Leading and Next-to-Leading Logarithmic Approximations in Quantum Electrodynamics

Abstrac

t—Methods for determination of QED radiative corrections in the leading and next-to- leading logarithmic approximations are described. Analytical solutions to the QED evolution equations for electron structure functions are presented. Some examples of applying the methods described to construction of the high-precision description of processes in high-energy accelerators are considered. A case of the calculation of higher-order corrections to the muon decay spectrum is considered in detail. Use of the approaches developed to justify methods of radiation return and direct measurement of the QED running constant is discussed.

Appendices

TABLE OF CONVOLUTION INTEGRALS

In many applications, in particular, in the case of a factorization into two or more functions, there arises a problem of finding an analytical expression for an integral of the product of two functions of a single argument, varying within the limits from 0 to 1. This situation is typical when considering perturbative expansions of evolution equations in the renormalization group approach.

The problem of a search for the integral convolution of a product of two functions is solved rather frequently by a method of moments using Mellin transformations. In the space of moments, a result of convolution, more specifically, its moment, is found as a product of the specified functions. Therefore, this method allows problems of the type under study to be solved systematically with the possibility of constructing a general algorithm and its implementation in the form of a computer program. However, this method contains a number of auxiliary steps, including, in particular, inverse Mellin transformations, which can be rather laborious. Moreover, in realistic cases it may be necessary to shift integration limits for separation of a certain region of the physical phase space. Additionally, in especially complicated functions, the convolutions completely or partially have to be found numerically, which does not fit well in the algorithm of the method of moments.

Here, an approach to finding convolutions by the direct integration method is presented, which works well for a wide class of functions, arising in problems, solved by the method of perturbation theory.

Let us consider two functions $f(x)$ and $g(y)$, defined on the interval $0 \leqslant x,\,\,y \leqslant 1.$ Their convolution is determined as follows:

$$\begin{gathered} \text{[}f \otimes g](z) = \int\limits_0^1 {{\text{d}}x} \int\limits_0^1 {{\text{d}}y} \delta (z - xy)f(x)g(y) \\ = \int\limits_z^1 {\frac{{{\text{d}}x}}{x}} f(x)g\left( {\frac{z}{x}} \right), \\ \end{gathered} $$

((A.1))

where $0 \leqslant z \leqslant 1.$

Sometimes it is necessary to find a convolution of special functions. In particular, the functions, defined using the so-called plus-prescription. This prescription specifies a rule of regularization of pole singularity at the point $x = 1$:

$$\begin{gathered} \int\limits_{{{x}_{{{\text{min}}}}}}^1 {{\text{d}}x} {{[f(x)]}_{ + }}g(x) \\ = \int\limits_0^1 {{\text{d}}x} f(x)[g(x)\Theta (x - {{x}_{{{\text{min}}}}}) - g(1)], \\ \Theta (x) = \left\{ {\begin{array}{*{20}{c}} {1\,\,\,\,{\text{for}}\,\,\,\,x \geqslant 0} \\ {0\,\,\,\,{\text{for}}\,\,\,\,x < 0} \end{array}} \right.,\,\,\,\,0 \leqslant {{x}_{{{\text{min}}}}} < 1. \\ \end{gathered} $$

((A.2))

Integrals of functions having poles at the points $x = z$ or $x = 1$ diverge. They can be regularized by introducing the small auxiliary parameter $\Delta \ll 1.$ Moreover, as a result of specific calculations, the terms depending on this parameter should cancel out. A physical meaning of the parameter, separating phase spaces of the soft and hard bremsstrahlung, can also be assigned to this parameter. However, there is an unambiguous correspondence between the methods of $\Delta$-regularization and plus-prescription (A.2):

$$\begin{gathered} {{[f(x)]}_{ + }} = \mathop {\lim}\limits_{\Delta \to 0} [\delta (1 - x){{f}_{\Delta }} + \Theta (1 - \Delta - x){{f}_{\Theta }}(x)], \\ {{f}_{\Delta }} = - \int\limits_0^{1 - \Delta } {{\text{d}}x} f(x),\,\,\,\,{{f}_{\Theta }}(x) = {{\left. {f(x)} \right|}_{{x < 1}}}. \\ \end{gathered} $$

((A.3))

We will call ${{f}_{\Delta }}$ and ${{f}_{\Theta }}(x)$ as the Δ- and Θ-parts of the special function $f(x),$ respectively. The second realization of the plus-prescription is convenient both in the case of analytical calculations and in the numerical calculation, where the first definition is certainly inapplicable directly, since it allows the appearance of zero in the denominator of the integrand. Below, it is shown how the $\Delta$-regularization of the divergent integral can be used systematically by making a check of the cancellation of the dependence on the auxiliary parameter after summing the contributions of the Δ- and Θ-parts, as, e.g., in (A.3).

