. These conjectures would be supported if it were possible to get $P(n)$ from the general integral representation obtained by the RIMS group \cite{JMMN}. \section{Acknowledgements} The authors would like to thank B. McCoy, A.~Razumov, M.~Shiroishi, Yu.~Stroganov, M.~Takahashi, L.Takhtajan for useful discussions. This research has been supported by the NSF grant PHY-9988566 and by INTAS Grant no. 01-561. \section{Appendix A} Here we discuss in detail the calculation of $P(3)$ performed by means of the general procedure described above. As was pointed out in the beginning of Section 2 the first step should be a reduction of the function $T_3(\l_1,\l_2,\l_3)$ to the form (\ref{T_n^c}) which we have called the "canonical" form. In comparison with the case $P(2)$ here we should reduce the power of the denominator in $T_3(\l_1,\l_2,\l_3)$. To do this we will use the formula (\ref{re1}) from the item II. Namely, \be T_3(\l_1,\l_2,\l_3)\;=\;{(\l_1+i)^2(\l_2+i)\l_2\l_3^2\over (\l_2-\l_1-i)(\l_3-\l_1-i)(\l_3-\l_2-i)}\;=\; I_1^{(3)}\;+\;I_2^{(3)}\;+\;I_3^{(3)} \label{T3} \ee where \be I_1^{(3)}\;=\;i{(\l_1+i)^2(\l_2+i)\l_2\l_3^2\over (\l_3-\l_1-i)(\l_3-\l_2-i)}, \label{I31} \ee \be I_2^{(3)}\;=\;i{(\l_1+i)^2(\l_2+i)\l_2\l_3^2\over (\l_2-\l_1-i)(\l_3-\l_1-i)}, \label{I32} \ee \be I_3^{(3)}\;=\;-i{(\l_1+i)^2(\l_2+i)\l_2\l_3^2\over (\l_2-\l_1-i)(\l_3-\l_2-i)}, \label{I33} \ee Due to the $1\leftrightarrow 2$ symmetry of the denominator the first term $I_1^{(3)}$ can be simplified as follows $$ I_1^{(3)}\;\sim\;-{(\l_1+i)(\l_2+i)\l_2\l_3^2\over (\l_3-\l_1-i)(\l_3-\l_2-i)}\;\sim\; {(\l_1+i)(\l_2+i)\l_3^2\over \l_3-\l_1-i}\;=\; $$ \be =\;(\l_1+i)(\l_2+i)(\l_1+\l_3+i)\;+\; {(\l_1+i)^3(\l_2+i)\over\l_3-\l_1-i}\;\sim\; \l_1^2\l_2\;-\;{(\l_1+i)^3(\l_3+i)\over\l_2-\l_1-i} \label{II31} \ee The denominator of the second term (\ref{I32}) has the symmetry under the transposition $2\leftrightarrow 3$. Therefore it can also be simplified $$ I_2^{(3)}\;\sim\;-{(\l_1+i)^2\l_2\l_3^2\over (\l_2-\l_1-i)(\l_3-\l_1-i)}\;\sim\; {(\l_1+i)^2\l_2\l_3\over \l_2-\l_1-i}\;=\; $$ \be =\;-(\l_1+i)^2\l_3\;-\; {(\l_1+i)^3\l_3\over\l_2-\l_1-i}\;\sim\; \l_1^2\l_2\;-\;{(\l_1+i)^3\l_3\over\l_2-\l_1-i} \label{II32} \ee The third term (\ref{I33}) is treated as follows $$ I_3^{(3)}\;=\; -i(\l_1+i)^2(\l_2+i)(\l_3+\l_2+i)\;- $$ $$ -\;i{(\l_1+i)^2(\l_2+i)^3\over\l_3-\l_2-i}\;+\; i{(\l_1+i)^3\l_3^2\over\l_2-\l_1-i}\;-\; i{(\l_1+i)^3\l_3^3\over(\l_2-\l_1-i)(\l_3-\l_2-i)}\;\sim $$ \be \sim\;-\l_1^2\l_2\;-\;{(\l_1+i)^3(\l_3+i)^2\over\l_2-\l_1-i}\;+\; i{(\l_1+i)^3\l_3^2\over\l_2-\l_1-i}\;-\; i{(\l_1+i)^3\l_3^3\over(\l_2-\l_1-i)(\l_3-\l_2-i)} \label{II33} \ee Now adding up all the three terms together we get \be T_3(\l_1,\l_2,\l_3)\;\sim\;\l_1^2\l_2\;-\; i{(\l_1+i)^3\l_3^3\over(\l_2-\l_1-i)(\l_3-\l_2-i)}\;\sim\; -\l_2\l_3^2\;-\; i{(\l_1+i)^3\l_3^3\over(\l_2-\l_1-i)(\l_3-\l_2-i)}. \label{T3a} \ee Let us note that up to this moment we have used only the symmetry property (\ref{transp}) from the item I, the formula (\ref{re1}) from the item II and a simple algebra. Now we would like to use the formula (\ref{l^m}) of the proposition 1 for $m=3$ and again apply the transposition formula (\ref{transp}) $$ T_3(\l_1,\l_2,\l_3)\;\sim\; -\l_2\l_3^2\;-\; i{\l_1^3(\l_3+i)^3\over(\l_2-\l_1)(\l_3-\l_2)}\;\sim\; $$ $$ \sim\;-\l_2\l_3^2\;+\; i{(\l_1+i)^3\l_3^3\over(\l_2-\l_1)(\l_3-\l_2)}\;=\; -\l_2\l_3^2\;+\; i{\l_1^3\l_3^3+3i\l_1^2\l_3^3-3\l_1\l_3^3-i\l_3^3\over (\l_2-\l_1)(\l_3-\l_2)}\;\sim\; $$ $$ \sim\;-\l_2\l_3^2\;-\; {3\l_1^2\l_3^2+3i\l_1\l_3(\l_3+\l_2)-\l_3^2-\l_3\l_2-\l_2^2\over \l_2-\l_1}\;\sim\; $$ \be \sim\;-\l_2\l_3^2\;-\; {3\l_1^2\l_3^2+3i\l_1\l_3^2+3i\l_1^2\l_3-\l_3^2-\l_3\l_1-\l_1^2\over \l_2-\l_1} \label{T3b} \ee Now we can reduce the power of $\l_1$ in the numerator of the second term (\ref{T3b}) by applying the formula (\ref{cor2}) of the corollary 2 from the item IV. Doing this we finally get the "canonical form" (\ref{T_n^c}-\ref{P_j}) of $T_3$ \be T_3(\l_1,\l_2,\l_3)\;\sim\;T_3^c(\l_1,\l_2,\l_3)\;=\; P_0^{(3)}\;+\;{P_1^{(3)}\over\l_2-\l_1} \label{T_3^c} \ee where the polynomials $P_0^{(3)}$ and $P_1^{(3)}$ are as follows \be P_0^{(3)}\;=\;-2\l_2\l_3^2,\quad P_1^{(3)}\;=\;{1\over 3}\;-\;i\l_1\;-\;i\l_3\;-\;2\l_1\l_3 \label{P013} \ee Let us note that if we express variables $\l_j$ through the real variables $x_j$ via $\l_j=x_j-i/2$ in order to get the polynomials $\tilde P_j^{(3)}$ ( see the formula (\ref{P_ja})) we get especially simple formulae, namely, \be \tilde P_0^{(3)}\;=\;-2x_2x_3^2,\quad \tilde P_1^{(3)}\;=\;-{1\over 6}\;-\;2x_1x_3. \label{P013a} \ee So, the function \be \tilde T_3^c(x_1,x_2,x_3)\;=\; \tilde P_0^{(3)}\;+\;{\tilde P_1^{(3)}\over x_2-x_1} \label{tildeT_3^c} \ee is odd i.e. $$ \tilde T_3^c(-x_1,-x_2,-x_3)\;=\;-\tilde