To find the Δ-parts of the specified function, it is possible to use a table of definite integrals over the interval $0 < x < 1 - \Delta {\kern 1pt} .$ For integrals of nonsingular functions, a full interval $0 < x < 1$ is taken, and the relevant values can be found in many tables, see, e.g., [279]. A convolution of two scalar functions, regularized using the plus-prescription, can be presented as

$$\begin{gathered} \text{[}{{[f]}_{ + }} \otimes {{[g]}_{ + }}](z) = \mathop {\lim}\limits_{\Delta \to 0} \left\{ {\int\limits_{{z \mathord{\left/ {\vphantom {z {(1}}} \right. \kern-0em} {(1}} - \Delta )}^{1 - \Delta } {\frac{{{\text{d}}x}}{x}} } \right.{{f}_{\Theta }}(x){{g}_{\Theta }}\left( {\frac{z}{x}} \right) \\ \left. {\frac{{^{{^{{}}}}}}{{_{{_{{}}}}}} + \,\,{{f}_{\Delta }}{{g}_{\Theta }}(z) + {{f}_{\Theta }}(z){{g}_{\Delta }}} \right\}. \\ \end{gathered} $$

((A.4))

We first consider integrals of singular functions, which we regularize by the parameter $\Delta \ll 1,$ leaving the dependence on it only in the logarithm argument:

$$\begin{gathered} \int\limits_{{z \mathord{\left/ {\vphantom {z {(1 - \Delta )}}} \right. \kern-0em} {(1 - \Delta )}}}^{1 - \Delta } {{\text{d}}x} \frac{{l{{n}^{n}}(1 - x)}}{{1 - x}} = \frac{1}{{n + 1}}l{{n}^{{n + 1}}}(1 - z) \\ - \,\,\frac{1}{{n + 1}}l{{n}^{{n + 1}}}\Delta ,\,\,\,\,n = 0,1,2, \ldots \\ \end{gathered} $$

((A.5))

$$\begin{gathered} \int\limits_{{z \mathord{\left/ {\vphantom {z {(1}}} \right. \kern-0em} {(1}} - \Delta )}^{1 - \Delta } {{\text{d}}x} \frac{1}{{x - z}} = - \ln\Delta + \ln(1 - z) - \ln{z}, \\ \int\limits_{{z \mathord{\left/ {\vphantom {z {(1}}} \right. \kern-0em} {(1}} - \Delta )}^{1 - \Delta } {{\text{d}}x} \frac{{\ln{x}}}{{x - z}} = - \ln\Delta \ln{z} + {\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right) \\ + \,\,\ln(1 - z)\ln{z} - \frac{1}{2}{{\ln}^{2}}z,\,\,\,\,\int\limits_{{z \mathord{\left/ {\vphantom {z {(1}}} \right. \kern-0em} {(1}} - \Delta )}^{1 - \Delta } {{\text{d}}x} \frac{{\ln(1 - x)}}{{x - z}} \\ = - \ln\Delta \ln(1 - z) - \ln(1 - z)\ln{z} + {{\ln}^{2}}(1 - z) - \zeta (2), \\ \int\limits_{{z \mathord{\left/ {\vphantom {z {(1}}} \right. \kern-0em} {(1}} - \Delta )}^{1 - \Delta } {{\text{d}}x} \frac{{{{\ln}^{2}}x}}{{x - z}} = - ln\Delta {{\ln}^{2}}z + 2{{{\text{S}}}_{{1,2}}}\left( {1 - z} \right) \\ + \,\,2{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)\ln{z} + \ln(1 - z){{\ln}^{2}}z - \frac{1}{3}{{\ln}^{3}}z, \\ \end{gathered} $$

$$\begin{gathered} \int\limits_{{z \mathord{\left/ {\vphantom {z {(1 - \Delta )}}} \right. \kern-0em} {(1 - \Delta )}}}^{1 - \Delta } {{\text{d}}x} \frac{{{{\ln}^{2}}(1 - x)}}{{x - z}} = - \ln\Delta {{\ln}^{2}}(1 - z) + {{\ln}^{3}}(1 - z) \\ - \,\,{{\ln}^{2}}(1 - z)\ln{z} - 2\zeta (2)\ln(1 - z) + 2\zeta (3), \\ \int\limits_{{z \mathord{\left/ {\vphantom {z {(1 - \Delta )}}} \right. \kern-0em} {(1 - \Delta )}}}^{1 - \Delta } {{\text{d}}x} \frac{{ln(1 - x)\ln{x}}}{{x - z}} = - ln\Delta ln(1 - z)\ln{z} \\ + \,\,2{{{\text{S}}}_{{1,2}}}\left( {1 - z} \right) - {\text{L}}{{{\text{i}}}_{3}}\left( {1 - z} \right) + {\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)ln(1 - z) \\ + \,\,{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)\ln{z} + {{\ln}^{2}}(1 - z)\ln{z} \\ - \,\,\frac{1}{2}\ln(1 - z){{\ln}^{2}}z - \zeta (2)\ln{z}, \\ \end{gathered} $$

$$\begin{gathered} \int\limits_{{z \mathord{\left/ {\vphantom {z {(1 - \Delta )}}} \right. \kern-0em} {(1 - \Delta )}}}^{1 - \Delta } {{\text{d}}x} \frac{{{\text{L}}{{{\text{i}}}_{2}}\left( {1 - x} \right)}}{{x - z}} = {\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)(\ln(1 - z) \\ - \,\,\ln{z} - \ln\Delta ) - {{{\text{S}}}_{{1,2}}}\left( {1 - z} \right) - {\text{L}}{{{\text{i}}}_{3}}\left( {1 - z} \right), \\ \int\limits_{{z \mathord{\left/ {\vphantom {z {(1 - \Delta )}}} \right. \kern-0em} {(1 - \Delta )}}}^{1 - \Delta } {{\text{d}}x} \frac{{{\text{L}}{{{\text{i}}}_{3}}\left( {1 - x} \right)}}{{x - z}} = {\text{L}}{{{\text{i}}}_{3}}\left( {1 - z} \right)(\ln(1 - z) \\ - \,\,\ln{z} - \ln\Delta ) - {{{\text{S}}}_{{2,2}}}\left( {1 - z} \right) - {\text{L}}{{{\text{i}}}_{4}}\left( {1 - z} \right), \\ \end{gathered} $$