T_3^c(x_1,x_2,x_3) $$ as it should be according to the formula (\ref{tildeT_n^ca}) from the beginning of Section 2. Now we are ready to calculate the integral in order to get the result for $P(3)$ \be P(3)\;=\;\prod_{j=1}^3\int_{C_{-1/2}}{d\l_j\over 2\pi i}\> U_3(\l_1,\l_2,\l_3)\>T_3(\l_1,\l_2,\l_3)\;=\; J_0^{(3)}\;+\;J_1^{(3)} \label{P3a} \ee where \be J_0^{(3)}=\prod_{j=1}^3\int_{C_{-1/2}}{d\l_j\over 2\pi i}\> U_3(\l_1,\l_2,\l_3)\>P_0^{(3)},\quad J_1^{(3)}=\prod_{j=1}^3\int_{C_{-1/2}}{d\l_j\over 2\pi i}\> U_3(\l_1,\l_2,\l_3)\> {P_1^{(3)}\over\l_2-\l_1} \label{J301} \ee Using the formulae (\ref{P013}),(\ref{F}),(\ref{Da}), (\ref{intl^a}-\ref{ro5}) we get $$ J_0^{(3)}\;=\;\prod_{j=1}^3\int_{C_{-1/2}}{d\l_j\over 2\pi i}\> U_3(\l_1,\l_2,\l_3)\>(-2)\l_2\l_3^2\;=\; $$ $$ =\;D^{(3)}\sum_{l_1=0}^{\infty}(-1)^{l_1} \sum_{l_2=0}^{\infty}(-1)^{l_2}\sum_{l_3=0}^{\infty}(-1)^{l_3} (-2)(i l_2+\e_2)(i l_3+\e_3)^2\;= $$ \be =\;D^{(3)}\sum_{l_1=0}^{\infty}(-1)^{l_1} \sum_{l_2=0}^{\infty}(-1)^{l_2}\sum_{l_3=0}^{\infty}(-1)^{l_3} (-2)\e_2\e_3^2\;=\; {1\over 2}\prod_{0\le k U_3(\l_1,\l_2,\l_3)\>{{1\over 3}\;-\;i\l_1\;-\;i\l_3\;-\;2\l_1\l_3 \over \l_2-\l_1}\;=\; $$ $$ =-D^{(3)}\sum_{l_1=1}^{\infty}(-1)^{l_1} \sum_{l_2=0}^{\infty}(-1)^{l_2}\sum_{l_3=0}^{\infty}(-1)^{l_3} {{1\over 3}\;-\;i(-il_1+\e_1)\;-\;i(il_3+\e_3)\;-\; 2(-il_1+\e_1)(il_3+\e_3) \over il_2+il_1+\e_2-\e_1}= $$ $$ =\;i\;D^{(3)}\sum_{l_1=1}^{\infty}(-1)^{l_1} \sum_{l_2=0}^{\infty}(-1)^{l_2}\sum_{l_3=0}^{\infty}(-1)^{l_3} {{1\over 3}\;-\;l_1\;+\;l_3\;-i\e_1\;-\;i\e_3\;-\; 2(l_1+i\e_1)(l_3-i\e_3) \over l_1+l_2+i(\e_1-\e_2)}\;=\; $$ $$ =\;i\;D^{(3)}\sum_{l_1=1}^{\infty}(-1)^{l_1} \sum_{l_2=0}^{\infty}(-1)^{l_2} {-{1\over 12}\;-\;i\e_3\;+l_1i\e_3\;-\;\e_1\e_3 \over l_1+l_2+i(\e_1-\e_2)}\;=\; $$ $$ =\;i\;D^{(3)}\int_0^1{ds\over s}s^{i(\e_1-\e_2)} \sum_{l_1=1}^{\infty}\sum_{l_2=0}^{\infty} (-{1\over 12}\;-\;i\e_3\;+l_1i\e_3\;-\;\e_1\e_3)(-s)^{l_1+l_2} \;=\; $$ $$ =\;i\;D^{(3)}\int_0^1{ds\over s}s^{i(\e_1-\e_2)} \biggl( {(-s)\over(1+s)^2}(-{1\over 12}\;-\;i\e_3\;-\;\e_1\e_3)\;+\; {(-s)\over(1+s)^3}i\e_3 \biggr)\;=\; $$ $$ =\;{i\over 2}\prod_{0\le k (\l_3\l_4^2+i\l_4^2-\l_4)\over\l_2-\l_1} \label{T4b} \ee The latter formula was obtained with the help of the formula (\ref{lpi^4}). In fact, it is nothing but the "canonical" form (\ref{T_n^c}) for the first term in (\ref{T4a}). Now let us treat the second term. Using the formulae (\ref{re1}) and (\ref{re2}) we can write it down as follows $$ -i{(\l_1+i)^4(\l_2+i)(\l_3+i)\l_3^3\l_4^3\over (\l_2-\l_1-i)(\l_3-\l_2-i)(\l_4-\l_1-i)(\l_4-\l_2-i)(\l_4-\l_3-i)}\;=\; $$ $$ \phantom{a}=\;-i(\l_1+i)^4(\l_2+i)(\l_3+i)\l_3^3\l_4^3\>\biggl\{ \phantom{a}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $$ $$ {1\over 2}{1\over(\l_3-\l_2-i)(\l_4-\l_1-i)(\l_4-\l_3-i)}\;-\; {1\over 2}{1\over(\l_2-\l_1-i)(\l_4-\l_1-i)(\l_4-\l_3-i)}\;- $$ $$ -\;{1\over 2}{1\over(\l_2-\l_1-i)(\l_3-\l_2-i)(\l_4-\l_1-i)}\;-\; {1\over 2}{1\over(\l_2-\l_1-i)(\l_3-\l_2-i)(\l_4-\l_3-i)}\;- $$ $$ -\;{1\over(\l_2-\l_1)(\l_3-\l_1)(\l_4-\l_1-i)}\;-\; {1\over(\l_3-\l_2-i)(\l_4-\l_1-i)(\l_4-\l_2-i)}\;- $$ $$ -\;{1\over(\l_3-\l_1)(\l_3-\l_2)(\l_4-\l_3-i)}\;+\; {1\over(\l_2-\l_1-i)(\l_4-\l_2-i)(\l_4-\l_3-i)}\;+ $$ $$ +\;{1\over(\l_2-\l_1)(\l_3-\l_2)(\l_4-\l_2-i)}\;+\; {1\over(\l_2-\l_1-i)(\l_3-\l_2-i)(\l_4-\l_2-i)} \>\biggr\} $$ Let us enumerate all ten terms as they enter here and make some appropriate transformations for each of them \be I_1^{(4)}= -{i (\l_1+i)^4(\l_2+i)(\l_3+i)\l_3^3\l_4^3 \over 2 (\l_3-\l_2-i)(\l_4-\l_1-i)(\l_4-\l_3-i)} \begin{array}{c} \quad\\ \quad\\ \sim\\ {\tiny \begin{array}{c} 1\rightarrow 4\\ 2\rightarrow 2\\ 3\rightarrow 1\\ 4\rightarrow 3 \end{array}} \end{array} -{i (\l_1+i)\l_1^3(\l_2+i)\l_3^3(\l_4+i)^4 \over 2 (\l_2-\l_1+i)(\l_3-\l_1-i)(\l_4-\l_3+i)} \label{I41} \ee \be I_2^{(4)}\;=\; {i\over 2}{(\l_1+i)^4(\l_2+i)(\l_3+i)\l_3^3\l_4^3 \over(\l_2-\l_1-i)(\l_4-\l_1-i)(\l_4-\l_3-i)}\; \begin{array}{c} \quad\\ \sim\\ {\tiny \begin{array}{c} 3\leftrightarrow 4 \end{array}} \end{array} {i\over 2}{(\l_1+i)^4(\l_2+i)\l_3^3(\l_4+i)\l_4^3 \over(\l_2-\l_1-i)(\l_3-\l_1-i)(\l_4-\l_3+i)} \label{I42} \ee \be I_3^{(4)}= {i\over 2}{(\l_1+i)^4(\l_2+i)(\l_3+i)\l_3^3\l_4^3 \over(\l_2-\l_1-i)(\l_3-\l_2-i)(\l_4-\l_1-i)} \begin{array}{c} \quad\\ \quad\\ \sim\\ {\tiny \begin{array}{c} 1\rightarrow 1\\ 2\rightarrow 3\\ 3\rightarrow 4\\ 4\rightarrow 2 \end{array}} \end{array} {i\over 2}{(\l_1+i)^4\l_2^3(\l_3+i)(\l_4+i)\l_4^3 \over(\l_2-\l_1-i)(\l_3-\l_1-i)(\l_4-\l_3-i)} \label{I43} \ee \be I_4^{(4)}\;=\; {i\over 2}{(\l_1+i)^4(\l_2+i)(\l_3+i)\l_3^3\l_4^3 \over(\l_2-\l_1-i)(\l_3-\l_2-i)(\l_4-\l_3-i)}\; \begin{array}{c} \quad\\ \sim\\ {\tiny \begin{array}{c} 1\leftrightarrow 2 \end{array}} \end{array} {i\over 2}{(\l_1+i)(\l_2+i)^4(\l_3+i)\l_3^3\l_4^3 \over(\l_2-\l_1+i)(\l_3-\l_1-i)(\l_4-\l_3-i)} \label{I44} \ee \be I_5^{(4)}\;=\; i{(\l_1+i)^4(\l_2+i)(\l_3+i)\l_3^3\l_4^3 \over(\l_2-\l_1)(\l_3-\l_1)(\l_4-\l_1-i)} \label{I45} \ee \be I_6^{(4)}= i{(\l_1+i)^4(\l_2+i)(\l_3+i)\l_3^3\l_4^3 \over(\l_3-\l_2-i)(\l_4-\l_1-i)(\l_4-\l_2-i)} \begin{array}{c} \quad\\ \quad\\ \sim\\ {\tiny \begin{array}{c} 1\rightarrow 4\\ 2\rightarrow 1\\ 3\rightarrow 2\\ 4\rightarrow 3 \end{array}} \end{array} i{(\l_1+i)(\l_2+i)\l_2^3\l_3^3(\l_4+i)^4 \over(\l_2-\l_1-i)(\l_3-\l_1-i)(\l_4-\l_3+i)} \label{I46} \ee \be I_7^{(4)}\;=\; i{(\l_1+i)^4(\l_2+i)(\l_3+i)\l_3^3\l_4^3 \over(\l_3-\l_1)(\l_3-\l_2)(\l_4-\l_3-i)}\; \begin{array}{c} \quad\\ \sim\\ {\tiny \begin{array}{c} 1\leftrightarrow 3 \end{array}} \end{array} -i{(\l_1+i)\l_1^3(\l_2+i)(\l_3+i)^4\l_4^3 \over(\l_2-\l_1)(\l_3-\l_1)(\l_4-\l_1-i)} \label{I47} \ee \be I_8^{(4)}= -{i(\l_1+i)^4(\l_2+i)(\l_3+i)\l_3^3\l_4^3 \over(\l_2-\l_1-i)(\l_4-\l_2-i)(\l_4-\l_3-i)} \begin{array}{c} \quad\\ \quad\\ \sim\\ {\tiny \begin{array}{c} 1\leftrightarrow 2\\ 3\leftrightarrow 4 \end{array}} \end{array} -{i(\l_1+i)(\l_2+i)^4\l_3^3(\l_4+i)\l_4^3 \over(\l_2-\l_1+i)(\l_3-\l_1-i)(\l_4-\l_3+i)} \label{I48} \ee \be I_9^{(4)}\;=\; -i{(\l_1+i)^4(\l_2+i)(\l_3+i)\l_3^3\l_4^3 \over(\l_2-\l_1)(\l_3-\l_2)(\l_4-\l_2-i)}\; \begin{array}{c} \quad\\ \quad\\ \sim\\ {\tiny \begin{array}{c} 1\rightarrow 3\\ 2\rightarrow 1\\ 3\rightarrow 2\\ 4\rightarrow 4\\ \end{array}} \end{array} i{(\l_1+i)(\l_2+i)\l_2^3(\l_3+i)^4\l_4^3 \over(\l_2-\l_1)(\l_3-\l_1)(\l_4-\l_1-i)} \label{I49} \ee \be I_{10}^{(4)}= -{i(\l_1+i)^4(\l_2+i)(\l_3+i)\l_3^3\l_4^3 \over(\l_2-\l_1-i)(\l_3-\l_2-i)(\l_4-\l_2-i)} \begin{array}{c} \quad\\ \sim\\ {\tiny \begin{array}{c} 1\leftrightarrow 2 \end{array}} \end{array} -{i(\l_1+i)(\l_2+i)^4(\l_3+i)\l_3^3\l_4^3 \over(\l_2-\l_1+i)(\l_3-\l_1-i)(\l_4-\l_1-i)} \label{I410} \ee Let us note that after these transformations $I_j^{(4)}$ have two kinds of the denominators, namely, $I_5^{(4)},I_7^{(4)},I_9^{(4)}, I_{10}^{(4)}$ have denominators of the form $$ {1\over(\l_2-\l_1-ia)(\l_3-\l_1-ib)(\l_4-\l_1-ic)} $$ with a set of integers $a,b,c$, while the denominators of the rest of them are of the form $$ {1\over(\l_2-\l_1-ia')(\l_3-\l_1-ib')(\l_4-\l_3-ic')} $$ with some other set of integers $a',b',c'$. Moreover, some of the terms $I_j^{(4)}$ have just coinciding denominators like, for example, $I_2^{(4)}$ and $I_6^{(4)}$. Nevertheless sometimes it will be more convenient to treat them separately. Let us start with the first group which is easier to treat. Since, the denominator of $I_5^{(4)}$ has the $2\leftrightarrow 3$ symmetry we can simplify it as follows \be I_5^{(4)}\;\sim\;i{(\l_1+i)^4(\l_2+i)(\l_3+i)(\l_3^2\,+\,\l_3\l_1 \,+\,\l_1^2)\l_4^3 \over(\l_2-\l_1)(\l_4-\l_1-i)} \label{I45a} \ee where we have used the trivial identity \be \l_3^3\;=\;(\l_3-\l_1)(\l_3^2\,+\,\l_3\l_1\,+\,\l_1^2)\;+\;\l_1^3 \label{l3^3} \ee and the fact that the second term in the r.h.s. of (\ref{l3^3}) does not contribute into $I_5^{(4)}$. Then summing up (\ref{I47}) and (\ref{I49}) we get $$ I_7^{(4)}\;+\;I_9^{(4)}\;\sim\; i{(\l_1+i)(\l_2+i)(\l_1^2\,+\,\l_1\l_2\,+\,\l_2^2)(\l_3+i)^4\l_4^3 \over(\l_3-\l_1)(\l_4-\l_1-i)} \begin{array}{c} \quad\\ \sim\\ {\tiny \begin{array}{c} 1\leftrightarrow 2 \end{array}} \end{array} $$ \be \sim\;-i{(\l_1+i)(\l_2+i)^4(\l_3+i) (\l_1^2\,+\,\l_1\l_3\,+\,\l_3^2)\l_4^3 \over(\l_2-\l_1)(\l_4-\l_1-i)} \label{I479} \ee Now adding the r.h.s. of (\ref{I45a}) to (\ref{I479}) we get $$ I_5^{(4)}\;+\;I_7^{(4)}\;+\;I_9^{(4)}\;\sim\; $$ $$ \sim\;-i{(\l_1+i)(\l_2+i)(\l_3+i) ((\l_1+i)^2\,+\,(\l_1+i)(\l_2+i)\,+\,(\l_2+i)^2) (\l_3^2+\l_3\l_1+\l_1^2)\l_4^3 \over\l_4-\l_1-i}\;\sim\; $$ $$ \sim\;3\;{(\l_1+i)(\l_2+i)(\l_3+i) (\l_1\,+\,\l_2\,+\,i)(\l_3^2+\l_3\l_1+\l_1^2)\l_4^3 \over\l_4-\l_1-i}\;=\; $$ $$ =\;-3\;(\l_2+i)(\l_3+i)(\l_1+\l_2+i)(\l_3^2+\l_3\l_1+\l_1^2)\l_4^3\;+\; $$ $$ \;+\;3\;{(\l_2+i)(\l_3+i) (\l_1\,+\,\l_2\,+\,i)(\l_3^2+\l_3\l_1+\l_1^2)\l_4^4 \over\l_4-\l_1-i}\;\sim\; $$ $$ \sim\;3\;{(\l_2+i)(\l_3+i) (\l_1\,+\,\l_2\,+\,i)(\l_3^2+\l_3\l_1+\l_1^2)\l_4^4 \over\l_4-\l_1-i}\;\sim\; $$ $$ \sim\;3\;{(\l_1+i)(\l_2+i)(\l_3+i)^2\l_3\l_4^4 \over\l_4-\l_1-i}\;+\; 3\;{(\l_2+i)(\l_3+i)\l_3(\l_2\l_3+1)\l_4^4 \over\l_4-\l_1-i}\;\sim\; $$ $$ \sim\;-3\l_2\l_3^2\l_4^3\;+\; 3\;{(\l_2+i)(\l_3+i)^2\l_3\l_4^5 \over\l_4-\l_1-i}\;+\; 3\;{(\l_2+i)(\l_3+i)\l_3(\l_2\l_3+1)\l_4^4 \over\l_4-\l_1-i}\;\sim\; $$ $$ \sim\;-3\l_2\l_3^2\l_4^3\;+\; 3\;{(\l_2+i)(\l_3+i)^2\l_3(\l_4+i)^5 \over\l_4-\l_1}\;+\; 3\;{(\l_2+i)(\l_3+i)\l_3(\l_2\l_3+1)(\l_4+i)^4 \over\l_4-\l_1} \begin{array}{c} \quad\\ \quad\\ \sim\\ {\tiny \begin{array}{c} 1\rightarrow 2\\ 2\rightarrow 3\\ 3\rightarrow 4\\ 4\rightarrow 1\\ \end{array}} \end{array} $$ $$ \sim\;-3\l_2\l_3^2\l_4^3\;-\; 3\;{(\l_1+i)^5(\l_3+i)(\l_4+i)^2\l_4 \over\l_2-\l_1}\;-\; 3\;{(\l_1+i)^4(\l_3+i)(\l_4+i)\l_4(\l_3\l_4+1) \over\l_2-\l_1}\;\sim\; $$ $$ \sim\;-{17\over 10}\l_2\l_3^2\l_4^3\;+\; {({\l_1\over 2}\,+\,{i\over 4})(\l_3+i)(\l_4+i)^2\l_4 \over\l_2-\l_1}\;+\; $$ $$ +\;{(3\l_1^2\,+\,3i\l_1\,-\,{9\over 10})(\l_3+i)(\l_4+i)\l_4(\l_3\l_4+1) \over\l_2-\l_1}\;\sim\; -{17\over 10}\l_2\l_3^2\l_4^3\;+\; $$ \be +\;{ ({\l_1\over 2}\,+\,{i\over 4}) (\l_3\l_4^3+i\l_4^3+2i\l_3\l_4^2-2\l_4^2-i\l_4)\:+\; (3\l_1^2\,+\,3i\l_1\,-\,{9\over 10}) (\l_3^2\l_4^3+i\l_3\l_4^3+i\l_4^2-\l_4) \over\l_2-\l_1} \label{I579} \ee In fact, it is the "canonical" form. To get this expression we have used symmetries, the formula (\ref{l^m}) for $m=4$ and $m=5$ of the proposition 1 and relations (\ref{lpi^4}) and (\ref{lpi^5}). Now let us treat $I_{10}^{(4)}$ $$ I_{10}^{(4)}\;=\; -i{(\l_1+i)(\l_2+i)^4(\l_3+i)\l_3^3\l_4^3 \over(\l_2-\l_1+i)(\l_3-\l_1-i)(\l_4-\l_1-i)}\;\sim\; -i{(\l_1+i)(\l_2+i)^4\l_3^3\l_4^3 \over(\l_2-\l_1+i)(\l_4-\l_1-i)}\;=\; $$ $$ =\;i{(\l_2+i)^4\l_3^3\l_4^3 \over\l_2-\l_1+i}\;-\; i{(\l_2+i)^4\l_3^3\l_4^4 \over(\l_2-\l_1+i)(\l_4-\l_1-i)}\;\sim\; -i{(\l_2+i)^4\l_3^3\l_4^4 \over(\l_2-\l_1+i)(\l_4-\l_1-i)}\;\sim\; $$ $$ \sim\;-i{\l_2^4\l_3^3(\l_4+i)^4 \over(\l_2-\l_1)(\l_4-\l_1)}\;=\; i{\l_3^3(\l_2^4\l_4^4+4i\l_2^3\l_4^4-6\l_2^2\l_4^4-4i\l_2\l_4^4+\l_4^4) \over(\l_2-\l_1)(\l_4-\l_1)}\;\sim\; $$ $$ \sim\;i{\l_3^3(4i\l_2^3\l_4^3-6\l_2^2\l_4^2(\l_4+\l_1)- 4i\l_2\l_4(\l_4^2+\l_4\l_1+\l_1^2)+\l_4^3+\l_4^2\l_1+\l_4\l_1^2+\l_1^3) \over\l_2-\l_1}\;\sim\; $$ $$ \sim\;i{\l_3^3(-6\l_1\l_2^2\l_4^2- 4i\l_1\l_2\l_4^2-4i\l_1^2\l_2\l_4+\l_1\l_4^2+\l_1^2\l_4+\l_1^3) \over\l_2-\l_1}\;=\; $$ $$ =\;i\l_3^3(-6\l_1^2\l_4^2-4i\l_1\l_4^2-4i\l_1^2\l_4)\;+\; i{\l_3^3(-6\l_1^3\l_4^2- 4i\l_1^2\l_4^2-4i\l_1^3\l_4+\l_1\l_4^2+\l_1^2\l_4+\l_1^3) \over\l_2-\l_1}\;\sim\; $$ $$ \sim\;i{\l_3^3(\l_4^2(5i\l_1^2-5\l_1-{3\over 2}i)+ \l_4(-5\l_1^2+4i\l_1+1)-{3\over 2}i\l_1^2+\l_1+{i\over 4}) \over\l_2-\l_1}\;+\; $$ $$ +\;\l_3^3(-3)\l_4^2i\l_2\;+\;i\l_3^3(-4)i\l_4 {1\over 2}\l_2^2 \;\sim\; $$ \be \sim\;-\l_2\l_3^2\l_4^3\;+\; {(5\l_1^2+5i\l_1-{3\over 2})\l_3^2\l_4^3+ (5i\l_1^2-4\l_1-i)\l_3\l_4^3+(-{3\over 2}\l_1^2-i\l_1+{1\over 4})\l_4^3 \over\l_2-\l_1} \label{I410a} \ee Here the symmetry, the formula (\ref{l^m}) for $m=4$ and the relation (\ref{l^3}). Now we shall treat the other six terms $I_1^{(4)}, I_2^{(4)}, I_3^{(4)}, I_4^{(4)}, I_6^{(4)}$ and $I_8^{(4)}$ given by the expressions (\ref{I41}),(\ref{I42}),(\ref{I43}),(\ref{I44}), (\ref{I46}) and (\ref{I48}) respectively. Here we shall proceed in two steps. As a first step we will reduce all these six terms to a following form \be I_j^{(4)}\;\sim\;{A_j\over (\l_2-\l_1)(\l_3-\l_1)(\l_4-\l_3)}\;+\; {B_j\over (\l_2-\l_1)(\l_3-\l_1)}\;+\; {C_j\over (\l_2-\l_1)(\l_4-\l_3)}\;+\; {D_j\over \l_2-\l_1} \label{I123468} \ee where $j=1,2,3,4,6,8$ and $A_j,B_j,C_j,D_j$ are some polynomials. Then we shall sum up all the six results and get the "canonical" form for the sum obtained. Let us start with the expression (\ref{I41}) for the term $I_1^{(4)}$ $$ I_1^{(4)}= -{i\over 2}{(\l_1+i)\l_1^3(\l_2+i)\l_3^3(\l_4+i)^4 \over(\l_2-\l_1+i)(\l_3-\l_1-i)(\l_4-\l_3+i)}\;=\; $$ $$ =\;-{i\over 2}{(\l_1^2+\l_1(\l_2+i)+(\l_2+i)^2)(\l_2+i)\l_3^3(\l_4+i)^4 \over\l_4-\l_3+i}\;+\; $$ $$ +\;{i\over 2}{(\l_1^2+\l_1(\l_2+i)+(\l_2+i)^2)(\l_2+i)\l_3^4(\l_4+i)^4 \over(\l_3-\l_1-i)(\l_4-\l_3+i)}\;+\; {i\over 2}{(\l_2+i)^4\l_3^3(\l_4+i)^3\over\l_2-\l_1+i}\;-\; $$ $$ -\;{i\over 2}{(\l_2+i)^4\l_3^4(\l_4+i)^3\over (\l_2-\l_1-i)(\l_3-\l_1+i)}\;-\; {i\over 2}{(\l_1+i)(\l_2+i)^4\l_3^4(\l_4+i)^3\over (\l_2-\l_1+i)(\l_3-\l_1-i)(\l_4-\l_3+i)}\;\sim\; $$ $$ \sim\;{i\over 2}{(\l_1^2+\l_1(\l_2+i)+(\l_2+i)^2)(\l_2+i)\l_3^3\l_4^4 \over\l_4-\l_3}\;-\; $$ $$ -\;{i\over 2} {(\l_1^2+\l_1(\l_2+i)+(\l_2+i)^2)(\l_2+i)(\l_3+i)^4(\l_4+i)^4 \over(\l_3-\l_1)(\l_4-\l_3)}\;-\; {i\over 2}{\l_2^4\l_3^3(\l_4+i)^3\over\l_2-\l_1}\;-\; $$ $$ -\;{i\over 2}{\l_2^4(\l_3+i)^4(\l_4+i)^3\over (\l_2-\l_1)(\l_3-\l_1)}\;-\; {i\over 2}{(\l_1+i)\l_2^4(\l_3+i)^4(\l_4+i)^3\over (\l_2-\l_1)(\l_3-\l_1)(\l_4-\l_3)}\;\sim\; $$ $$ \sim\;{i\over 2}{\l_1^3\l_2^4(\l_3^2+\l_3(\l_4+i)+(\l_4+i)^2)(\l_4+i) \over\l_2-\l_1}\;+\; $$ $$ +\;{i\over 2} {(\l_1+i)^4(\l_2+i)^4(\l_3^2+\l_3(\l_4+i)+(\l_4+i)^2)(\l_4+i) \over(\l_2-\l_1)(\l_3-\l_1)}\;-\; {i\over 2}{\l_2^4\l_3^3(\l_4+i)^3\over\l_2-\l_1}\;-\; $$ \be -\;{i\over 2}{\l_2^4(\l_3+i)^4(\l_4+i)^3\over (\l_2-\l_1)(\l_3-\l_1)}\;-\; {i\over 2}{(\l_1+i)\l_2^4(\l_3+i)^4(\l_4+i)^3\over (\l_2-\l_1)(\l_3-\l_1)(\l_4-\l_3)} \label{I41a} \ee where we have used symmetries and the formula (\ref{l^m}) with $m=4$. For the next term $I_2^{(4)}$ we start with the expression (\ref{I42}) $$ I_2^{(4)}\;\sim\; {i\over 2}{(\l_1+i)^4(\l_2+i)\l_3^3(\l_4+i)\l_4^3 \over(\l_2-\l_1-i)(\l_3-\l_1-i)(\l_4-\l_3+i)}\;=\; $$ $$ =\;-{i\over 2}{(\l_1+i)^4(\l_2+i) (\l_3^2+\l_3(\l_4+i)+(\l_4+i)^2)(\l_4+i)\l_4^3 \over(\l_2-\l_1-i)(\l_3-\l_1-i)}\;+\; $$ $$ +\;{i\over 2}{(\l_1+i)^4(\l_2+i)(\l_4+i)^4\l_4^3 \over(\l_2-\l_1-i)(\l_3-\l_1-i)(\l_4-\l_3+i)}\;\sim\; $$ \be \sim\;{i\over 2}{\l_1^4(\l_2+i) (\l_3^2+\l_3(\l_4+i)+(\l_4+i)^2)(\l_4+i)\l_4^3 \over(\l_2-\l_1)(\l_3-\l_1)}\;+\; {i\over 2}{\l_1^4(\l_2+i)\l_4^4(\l_4-i)^3 \over(\l_2-\l_1)(\l_3-\l_1)(\l_4-\l_3)} \label{I42a} \ee The term $I_3^{(4)}$ given by (\ref{I43}) is also simple to treat $$ I_3^{(4)}= {i\over 2}{(\l_1+i)^4\l_2^3(\l_3+i)(\l_4+i)\l_4^3 \over(\l_2-\l_1-i)(\l_3-\l_1-i)(\l_4-\l_3-i)}\;=\; $$ $$ =\;-{i\over 2}{(\l_1+i)^4\l_2^3(\l_4+i)\l_4^3 \over(\l_2-\l_1-i)(\l_3-\l_1-i)}\;+\; {i\over 2}{(\l_1+i)^4\l_2^3(\l_4+i)\l_4^4 \over(\l_2-\l_1-i)(\l_3-\l_1-i)(\l_4-\l_3-i)}\;\sim\; $$ \be \sim\;{i\over 2}{\l_1^4\l_2^3(\l_4+i)\l_4^3 \over(\l_2-\l_1)(\l_3-\l_1)}\;+\; {i\over 2}{\l_1^4\l_2^3(\l_4+2i)(\l_4+i)^4 \over(\l_2-\l_1)(\l_3-\l_1)(\l_4-\l_3)} \label{I43a} \ee The fourth term $I_4^{(4)}$ (\ref{I44}) demands more work in order to reduce it to the form (\ref{I123468}) $$ I_4^{(4)}\;\sim\; {i\over 2}{(\l_1+i)(\l_2+i)^4(\l_3+i)\l_3^3\l_4^3 \over(\l_2-\l_1+i)(\l_3-\l_1-i)(\l_4-\l_3-i)}\;=\; $$ $$ =\;-{i\over 2}{(\l_1+i)(\l_2+i)^4\l_3^3\l_4^3 \over(\l_2-\l_1+i)(\l_3-\l_1-i)}\;+\; {i\over 2}{(\l_1+i)(\l_2+i)^4\l_3^3\l_4^4 \over(\l_2-\l_1+i)(\l_3-\l_1-i)(\l_4-\l_3-i)}\;=\; $$ $$ =\;{i\over 2}{(\l_2+i)^4\l_3^3\l_4^3 \over(\l_2-\l_1+i)}\;-\; {i\over 2}{(\l_2+i)^4\l_3^4\l_4^3 \over(\l_2-\l_1+i)(\l_3-\l_1-i)}\;+\; $$ $$ +{i (\l_1+i)(\l_2+i)^4(\l_3^2+\l_3(\l_1+i)+(\l_1+i)^2)\l_4^4 \over 2 (\l_2-\l_1+i)(\l_4-\l_3-i)}+ {i (\l_1+i)^4(\l_2+i)^4\l_4^4 \over 2 (\l_2-\l_1+i)(\l_3-\l_1-i)(\l_4-\l_3-i)}\sim $$ $$ \sim\;-{i\over 2}{\l_2^4(\l_3+i)^4\l_4^3 \over(\l_2-\l_1)(\l_3-\l_1)}\;+\; {i\over 2}{(\l_1+i)\l_2^4(\l_3^2+\l_3(\l_1+i)+(\l_1+i)^2)(\l_4+i)^4 \over(\l_2-\l_1)(\l_4-\l_3)}\;-\; $$ $$ -\;{i\over 2}{\l_1^4\l_2^4(\l_4+i)^4 \over(\l_2-\l_1+i)(\l_3-\l_1)(\l_4-\l_3)}\;\sim\; $$ $$ \sim\;-{i\over 2}{\l_2^4(\l_3+i)^4\l_4^3 \over(\l_2-\l_1)(\l_3-\l_1)}\;+\; {i\over 2}{(\l_1+i)\l_2^4(\l_3^2+\l_3(\l_1+i)+(\l_1+i)^2)(\l_4+i)^4 \over(\l_2-\l_1)(\l_4-\l_3)}\;+\; $$ $$ +{i (\l_1^3+\l_1^2(\l_2+i)+\l_1(\l_2+i)^2+(\l_2+i)^3) \l_2^4(\l_4+i)^4 \over 2 (\l_3-\l_1)(\l_4-\l_3)}- {i (\l_2+i)^4\l_2^4(\l_4+i)^4 \over 2 (\l_2-\l_1+i)(\l_3-\l_1)(\l_4-\l_3)}\sim $$ $$ \sim\;-{i\over 2}{\l_2^4(\l_3+i)^4\l_4^3 \over(\l_2-\l_1)(\l_3-\l_1)}\;+\; {i\over 2}{(\l_1+i)\l_2^4(\l_3^2+\l_3(\l_1+i)+(\l_1+i)^2)(\l_4+i)^4 \over(\l_2-\l_1)(\l_4-\l_3)}\;-\; $$ \be -\;{i\over 2}{(\l_2+i)^4\l_4^4 (\l_3^3+\l_3^2(\l_4+i)+\l_3(\l_4+i)^2+(\l_4+i)^3) \over(\l_2-\l_1)(\l_3-\l_1)}\;+\; {i\over 2}{\l_2^4(\l_2-i)^4(\l_4+i)^4 \over(\l_2-\l_1)(\l_3-\l_1)(\l_4-\l_3)} \label{I44a} \ee Now we treat $I_6^{(4)}$ given by (\ref{I46}) as follows $$ I_6^{(4)}\;\sim\; i{(\l_1+i)(\l_2+i)\l_2^3\l_3^3(\l_4+i)^4 \over(\l_2-\l_1-i)(\l_3-\l_1-i)(\l_4-\l_3+i)}\;=\; $$ $$ =\;i{(\l_1+i)(\l_2+i)(\l_2^2+\l_2(\l_1+i)+(\l_1+i)^2)\l_3^3(\l_4+i)^4 \over(\l_3-\l_1-i)(\l_4-\l_3+i)}\;+\; $$ $$ +\;i{(\l_1+i)^4(\l_2+i)\l_3^3(\l_4+i)^4 \over(\l_2-\l_1-i)(\l_3-\l_1-i)(\l_4-\l_3+i)}\;=\; $$ $$ =\;-i{(\l_2+i)(\l_2^2+\l_2(\l_1+i)+(\l_1+i)^2)\l_3^3(\l_4+i)^4 \over\l_4-\l_3+i}\;+\; $$ $$ +\;i{(\l_2+i)(\l_2^2+\l_2(\l_1+i)+(\l_1+i)^2)\l_3^4(\l_4+i)^4 \over(\l_3-\l_1-i)(\l_4-\l_3+i)}\;+\; i{(\l_1+i)^4(\l_2+i)\l_3^3(\l_4+i)^4 \over(\l_2-\l_1-i)(\l_3-\l_1-i)(\l_4-\l_3+i)}\;=\; $$ $$ =\;i{(\l_2+i)(\l_2^2+\l_2(\l_1+i)+(\l_1+i)^2)\l_3^3\l_4^4 \over\l_4-\l_3}\;+\; $$ $$ +\;i{(\l_2+i)(\l_2^2+\l_2(\l_1+i)+(\l_1+i)^2)(\l_3+i)^4(\l_4+i)^4 \over(\l_3-\l_1)(\l_4-\l_3)}\;+\; i{\l_1^4(\l_2+i)\l_3^3\l_4^4 \over(\l_2-\l_1)(\l_3-\l_1)(\l_4-\l_3)}\;\sim\; $$ $$ \sim\;-i{\l_1^3\l_2^4(\l_3+i)(\l_3^2+\l_3(\l_4+i)+(\l_4+i)^2) \over\l_2-\l_1}\;-\; $$ \be -\;i{(\l_1+i)^4(\l_2+i)^4(\l_4^2+\l_4(\l_3+i)+(\l_3+i)^2)(\l_4+i) \over(\l_2-\l_1)(\l_3-\l_1)}\;+\; i{\l_1^4(\l_2+i)\l_3^3\l_4^4 \over(\l_2-\l_1)(\l_3-\l_1)(\l_4-\l_3)} \label{I46a} \ee The last term $I_8^{(4)}$ (\ref{I48}) is more simple $$ I_8^{(4)}\;\sim\; -i{(\l_1+i)(\l_2+i)^4\l_3^3(\l_4+i)\l_4^3 \over(\l_2-\l_1+i)(\l_3-\l_1-i)(\l_4-\l_3+i)}\;=\; $$ $$ =\;i{(\l_2+i)^4\l_3^3(\l_4+i)\l_4^3 \over(\l_2-\l_1+i)(\l_4-\l_3+i)}\;-\; i{(\l_2+i)^4\l_3^4(\l_4+i)\l_4^3 \over(\l_2-\l_1+i)(\l_3-\l_1-i)(\l_4-\l_3+i)}\;=\; $$ $$ =\;i{(\l_2+i)^4\l_3^3\l_4^3 \over\l_2-\l_1+i}\;+\; i{(\l_2+i)^4\l_3^4\l_4^3 \over(\l_2-\l_1+i)(\l_4-\l_3+i)}\;-\; i{(\l_2+i)^4\l_3^4(\l_4+i)\l_4^3 \over(\l_2-\l_1+i)(\l_3-\l_1-i)(\l_4-\l_3+i)}\;\sim\; $$ \be \sim\;i{\l_2^4(\l_3+i)^4\l_4^3 \over(\l_2-\l_1)(\l_4-\l_3)}\;-\; i{\l_2^4(\l_3+i)^4(\l_4+i)\l_4^3 \over(\l_2-\l_1)(\l_3-\l_1)(\l_4-\l_3)} \label{I48a} \ee Now we are prepared to perform the next our step. Namely, we will gather all the