((A.6))

$$\begin{gathered} \int\limits_{{z \mathord{\left/ {\vphantom {z {(1 - \Delta )}}} \right. \kern-0em} {(1 - \Delta )}}}^{1 - \Delta } {{\text{d}}x} \frac{{{{{\text{S}}}_{{1,2}}}\left( {1 - x} \right)}}{{x - z}} = {{{\text{S}}}_{{1,2}}}\left( {1 - z} \right)(\ln(1 - z) - \ln{z} \\ - \ln\Delta ) + {{{\text{S}}}_{{2,2}}}\left( {1 - z} \right) - 2{{{\text{S}}}_{{1,3}}}\left( {1 - z} \right) - \frac{1}{2}{{\left( {{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)} \right)}^{2}}, \\ \int\limits_{{z \mathord{\left/ {\vphantom {z {(1 - \Delta )}}} \right. \kern-0em} {(1 - \Delta )}}}^{1 - \Delta } {{\text{d}}x} \frac{{{\text{L}}{{{\text{i}}}_{2}}\left( {1 - x} \right)ln(1 - x)}}{{x - z}} = {\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)\ln(1 - z) \\ \times \,\,(\ln(1 - z) - \ln{z} - \ln\Delta ) + {\text{L}}{{{\text{i}}}_{4}}\left( {1 - z} \right) - 2{{{\text{S}}}_{{2,2}}}\left( {1 - z} \right) \\ + \,\,\frac{1}{2}{{\left( {{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)} \right)}^{2}} - {\text{L}}{{{\text{i}}}_{3}}\left( {1 - z} \right)\ln(1 - z) \\ - \,\,{{{\text{S}}}_{{1,2}}}\left( {1 - z} \right)\ln(1 - z) - \zeta (2){\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right), \\ \end{gathered} $$

$$\begin{gathered} \int\limits_{{z \mathord{\left/ {\vphantom {z {(1 - \Delta )}}} \right. \kern-0em} {(1 - \Delta )}}}^{1 - \Delta } {{\text{d}}x} \frac{{{\text{L}}{{{\text{i}}}_{2}}\left( {1 - x} \right)\ln{x}}}{{x - z}} = {\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)\ln{z}(\ln(1 - z) \\ - \ln{z} - \ln\Delta ) + 4{{{\text{S}}}_{{1,3}}}\left( {1 - z} \right) - 4{{{\text{S}}}_{{2,2}}}\left( {1 - z} \right) \\ + \,\,{{\left( {{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)} \right)}^{2}} - {\text{L}}{{{\text{i}}}_{3}}\left( {1 - z} \right)\ln{z} \\ + \,\,{{{\text{S}}}_{{1,2}}}\left( {1 - z} \right)\ln{z} + \frac{1}{2}{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right){{\ln}^{2}}z, \\ \end{gathered} $$

$$\begin{gathered} \int\limits_{{z \mathord{\left/ {\vphantom {z {(1 - \Delta )}}} \right. \kern-0em} {(1 - \Delta )}}}^{1 - \Delta } {{\text{d}}x} \frac{{{{\ln}^{2}}(1 - x)\ln{x}}}{{x - z}} = - \ln\Delta {{\ln}^{2}}(1 - z)\ln{z} \\ + \,\,2{\text{L}}{{{\text{i}}}_{4}}\left( {1 - z} \right) - {{\left( {{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)} \right)}^{2}} - 2{\text{L}}{{{\text{i}}}_{3}}\left( {1 - z} \right) \\ \times \,\,(\ln(1 - z) + \ln{z}) + 4{{{\text{S}}}_{{1,2}}}\left( {1 - z} \right)\ln(1 - z) \\ + \,\,2{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)\ln(1 - z)\ln{z} + {\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right){{\ln}^{2}}(1 - z) \\ + \,\,{{\ln}^{3}}(1 - z)\ln{z} - \frac{1}{2}{{\ln}^{2}}(1 - z){{\ln}^{2}}z \\ - \,\,2\zeta (2)\ln(1 - z)\ln{z} + 2\zeta (3)\ln{z}, \\ \end{gathered} $$

$$\begin{gathered} \int\limits_{{z \mathord{\left/ {\vphantom {z {(1 - \Delta )}}} \right. \kern-0em} {(1 - \Delta )}}}^{1 - \Delta } {{\text{d}}x} \frac{{\ln(1 - x){{\ln}^{2}}x}}{{x - z}} = - \ln\Delta \ln(1 - z){{\ln}^{2}}z \\ + \,\,2{{{\text{S}}}_{{2,2}}}\left( {1 - z} \right) - {{\left( {{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)} \right)}^{2}} - 2{\text{L}}{{{\text{i}}}_{3}}\left( {1 - z} \right)\ln{z} \\ + \,\,2{{{\text{S}}}_{{1,2}}}\left( {1 - z} \right)(ln(1 - z) + \ln{z}) + 2{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right) \\ \times \,\,\ln(1 - z)\ln{z} + {\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right){{\ln}^{2}}z + {{\ln}^{2}}(1 - z){{\ln}^{2}}z \\ - \,\,\frac{1}{3}\ln(1 - z){{\ln}^{3}}z - \zeta (2){{\ln}^{2}}z, \\ \end{gathered} $$

$$\begin{gathered} \int\limits_{{z \mathord{\left/ {\vphantom {z {(1 - \Delta )}}} \right. \kern-0em} {(1 - \Delta )}}}^{1 - \Delta } {{\text{d}}x} \frac{{{{\ln}^{3}}(1 - x)}}{{x - z}} = {{\ln}^{3}}(1 - z)(\ln(1 - z) - \ln{z} - \ln\Delta ) \\ - \,\,3\zeta (2){{\ln}^{2}}(1 - z) + 6\zeta (3)\ln(1 - z) - 6\zeta (4), \\ \int\limits_{{z \mathord{\left/ {\vphantom {z {(1 - \Delta )}}} \right. \kern-0em} {(1 - \Delta )}}}^{1 - \Delta } {{\text{d}}x} \frac{{{{\ln}^{3}}x}}{{x - z}} = - \ln\Delta {{\ln}^{3}}z + 6{{{\text{S}}}_{{1,3}}}\left( {1 - z} \right) \\ + \,\,6{{{\text{S}}}_{{1,2}}}\left( {1 - z} \right)\ln{z} + 3{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right){{\ln}^{2}}z \\ + \,\,\ln(1 - z){{\ln}^{3}}z - \frac{1}{4}l{{n}^{4}}z. \\ \end{gathered} $$

The following set of integrals of convolution of nonsingular functions was used in [45, 68, 93, 110] and other works in obtaining analytical expressions for higher-order corrections in LLA and NLLA:

$$\begin{gathered} \int\limits_z^1 {{\text{d}}x{{x}^{n}}} = \frac{1}{{n + 1}}(1 - {{z}^{{n + 1}}}),\,\,\,\,n \ne - 1, \\ \int\limits_z^1 {{\text{d}}x} \frac{{l{{n}^{n}}x}}{x} = - \frac{1}{{n + 1}}l{{n}^{{n + 1}}}z,\,\,\,\,n = 0,1,2, \ldots \\ \int\limits_z^1 {{\text{d}}x} \frac{{l{{n}^{n}}(1 - x)}}{x} = {{( - 1)}^{n}}n![\zeta (n + 1) - {{{\text{S}}}_{{1,n}}}\left( z \right)], \\ n = 1,2,3, \ldots \\ \int\limits_z^1 {{\text{d}}x} \frac{{\ln(1 - x)\ln{x}}}{x} = {{{\text{S}}}_{{1,2}}}\left( {1 - z} \right) - \frac{1}{2}\ln(1 - z){{\ln}^{2}}z, \\ \int\limits_z^1 {{\text{d}}x} \int\limits_{}^{} {{x}^{n}}\ln{x} = - \frac{{{{z}^{{n + 1}}}}}{{n + 1}}\ln{z} - \frac{1}{{{{{(n + 1)}}^{2}}}}(1 - {{z}^{{n + 1}}}), \\ n = 0,1,2, \ldots \\ \end{gathered} $$

$$\begin{gathered} \int\limits_z^1 {{\text{d}}x} \frac{{\ln{x}}}{{{{x}^{n}}}} = \frac{1}{{{{z}^{{n - 1}}}{{{(n - 1)}}^{2}}}}((n - 1)\ln{z} + 1 - {{z}^{{n - 1}}}), \\ n = 2,3,4, \ldots \\ \int\limits_z^1 {{\text{d}}x} \frac{{l{{n}^{n}}x}}{{1 - x}} = {{( - 1)}^{n}}n!{{{\text{S}}}_{{1,n}}}\left( {1 - z} \right),\,\,\,\,n = 1,2,3, \ldots \\ \int\limits_z^1 {{\text{d}}x} \frac{{\ln(1 - x)\ln{x}}}{{1 - x}} = {\text{L}}{{{\text{i}}}_{3}}\left( {1 - z} \right) - {\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)\ln(1 - z), \\ \int\limits_z^1 {{\text{d}}x} {{x}^{n}}\ln(1 - x) = \frac{{1 - {{z}^{{n + 1}}}}}{{n + 1}}\ln(1 - z) \\ + \,\,\frac{1}{{n + 1}}\left( {\sum\limits_{k = 1}^{n + 1} {\frac{{{{z}^{k}}}}{k}} - {{{\text{S}}}_{1}}\left( {n + 1} \right)} \right),\,\,\,\,n = 0,1,2, \ldots \\ \end{gathered} $$

$$\begin{gathered} \int\limits_z^1 {{\text{d}}x} \frac{{\ln(1 - x)}}{{{{x}^{2}}}} = \ln{z} + \frac{{1 - z}}{z}\ln(1 - z), \\ \int\limits_z^1 {{\text{d}}x} \frac{{\ln(1 - x)}}{{{{x}^{n}}}} = \frac{1}{{n - 1}}\left( {\ln{z} + \frac{{1 - {{z}^{{n - 1}}}}}{{{{z}^{{n - 1}}}}}} \right.\ln(1 - z) \\ - \,\,\sum\limits_{k = 1}^{n - 2} {\frac{1}{{k{{z}^{k}}}}} + {{{\text{S}}}_{1}}\left( {n - 2} \right)),\,\,\,\,n = 3,4,5, \ldots \\ \int\limits_z^1 {{\text{d}}x} {{x}^{n}}{{\ln}^{2}}x = \frac{2}{{{{{(n + 1)}}^{3}}}}(1 - {{z}^{{n + 1}}}) \\ + \,\,\frac{{2{{z}^{{n + 1}}}}}{{{{{(n + 1)}}^{2}}}}lnz - \frac{{{{z}^{{n + 1}}}}}{{n + 1}}{{\ln}^{2}}z,\,\,\,\,n \ne - 1, \\ \end{gathered} $$