six results (\ref{I41a}-\ref{I48a}) into the form like (\ref{I123468}) $$ I_1^{(4)}\;+\;I_2^{(4)}\;+\;I_3^{(4)}\;+\; I_4^{(4)}\;+\;I_6^{(4)}\;+\;I_8^{(4)}\;\sim\; $$ \be \sim\;{A\over (\l_2-\l_1)(\l_3-\l_1)(\l_4-\l_3)}\;+\; {B\over (\l_2-\l_1)(\l_3-\l_1)}\;+\; {C\over (\l_2-\l_1)(\l_4-\l_3)}\;+\; {D\over \l_2-\l_1} \label{I123468a} \ee where $$ A\;=\;-\;{i\over 2}(\l_1+i)\l_2^4(\l_3+i)^4(\l_4+i)^3\;+\; {i\over 2}\l_1^4(\l_2+i)\l_4^4(\l_4-i)^3\;+\; {i\over 2}\l_1^4\l_2^3(\l_4+2i)(\l_4+i)^4\;+\; $$ \be +\;{i\over 2}\l_2^4(\l_2-i)(\l_4-i)^4\;+\; i\l_1^4(\l_2+i)\l_3^3\l_4^4\;-\; i\l_2^4(\l_3+i)^4(\l_4+i)\l_4^3 \label{A} \ee $$ B\;={i\over 2}(\l_1+i)^4(\l_2+i)^4(\l_3^2+\l_3(\l_4+i)+(\l_4+i)^2) (\l_4+i)\;-\;i\l_2^4(\l_3+i)^4(\l_4+i)^3\;+\; $$ $$ +\;{i\over 2}\l_1^4(\l_2+i)(\l_3^2+\l_3(\l_4+i)+(\l_4+i)^2) (\l_4+i)\l_4^3\;+\; i\l_1^4\l_2^3(\l_4+i)\l_4^3\;+\; $$ \be +\;i(\l_1+i)^4(\l_2+i)^4(\l_4^2+\l_4(\l_3+i)+(\l_3+i)^2) (\l_4+i) \label{B} \ee \be C\;={i\over 2}(\l_1+i)\l_2^4(\l_3^2+\l_3(\l_1+i)+(\l_1+i)^2) (\l_4+i)^4\;+\;i\l_2^4(\l_3+i)^4\l_4^3 \label{C} \ee $$ D\;={i\over 2}\l_1^3\l_2^4(\l_3^2+\l_3(\l_4+i)+(\l_4+i)^2) (\l_4+i)\;-\;i\l_2^4\l_3^3(\l_4+i)^3\;-\; $$ \be -\;i\l_1^3\l_2^4(\l_3^2+\l_3(\l_4+i)+(\l_4+i)^2) (\l_3+i) \label{DD} \ee Now we want to get the "canonical" form (\ref{T_n^c}) for the expression (\ref{I123468a}). To do this we actively used the program {\it MATHEMATICA} because the calculations are straightforward but become more cumbersome. Let us outline our further actions. It is more convenient to start with the first term in the r.h.s. of the formula (\ref{I123468a}) \be {A\over (\l_2-\l_1)(\l_3-\l_1)(\l_4-\l_3)} \label{Aa} \ee We shall use the fact that the denominator is antisymmetric under the substitution \beq \l_1\leftrightarrow\l_3\nonumber\\ \l_2\leftrightarrow\l_4\nonumber \eeq Since $A$ is a polynomial given by (\ref{A}) then the simplification procedure of the term (\ref{Aa}) is as follows. If in the expression (\ref{A}) one faces a monomial \be \l_1^{i_1}\l_2^{i_2}\l_3^{i_3}\l_4^{i_4} \label{mon1234} \ee where without loss of generality $i_4\ge i_2$ one can apply the evident identity \be \l_4^{i_4}\;=\;\l_4^{i_2}\l_3^{i_4-i_2}\;+\; \cases{(\l_4-\l_3)\sum_{k=0}^{i_4-i_2-1} \l_4^{i_2+k}\l_3^{i_4-i_2-1-k}, & $i_4>i_2;$\cr 0 \quad,& $i_4=i_2$\cr} \label{l4i4} \ee Therefore if $i_4>i_2$ then \be {\l_1^{i_1}\l_2^{i_2}\l_3^{i_3}\l_4^{i_4}\over (\l_2-\l_1)(\l_3-\l_1)(\l_4-\l_3)}\;=\; {\l_1^{i_1}\l_2^{i_2}\l_3^{i_4-i_2+i_3}\l_4^{i_2}\over (\l_2-\l_1)(\l_3-\l_1)(\l_4-\l_3)}\;+\; \sum_{k=0}^{i_4-i_2-1} {\l_1^{i_1}\l_2^{i_2}\l_3^{i_4-i_2+i_3-1-k}\l_4^{i_2+k}\over (\l_2-\l_1)(\l_3-\l_1)} \label{l4mon} \ee Let us note that the second term in (\ref{l4mon}) gives rise into the second term $"B"$ in the r.h.s. of the formula (\ref{I123468a}). If $i_4=i_2$ then only the first term in (\ref{l4mon}) survives. The first term in (\ref{l4mon}) is symmetric under the transposition $\l_2\leftrightarrow\l_4$. If $i_1+i_2=i_3+i_4$ then it is also symmetric under the transposition $\l_1\leftrightarrow\l_3$ and the first term is "weakly" equivalent to zero according to the formula (\ref{int}). If $i_1+i_2i_3+i_4$ the sum in (\ref{l31}) should be substituted by \be -{1\over 2}\sum_{k=0}^{i_1+i_2-i_3-i_4-1} {\l_1^{i_1-1-k}\l_2^{i_2}\l_3^{i_3+i_4-i_2+k}\l_4^{i_2}\over (\l_2-\l_1)(\l_4-\l_3)} \label{l31a} \ee In both cases, namely, if $i_1+i_2\ne i_3+i_4$ the sum in (\ref{l31}) or (\ref{l31a}) gives rise into the third term "C" in the r.h.s. of (\ref{I123468a}). Performing this procedure for all monomials of the form (\ref{mon1234}) participating in the polynomial $A$ given by the formula (\ref{A}) one can arrive at the formula \be I_1^{(4)}\;+\;I_2^{(4)}\;+\;I_3^{(4)}\;+\; I_4^{(4)}\;+\;I_6^{(4)}\;+\;I_8^{(4)}\;\sim\; {B'\over (\l_2-\l_1)(\l_3-\l_1)}\;+\; {C'\over (\l_2-\l_1)(\l_4-\l_3)}\;+\; {D\over \l_2-\l_1} \label{I123468b} \ee with some other polynomials $B'$ and $C'$. Due to the $2\leftrightarrow 3$ symmetry of the denominator of the first term in (\ref{I123468b}) one can treat any monomial $\l_1^{j_1}\l_2^{j_2}\l_3^{j_3}\l_4^{j_4}$ participating in $B'$ as follows \be {\l_1^{j_1}\l_2^{j_2}\l_3^{j_3}\l_4^{j_4}\over (\l_2-\l_1)(\l_3-\l_1)}\;\sim\; \sum_{k=0}^{j_3-j_2-1} {\l_1^{j_1}\l_2^{j_2+k}\l_3^{j_3-k-1}\l_4^{j_4}\over \l_2-\l_1} \label{monB'} \ee where without loss of generality it is implied that $j_3>j_2$ because if $j_2=j_3$ then due to the $2\leftrightarrow 3$ symmetry and the formula (\ref{int}) this term would make zero contribution. So the r.h.s. of the (\ref{monB'}) gives rise into the third term in (\ref{I123468b}). Using this one can treat the whole first term in (\ref{I123468b}). For any monomial $\l_1^{k_1}\l_2^{k_2}\l_3^{k_3}\l_4^{k_4}$ participating in $C'$ of the second term