$$\begin{gathered} \int\limits_z^1 {{\text{d}}x} {{x}^{n}}{{\ln}^{2}}(1 - x) = \frac{{1 - {{z}^{{n + 1}}}}}{{n + 1}}{{\ln}^{2}}(1 - z) \\ + \,\,\sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{c}} n \\ k \end{array}} \right)} {{( - 1)}^{k}} \times \,\,\frac{{{{{(1 - z)}}^{{k + 1}}}}}{{{{{(k + 1)}}^{2}}}}\left( {\frac{2}{{k + 1}} - 2\ln(1 - z)} \right), \\ n = 0,1,2, \ldots \\ \int\limits_z^1 {{\text{d}}x} {{x}^{n}}\ln(1 - x)\ln{x} = - \frac{1}{{n + 1}}{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right) \\ - \,\,\frac{{{{z}^{{n + 1}}}}}{{n + 1}}\ln(1 - z)\ln{z} - \frac{{1 - {{z}^{{n + 1}}}}}{{{{{(n + 1)}}^{2}}}}\ln(1 - z) \\ + \,\,\frac{1}{{n + 1}}\ln{z}\sum\limits_{k = 1}^{n + 1} {\frac{{{{z}^{k}}}}{k}} + \frac{1}{{{{{(n + 1)}}^{2}}}} \\ \times \,\,\sum\limits_{k = 1}^{n + 1} {(1 - {{z}^{k}})} \frac{{n + k + 1}}{{{{k}^{2}}}}, \\ n = 0,1,2, \ldots \\ \end{gathered} $$

((A.7))

$$\begin{gathered} \int\limits_z^1 {{\text{d}}x} \frac{{\ln{x}\ln(1 - x)}}{{{{x}^{n}}}} = \frac{1}{{n - 1}}{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right) + \frac{{\ln{z}ln(1 - z)}}{{{{z}^{{n - 1}}}(n - 1)}} \\ + \,\,\ln{z}(\frac{1}{{{{{(n - 1)}}^{2}}}} - \frac{1}{{n - 1}}\left. {\sum\limits_{k = 1}^{n - 2} {\frac{1}{{{{z}^{k}}k}}} } \right) + \frac{{1 - {{z}^{{n - 1}}}}}{{{{z}^{{n - 1}}}{{{(n - 1)}}^{2}}}}\ln(1 - z) \\ + \,\,\frac{{{{\ln}^{2}}z}}{{2(n - 1)}} - \frac{1}{{{{{(n - 1)}}^{2}}}}\sum\limits_{k = 1}^{n - 2} {\frac{{(1 - {{z}^{k}})(n + k - 1)}}{{{{z}^{k}}{{k}^{2}}}}} , \\ n = 2,3,4, \ldots \\ \end{gathered} $$

$$\begin{gathered} \int\limits_z^1 {{\text{d}}x} \frac{{{{\ln}^{2}}(1 - x)}}{{{{x}^{2}}}} = \frac{{1 - z}}{z}{{\ln}^{2}}(1 - z) + 2{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right) \\ + \,\,2\ln(1 - z)\ln{z}, \\ \int\limits_z^1 {{\text{d}}x} \frac{{{{\ln}^{2}}(1 - x)}}{{{{x}^{n}}}} = \frac{{1 - {{z}^{{n - 1}}}}}{{(n - 1){{z}^{{n - 1}}}}}{{\ln}^{2}}(1 - z) + \frac{2}{{n - 1}} \\ \times \,\,({\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right) + \ln(1 - z)\ln{z}) \\ - \,\,\frac{2}{{n - 1}}\sum\limits_{k = 2}^{n - 1} {\int\limits_z^1 {{\text{d}}x} } \frac{{\ln(1 - x)}}{{{{x}^{k}}}},\,\,\,\,n = 3,4,5, \ldots \\ \end{gathered} $$

$$\begin{gathered} \int\limits_z^1 {{\text{d}}x} {{\ln}^{2}}(1 - x)\ln{x} \\ = 2{\text{L}}{{{\text{i}}}_{3}}\left( {1 - z} \right) - 2{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)\ln(1 - z) \\ - \,\,z{{\ln}^{2}}(1 - z)\ln{z} \\ + 2{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right) + 2z\ln(1 - z)\ln{z} \\ - \,\,(1 - z){{\ln}^{2}}(1 - z) + 4(1 - z) \\ \times \,\,\ln(1 - z) - 2z\ln{z} + 6z - 6, \\ \end{gathered} $$

$$\begin{gathered} \int\limits_z^1 {{\text{d}}x} \frac{{{{\ln}^{2}}(1 - x)\ln{x}}}{x} = 2{{{\text{S}}}_{{1,2}}}\left( {1 - z} \right)\ln(1 - z) \\ - \,\,2{{{\text{S}}}_{{2,2}}}\left( {1 - z} \right) - \frac{1}{2}{{\ln}^{2}}(1 - z){{\ln}^{2}}z, \\ \int\limits_z^1 {{\text{d}}x} \frac{{{{\ln}^{2}}(1 - x)\ln{x}}}{{{{x}^{2}}}} = - 2{{{\text{S}}}_{{1,2}}}\left( {1 - z} \right) - 2{\text{L}}{{{\text{i}}}_{3}}\left( {1 - z} \right) \\ + \,\,2{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)\ln(1 - z) + \frac{1}{z}{{\ln}^{2}}(1 - z)\ln{z} \\ + \,\,\ln(1 - z){{\ln}^{2}}z + 2{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right) + 2\ln(1 - z)\ln{z} \\ + \,\,\frac{{1 - z}}{z}{{\ln}^{2}}(1 - z), \\ \end{gathered} $$

$$\begin{gathered} \int\limits_z^1 {{\text{d}}x} \frac{{{{\ln}^{2}}(1 - x)\ln{x}}}{{1 - x}} = - 2{\text{L}}{{{\text{i}}}_{4}}\left( {1 - z} \right) + 2{\text{L}}{{{\text{i}}}_{3}}\left( {1 - z} \right) \\ \times \,\,\ln(1 - z) - {\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right){{\ln}^{2}}(1 - z), \\ \int\limits_z^1 {{\text{d}}x} \ln(1 - x){{\ln}^{2}}x = 2{{{\text{S}}}_{{1,2}}}\left( {1 - z} \right) - z\ln(1 - z) \\ \times \,\,{{\ln}^{2}}z + 2{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right) + 2z\ln(1 - z)\ln{z} \\ + \,\,z{{\ln}^{2}}z + 2(1 - z)\ln(1 - z) - 4z\ln{z} + 6z - 6, \\ \end{gathered} $$

$$\begin{gathered} \int\limits_z^1 {{\text{d}}x} \frac{{\ln(1 - x){{\ln}^{2}}x}}{x} = - 2{{{\text{S}}}_{{1,3}}}\left( {1 - z} \right) - \frac{1}{3}\ln(1 - z){{\ln}^{3}}z, \\ \int\limits_z^1 {{\text{d}}x} \frac{{\ln(1 - x){{\ln}^{2}}x}}{{{{x}^{2}}}} = - 2{{{\text{S}}}_{{1,2}}}\left( {1 - z} \right) + \frac{1}{z}\ln(1 - z) \\ \times \,\,{{\ln}^{2}}z + \frac{1}{3}{{\ln}^{3}}z + 2{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right) + \frac{2}{z}\ln(1 - z)\ln{z} \\ + \,\,{{\ln}^{2}}z + 2\frac{{1 - z}}{z}\ln(1 - z) + 2\ln{z}, \\ \int\limits_z^1 {{\text{d}}x} \frac{{\ln(1 - x){{\ln}^{2}}x}}{{1 - x}} \\ = - 2{{{\text{S}}}_{{2,2}}}\left( {1 - z} \right) + 2{{{\text{S}}}_{{1,2}}}\left( {1 - z} \right)\ln(1 - z), \\ \int\limits_z^1 {{\text{d}}x} {{\ln}^{3}}(1 - x) = (1 - z)({{\ln}^{3}}(1 - z) - 3{{\ln}^{2}}(1 - z) \\ + \,\,6ln(1 - z) - 6), \\ \end{gathered} $$

$$\begin{gathered} \int\limits_z^1 {{\text{d}}x} \frac{{{{\ln}^{3}}(1 - x)}}{{{{x}^{2}}}} = - 6{\text{L}}{{{\text{i}}}_{3}}\left( {1 - z} \right) + 6{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)\ln(1 - z) \\ + \,\,3{{\ln}^{2}}(1 - z)\ln{z} + \frac{{1 - z}}{z}{{\ln}^{3}}(1 - z), \\ \end{gathered} $$

$$\begin{gathered} \int\limits_z^1 {{\text{d}}x} {{\ln}^{3}}x = - z{{\ln}^{3}}z + 3z{{\ln}^{2}}z - 6z\ln{z} + 6z - 6, \\ \int\limits_z^1 {{\text{d}}x} \frac{{{{\ln}^{3}}x}}{{{{x}^{2}}}} = \frac{1}{z}({{\ln}^{3}}z + 3{{\ln}^{2}}z + 6\ln{z} - 6z + 6). \\ \end{gathered} $$

By means of identical transformations (see Appendix B), we can simplify arguments of polylogarithmic functions to the form $(1 - x)$. In the integrands, we perform the same simplification. We note that there are certain arguments in favor of selection of the form $(1 - x)$ for the argument, which seems, at first sight, less convenient than simply $x$. In most applications, the point $x = 1$ corresponds to the singularity (regularized somehow) of structure functions or fragmentation functions. We recall, e.g., the nonsinglet splitting function in the lower order

$${{P}^{{(0)}}}(x) = \mathop {\left[ {\frac{{1 + {{x}^{2}}}}{{1 - x}}} \right]}\nolimits_ + .$$

((A.8))

Thus, in analytical results obtained by convolution the limit $x \to 1$ has a fundamental value. Our chosen method of reducing the arguments of polylogarithmic functions simplifies the analysis of the behavior of the result in this limit. In any case, the transformation of functions depending on $(1 - x)$ to those having $x$ as an argument can be done using the formulas of the same set, presented in Appendix B.