of the expression (\ref{I123468b}) one can write \be {\l_1^{k_1}\l_2^{k_2}\l_3^{k_3}\l_4^{k_4} \over (\l_2-\l_1)(\l_4-\l_3)}\;=\; \sum_{l=0}^{k_2-1} {\l_1^{k_1+l}\l_2^{k_2}\l_3^{k_3-1-l}\l_4^{k_4} \over \l_4-\l_3}\;+\; \sum_{l=0}^{k_4-1} {\l_1^{k_1+k_2}\l_3^{k_3+l}\l_4^{k_4-1-l} \over \l_2-\l_1}\;+\; {\l_1^{k_1+k_2}\l_3^{k_3+k_4}\over (\l_2-\l_1)(\l_4-\l_3)}\;\sim\; \label{l12a} \ee \be \sim\;\sum_{l=0}^{k_2-1} {\l_1^{k_1}\l_2^{k_4}\l_3^{k_1+l}\l_4^{k_2-1-l} \over \l_2-\l_1}\;+\; \sum_{l=0}^{k_4-1} {\l_1^{k_1+k_2}\l_3^{k_3+l}\l_4^{k_4-1-l} \over \l_2-\l_1}\;+\; {\l_1^{k_1+k_2}\l_3^{k_3+k_4}\over (\l_2-\l_1)(\l_4-\l_3)} \label{l12b} \ee where for the first sum of (\ref{l12a}) we have applied transformation \beq \l_1\leftrightarrow\l_3\nonumber\\ \l_2\leftrightarrow\l_4\nonumber \eeq So the both the first term and the second term in (\ref{l12b}) give rise into the third "D" term in (\ref{I123468b}) while the third term in (\ref{l12b}) gives rise to the second term "C" in (\ref{I123468b}). Proceeding this way one can treat the whole expression $$ {C'\over (\l_2-\l_1)(\l_4-\l_3)}. $$ As a result of performing this scheme one can arrive at the expression $$ I_1^{(4)}\;+\;I_2^{(4)}\;+\;I_3^{(4)}\;+\; I_4^{(4)}\;+\;I_6^{(4)}\;+\;I_8^{(4)}\;\sim\; $$ \be \sim\;{C''\over (\l_2-\l_1)(\l_4-\l_3)}\;+\; {D'\over \l_2-\l_1} \label{I123468c} \ee where $C''$ is a polynomial of two variable $\l_1$ and $\l_3$. Now with the help of the identity \be \l_1^{i_1}\l_2^{i_2}\;=\;\l_1^{i_1+i_2}\;+\; \cases{(\l_2-\l_1)\sum_{k=0}^{i_2-1} \l_1^{i_1+k}\l_2^{i_2-1-k}, & $i_2>0;$\cr 0 \quad,& $i_2=0$\cr} \label{l12} \ee one can reduce the second term in (\ref{I123468c}) to the form $$ {D'\over \l_2-\l_1}\;\sim\;{D''\over \l_2-\l_1}\;+\;E $$ where $D''$ is a polynomial of $\l_1,\l_3$ and $\l_4$ and $E$ is some polynomial. What is left now is to reduce the power of the polynomials $C'',D''$ and $E$ with the help of the formulae (\ref{l^m}), (\ref{cor1}), (\ref{cor2}) or (\ref{l^3}-\ref{l^6}). We should also use the fact that if we do the substitution $\l_j\rightarrow x_j-i/2$ there is a restriction that the function should be even under the transformation $\{x_1,x_2,x_3,x_4\}\rightarrow \{-x_1,-x_2,-x_3,-x_4\}$ according to the formulae (\ref{tildeU}), (\ref{tildeF}) and (\ref{P_ja}). As a result of all these actions described above we get the "canonical" form for the sum \be I_1^{(4)}\;+\;I_2^{(4)}\;+\;I_3^{(4)}\;+\; I_4^{(4)}\;+\;I_6^{(4)}\;+\;I_8^{(4)}\;\sim\; {C'''\over (\l_2-\l_1)(\l_4-\l_3)}\;+\; {D'''\over \l_2-\l_1}\;+\;E' \label{I123468d} \ee where \be C'''\;=\;2\l_1^2\l_3^2\;+\;4i\l_1\l_3^2\;-\;{3\over 2}\l_3^2 \;-\;{3\over 2}\l_1\l_3\;-\;i\l_3\;+\;{1\over 5} \label{C'''} \ee $$ D'''\;=\; \l_1^2\,(22\l_3^2\l_4^3\;+\;22i\l_3\l_4^3\;-\;{29\over 2}\l_4^3\;+\; 19\l_3\l_4^2\;-\;{5\over 4}i\l_4^2)\;+\; $$ $$ +\;\l_1\,(22i\l_3^2\l_4^3\;-\;{47\over 2}\l_3\l_4^3\;-\; {61\over 4}i\l_4^3\;+\;{67\over 4}i\l_3\l_4^2\;-\; {3\over 2}\l_4^2\;+\;{23\over 4}i\l_4)\;-\; $$ \be -\;{88\over 5}\l_3^2\l_4^3\;-\;{367\over 20}i\l_3\l_4^3\;+\; {427\over 40}\l_4^3\;-\;{303\over 40}\l_3\l_4^2\;+\; {37\over 40}i\l_4^2\;-\;{97\over 20}\l_4) \label{D'''} \ee \be E'\;=\;-\;{57\over 10}\l_2\l_3^2\l_4^3 \label{E'} \ee Now we have to sum up the four contributions (\ref{T4b}), (\ref{I579}), (\ref{I410a}) and (\ref{I123468d}) and get the "canonical" form for $T_4(\l_1,\l_2,\l_3,\l_4)$ \be T_4(\l_1,\l_2,\l_3,\l_4)\;\sim\; P_0^{(4)}\;+\;{P_1^{(4)}\over\l_2-\l_1}\;+\; {P_2^{(4)}\over(\l_2-\l_1)(\l_4-\l_3)} \label{T_4^c} \ee where \be P_0^{(4)}\;=\;-{34\over 5}\l_2\l_3^2\l_4^3 \label{P40} \ee $$ P_1^{(4)}\;=\; \l_1^2\,(30\,\l_3^2\l_4^3\;+\;30\,i\l_3\l_4^3\;-\;16\,\l_4^3\;+\; 18\,\l_3\l_4^2\;+\;8\,\l_4)\;+\; $$ $$ +\;\l_1\,(30\,i\l_3^2\l_4^3\;+\;30\,\l_3\l_4^3\;-\; 16\,i\l_4^3\;+\;18\,i\l_3\l_4^2\;-\; 4\,\l_4^2\;+\;4\,i\l_4)\;-\; $$ \be -\;20\,\l_3^2\l_4^3\;-\;20\,i\l_3\l_4^3\;+\; {54\over 5}\l_4^3\;-\;{42\over 5}\l_3\l_4^2\;-\; {43\over 10}i\l_4 \label{P41} \ee \be P_2^{(4)}\;=\;2\l_1^2\l_3^2\;+\;4i\l_1\l_3^2\;-\;{3\over 2}\l_3^2 \;-\;{3\over 2}\l_1\l_3\;-\;i\l_3\;+\;{1\over 5} \label{P42} \ee Let us note that in terms of the real variables $x_j$ the polynomials $\tilde P_j^{(4)}$ (see (\ref{P_ja})) look a little bit simpler \be \tilde P_0^{(4)}\;=\;-{34\over 5}x_2x_3^2x_4^3 \label{P40a} \ee $$ \tilde P_1^{(4)}\;=\;x_1^2\;(30\,x_3^2x_4^3\;-\;{17\over 2}x_4^3 \;+\;{81\over 2}x_3x_4^2\;+\;{79\over 8}x_4)\;-\;4\,x_1x_4^2\;-\; $$ \be -\;{25\over 2}x_3^2x_4^3 \;+\;{147\over 40}x_4^3\;-\;{531\over 40}x_3x_4^2\;-\; {653\over 160}x_4 \label{P41a} \ee \be \tilde P_2^{(4)}\;=\;2\,x_1^2x_3^2\;+\;{1\over 2}x_1x_3\;-\; {1\over 2}x_3^2\;+\;{3\over 40} \label{P42a} \ee So, the function \be \tilde T_4^c(x_1,x_2,x_3,x_4)\;=\; \tilde P_0^{(4)}\;+\;{\tilde P_1^{(4)}\over