The main integrals of polylogarithmic functions include the following:

$$\begin{gathered} \int\limits_z^1 {{\text{d}}x} {{x}^{n}}{\text{L}}{{{\text{i}}}_{2}}\left( {1 - x} \right) = \frac{{1 - {{z}^{{n + 1}}}}}{{n + 1}}{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right) - \frac{1}{{n + 1}} \\ \times \,\,\sum\limits_{k = 1}^{n + 1} {\left( {\frac{{{{z}^{k}}}}{k}\ln{z} + \frac{{1 - {{z}^{k}}}}{{{{k}^{2}}}}} \right)} ,\,\,\,\,n = 0,1,2, \ldots \\ \int\limits_z^1 {{\text{d}}x} \frac{{{\text{L}}{{{\text{i}}}_{2}}\left( {1 - x} \right)}}{{1 - x}} = {\text{L}}{{{\text{i}}}_{3}}\left( {1 - z} \right), \\ \int\limits_z^1 {{\text{d}}x} \frac{{{\text{L}}{{{\text{i}}}_{2}}\left( {1 - x} \right)}}{x} = - 2{{{\text{S}}}_{{1,2}}}\left( {1 - z} \right) - {\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)\ln{z}, \\ \end{gathered} $$

$$\begin{gathered} \int\limits_z^1 {{\text{d}}x} \frac{{{\text{L}}{{{\text{i}}}_{2}}\left( {1 - x} \right)}}{{{{x}^{n}}}} = \frac{{1 - {{z}^{{n - 1}}}}}{{{{z}^{{n - 1}}}(n - 1)}}{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right) + \frac{1}{{n - 1}} \\ \times \,\,\left[ {\sum\limits_{k = 1}^{n - 2} {\left( {\frac{{1 - {{z}^{k}}}}{{{{z}^{k}}{{k}^{2}}}} + \frac{{\ln{z}}}{{{{z}^{k}}k}}} \right)} - \frac{1}{2}{{\ln}^{2}}z} \right],\,\,\,\,n = 2,3,4, \ldots \\ \int\limits_z^1 {{\text{d}}x} {\text{L}}{{{\text{i}}}_{3}}\left( {1 - x} \right) = (1 - z){\text{L}}{{{\text{i}}}_{3}}\left( {1 - z} \right) \\ - \,\,(1 - z){\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right) + z\ln{z} + 1 - z, \\ \end{gathered} $$

$$\begin{gathered} \int\limits_z^1 {{\text{d}}x} \frac{{{\text{L}}{{{\text{i}}}_{3}}\left( {1 - x} \right)}}{x} = - {\text{L}}{{{\text{i}}}_{3}}\left( {1 - z} \right)\ln{z} - \frac{1}{2}{{\left( {{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)} \right)}^{2}}, \\ \int\limits_z^1 {{\text{d}}x} \frac{{{\text{L}}{{{\text{i}}}_{3}}\left( {1 - x} \right)}}{{{{x}^{2}}}} = \frac{{1 - z}}{z}{\text{L}}{{{\text{i}}}_{3}}\left( {1 - z} \right) \\ + \,\,2{{{\text{S}}}_{{1,2}}}\left( {1 - z} \right) + {\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)\ln{z}, \\ \int\limits_z^1 {{\text{d}}x} \frac{{{\text{L}}{{{\text{i}}}_{3}}\left( {1 - x} \right)}}{{1 - x}} = {\text{L}}{{{\text{i}}}_{4}}\left( {1 - z} \right), \\ \end{gathered} $$

$$\begin{gathered} \int\limits_z^1 {{\text{d}}x} {{{\text{S}}}_{{1,2}}}\left( {1 - x} \right) = (1 - z){{{\text{S}}}_{{1,2}}}\left( {1 - z} \right) \\ + \,\,\frac{z}{2}{{\ln}^{2}}z - z\ln{z} + z - 1, \\ \int\limits_z^1 {{\text{d}}x} \frac{{{{{\text{S}}}_{{1,2}}}\left( {1 - x} \right)}}{x} = - 3{{{\text{S}}}_{{1,3}}}\left( {1 - z} \right) - {{{\text{S}}}_{{1,2}}}\left( {1 - z} \right)\ln{z}, \\ \int\limits_z^1 {{\text{d}}x} \frac{{{{{\text{S}}}_{{1,2}}}\left( {1 - x} \right)}}{{{{x}^{2}}}} = \frac{{1 - z}}{z}{{{\text{S}}}_{{1,2}}}\left( {1 - z} \right) + \frac{1}{6}{{\ln}^{3}}z, \\ \end{gathered} $$

((A.9))

$$\begin{gathered} \int\limits_z^1 {{\text{d}}x} \frac{{{{{\text{S}}}_{{1,2}}}\left( {1 - x} \right)}}{{1 - x}} = {{{\text{S}}}_{{2,2}}}\left( {1 - z} \right), \\ \int\limits_z^1 {{\text{d}}x} {\text{L}}{{{\text{i}}}_{2}}\left( {1 - x} \right)ln(1 - x) \\ = (1 - z){\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)\ln(1 - z) + (z - 2){\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right) \\ - \,\,z\ln(1 - z)\ln{z} - (1 - z)\ln(1 - z) + 2z\ln{z} + 3 - 3z, \\ \end{gathered} $$

$$\begin{gathered} \int\limits_z^1 {{\text{d}}x} \frac{{{\text{L}}{{{\text{i}}}_{2}}\left( {1 - x} \right)\ln(1 - x)}}{x} \\ = 2{{{\text{S}}}_{{2,2}}}\left( {1 - z} \right) - \frac{1}{2}{{\left( {{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)} \right)}^{2}} \\ - \,\,2{{{\text{S}}}_{{1,2}}}\left( {1 - z} \right)\ln(1 - z) \\ - \,\,{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)\ln(1 - z)\ln{z}, \\ \int\limits_z^1 {{\text{d}}x} \frac{{{\text{L}}{{{\text{i}}}_{2}}\left( {1 - x} \right)\ln(1 - x)}}{{{{x}^{2}}}} \\ = 3{{{\text{S}}}_{{1,2}}}\left( {1 - z} \right) + {\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)\ln{z} \\ + \,\,\frac{{1 - z}}{z}{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)\ln(1 - z) - \frac{1}{2}\ln(1 - z){{\ln}^{2}}z, \\ \end{gathered} $$