x_2-x_1} \;+\;{\tilde P_2^{(4)}\over (x_2-x_1)(x_4-x_3)} \label{tildeT_4^c} \ee is even i.e. $$ \tilde T_4^c(-x_1,-x_2,-x_3,-x_4)\;=\;\tilde T_4^c(x_1,x_2,x_3,x_4) $$ as it should be according to the formula (\ref{tildeT_n^ca}) with $n=4$. Now let us start the second step of our general scheme, namely, the performing of the integration of the "canonical" form (\ref{T_4^c}) \be P(4)\;=\;\prod_{j=1}^4\int_{C_{-1/2}}{d\l_j\over 2\pi i}\> U_4(\l_1,\l_2,\l_3,\l_4)\>T_4(\l_1,\l_2,\l_3,\l_4)\;=\; J_0^{(4)}\;+\;J_1^{(4)}\;+\;J_2^{(4)} \label{P4a} \ee where \be J_0^{(4)}=\prod_{j=1}^4\int_{C_{-1/2}}{d\l_j\over 2\pi i}\> U_4(\l_1,\l_2,\l_3,\l_4)\>P_0^{(4)} \label{J40} \ee \be J_1^{(4)}=\prod_{j=1}^4\int_{C_{-1/2}}{d\l_j\over 2\pi i}\> U_4(\l_1,\l_2,\l_3,\l_4)\> {P_1^{(4)}\over\l_2-\l_1} \label{J41} \ee \be J_1^{(4)}=\prod_{j=1}^4\int_{C_{-1/2}}{d\l_j\over 2\pi i}\> U_4(\l_1,\l_2,\l_3,\l_4)\> {P_2^{(4)}\over(\l_2-\l_1)(\l_4-\l_3)} \label{J42} \ee Using the formulae (\ref{J40}-\ref{J42}), (\ref{P40}-\ref{P42}),(\ref{F}),(\ref{Da}), (\ref{intl^a}-\ref{ro5}) we get $$ J_0^{(4)}\;=\;\prod_{j=1}^4\int_{C_{-1/2}}{d\l_j\over 2\pi i}\> U_4(\l_1,\l_2,\l_3,\l_4)\>(-{34\over 5})\l_2\l_3^2\l_4^3\;=\; $$ $$ =\;D^{(4)}\sum_{l_1=0}^{\infty}(-1)^{l_1} \sum_{l_2=0}^{\infty}(-1)^{l_2}\sum_{l_3=0}^{\infty}(-1)^{l_3} \sum_{l_4=0}^{\infty}(-1)^{l_4} (-{34\over 5})(i l_2+\e_2)(i l_3+\e_3)^2(i l_4+\e_4)^2\;= $$ $$ =\;D^{(4)}\sum_{l_1=0}^{\infty}(-1)^{l_1} \sum_{l_2=0}^{\infty}(-1)^{l_2}\sum_{l_3=0}^{\infty}(-1)^{l_3} \sum_{l_4=0}^{\infty}(-1)^{l_4} (-{34\over 5})\e_2\e_3^2\e_4^3\;=\; $$ \be =\;{1\over 12}\prod_{0\le k U_4(\l_1,\l_2,\l_3,\l_4)\>{P_1^{(4)}(\l_1|\l_3,\l_4) \over \l_2-\l_1}\;=\; $$ $$ =\;D^{(4)}\sum_{l_1=1}^{\infty}(-1)^{l_1} \sum_{l_2=0}^{\infty}(-1)^{l_2}\sum_{l_3=1}^{\infty}(-1)^{l_3} \sum_{l_4=0}^{\infty}(-1)^{l_4} {P_1^{(4)}(-il_1+\e_1|-il_3+\e_3,il_4+\e_4) \over il_2+il_1+\e_2-\e_1}\;=\; $$ $$ =\;-\;i D^{(4)}\int_0^1{ds\over s}\sum_{l_1=1}^{\infty}(-1)^{l_1} \sum_{l_2=0}^{\infty}(-1)^{l_2}\sum_{l_3=1}^{\infty}(-1)^{l_3} \sum_{l_4=0}^{\infty}(-1)^{l_4} $$ $$ P_1^{(4)}(-il_1+\e_1|-il_3+\e_3,il_4+\e_4) s^{l_1+l_2+i(\e_1-\e_2)}\;= $$ $$ =\;\int_0^1 ds \Bigl(-{15\over 2}{(s-1)\over(1+s)^3}\;+\; {3\over 4}{(7-26s+7s^2)\over(1+s)^4}\ln{s}\;+\; {21\over 8}{(s-1)\over(1+s)^3}\ln^2{s}\;-\; {7\over 10}{(2-s+2s^2)\over(1+s)^4}\ln^3{s}\;+\; $$ \be +\;{5\over 48}{(s-1)\over(1+s)^3}\ln^4{s}\;+\; {1\over 240}{(1+22s+s^2)\over(1+s)^4}\ln^5{s}\Bigr)\;=\; {5\over 8}\;-\;2\ln{2}\;+\;{61\over 20}\zeta(3)\;-\; {65\over 32}\zeta(5) \label{J41a} \ee The last term $J_2^{(4)}$ can be calculated as follows $$ J_2^{(4)}\;=\;\prod_{j=1}^4\int_{C_{-1/2}}{d\l_j\over 2\pi i}\> U_4(\l_1,\l_2,\l_3,\l_4)\>{P_2^{(4)}(\l_1,\l_3) \over (\l_2-\l_1)(\l_4-\l_3)}\;=\; $$ $$ =\;D^{(4)}\sum_{l_1=1}^{\infty}(-1)^{l_1} \sum_{l_2=0}^{\infty}(-1)^{l_2}\sum_{l_3=1}^{\infty}(-1)^{l_3} \sum_{l_4=0}^{\infty}(-1)^{l_4} {P_2^{(4)}(-il_1+\e_1,-il_3+\e_3) \over (il_2+il_1+\e_2-\e_1)(il_4+il_3+\e_4-\e_3)}\;=\; $$ $$ =\;-\;D^{(4)}\int_0^1{ds\over s} \int_0^1{dt\over t}\sum_{l_1=1}^{\infty}(-1)^{l_1} \sum_{l_2=0}^{\infty}(-1)^{l_2}\sum_{l_3=1}^{\infty}(-1)^{l_3} \sum_{l_4=0}^{\infty}(-1)^{l_4} $$ $$ P_2^{(4)}(-il_1+\e_1,-il_3+\e_3) s^{l_1+l_2+i(\e_1-\e_2)}t^{l_3+l_4+i(\e_3-\e_4)}\;=\; $$ $$ =\;\int_0^1 ds\int_0^1 dt \Bigl\{4{\ln{s}(\ln{s}-\ln{t})\over(1+s)^2(1+t)^2}\;+\; (-{4\over 3}(-3-s+t+3st)\ln^3{s}\;+\;4(-1+3s-3t+st)\ln^2{s}\ln{t}\;-\; $$ $$ -\;{3\over 4}(3-13s+3t+3st)\ln^2{s}\ln^2{t}\;+\; {5\over 12}(3-5s-5t+3st)\ln^4{s}){1\over(1+s)^3(1+t)^3}\;-\; $$ $$ -\;{1\over 6}{(-1+5s-2s^2+2t-5st+s^2t)\over (1+s)^4(1+t)^3} \ln{s}(\ln^2{s}-\ln^2{t}) (\ln^2{s}-5\ln^2{t})\;+\; $$ $$ +\;({1\over 3}(3+30s-5s^2-34t-4st-34s^2t-5t^2+30st^2+3s^2t^2) \ln^3{s}\ln{t}\;+\; $$ $$ +\;{1\over 30}(4-17s+9s^2-17t+66st-17s^2t+9t^2-17st^2+4s^2t^2) \ln^5{s}\ln{t}\;-\; $$ $$ -\;{1\over 15}(2-31s+7s^2+14t+33st-21s^2t+2t^2+4st^2+2s^2t^2) \ln^3{s}\ln^3{t}){1\over(1+s)^4(1+t)^4}\Bigl\} $$ Taking the integral we come to an answer for $J_2^{(4)}$ \be J_2^{(4)}\;=\;-{1\over 6}\zeta(3)\;-\;{11\over 6}\zeta(3)\ln{2}\;-\; {51\over 80}\zeta(3)^2\;-\;{25\over 96}\zeta(5)\;+\; {85\over 24}\zeta(5)\ln{2} \label{J42a} \ee Summing up all the three results (\ref{J40a}), (\ref{J41a}) and (\ref{J42a}) we finally get our main formula (\ref{P4}) $$ P(4)\;=\;J_0^{(4)}\;+\;J_1^{(4)}\;+\;J_2^{(4)}\;=\; $$ \be {1\over 5}\; -\; 2\ln{2} \; + \; {173\over 60}\,\zeta(3) \; -\; {11\over 6} \,\zeta(3)\, \ln{2}\;-\; {51\over 80} \, \zeta^2(3) \;-\; {55\over 24}\,\zeta(5) \; + \; {85\over 24}\,\zeta(5)\, \ln{2} \label{P4b} \ee \begin{thebibliography}{**} \bibitem{KR} D. ~Kreimer, {\it ``Knots and Feynman Diagrams''}, Cambridge University Press, 2000. \bibitem{BK} H.E. 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