$$\begin{gathered} \int\limits_z^1 {{\text{d}}x} \frac{{{\text{L}}{{{\text{i}}}_{2}}\left( {1 - x} \right)\ln(1 - x)}}{{1 - x}} \\ = - {\text{L}}{{{\text{i}}}_{4}}\left( {1 - z} \right) + {\text{L}}{{{\text{i}}}_{3}}\left( {1 - z} \right)\ln(1 - z), \\ \int\limits_z^1 {{\text{d}}x} {\text{L}}{{{\text{i}}}_{2}}\left( {1 - x} \right)\ln{x} = - 2{{{\text{S}}}_{{1,2}}}\left( {1 - z} \right) \\ - \,\,(z\ln{z} - z + 1){\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right) - z{{\ln}^{2}}z \\ + \,\,3z\ln{z} - 3z + 3, \\ \end{gathered} $$

$$\begin{gathered} \int\limits_z^1 {{\text{d}}x} \frac{{{\text{L}}{{{\text{i}}}_{2}}\left( {1 - x} \right)\ln{x}}}{x} = 3{{{\text{S}}}_{{1,3}}}\left( {1 - z} \right) - \frac{1}{2}{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right){{\ln}^{2}}z, \\ \int\limits_z^1 {{\text{d}}x} \frac{{{\text{L}}{{{\text{i}}}_{2}}\left( {1 - x} \right)\ln{x}}}{{{{x}^{2}}}} = 2{{{\text{S}}}_{{1,2}}}\left( {1 - z} \right) \\ + \,\,\frac{1}{z}(\ln{z} - z + 1){\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right) - \frac{1}{3}{{\ln}^{3}}z - \frac{1}{2}{{\ln}^{2}}z, \\ \int\limits_z^1 {{\text{d}}x} \frac{{{\text{L}}{{{\text{i}}}_{2}}\left( {1 - x} \right)\ln{x}}}{{1 - x}} = - \frac{1}{2}{{\left( {{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)} \right)}^{2}}. \\ \end{gathered} $$

Sometimes, the integrals of functions having $(1 + x)$ in the argument arise (we consider only real parts of these functions):

$$\begin{gathered} \int\limits_z^1 {{\text{d}}x} \frac{1}{{1 + x}} = \ln{2} - \ln(1 + z), \\ \int\limits_z^1 {{\text{d}}x} \frac{{\ln{x}}}{{1 + x}} = {\text{L}}{{{\text{i}}}_{2}}\left( {1 + z} \right) - \frac{3}{2}\zeta (2), \\ \int\limits_z^1 {{\text{d}}x} \frac{{{{\ln}^{2}}x}}{{1 + x}} = \frac{7}{2}\zeta (3) - 2{{{\text{S}}}_{{1,2}}}\left( {1 + z} \right), \\ \int\limits_z^1 {{\text{d}}x} {\text{L}}{{{\text{i}}}_{2}}\left( {1 + x} \right) = - (1 + z){\text{L}}{{{\text{i}}}_{2}}\left( {1 + z} \right) \\ - \,\,z\ln{z} - 1 + z + 3\zeta (2), \\ \end{gathered} $$

$$\begin{gathered} \int\limits_z^1 {{\text{d}}x} \frac{{{\text{L}}{{{\text{i}}}_{2}}\left( {1 + x} \right)}}{{1 + x}} \\ = - {\text{L}}{{{\text{i}}}_{3}}\left( {1 + z} \right) + \frac{7}{8}\zeta (3) + \frac{3}{2}\zeta (2)\ln{2}, \\ \int\limits_z^1 {{\text{d}}x} \frac{{{\text{L}}{{{\text{i}}}_{2}}\left( {1 + x} \right)}}{x} = \frac{7}{2}\zeta (3) \\ - \,\,2{{{\text{S}}}_{{1,2}}}\left( {1 + z} \right) - {\text{L}}{{{\text{i}}}_{2}}\left( {1 + z} \right)\ln{z}, \\ \end{gathered} $$

((A.10))

$$\begin{gathered} \int\limits_z^1 {{\text{d}}x} \frac{{{\text{L}}{{{\text{i}}}_{2}}\left( {1 + x} \right)}}{{{{x}^{n}}}} = - \frac{3}{{2(n - 1)}}\zeta (2) \\ + \,\,\frac{1}{{{{z}^{{n - 1}}}(n - 1)}}{\text{L}}{{{\text{i}}}_{2}}\left( {1 + z} \right) - \frac{{{{{( - 1)}}^{{n - 1}}}}}{{n - 1}} \\ \times \,\,({\text{L}}{{{\text{i}}}_{2}}\left( {1 + z} \right) + \frac{1}{2}{{\ln}^{2}}z - \frac{3}{2}\zeta (2)) - \frac{1}{{n - 1}} \\ \times \,\,\sum\limits_{k = 1}^{n - 2} {{{{( - 1)}}^{{n + k}}}} \left( {\frac{{1 - {{z}^{k}}}}{{{{z}^{k}}{{k}^{2}}}} + \frac{{\ln{z}}}{{{{z}^{k}}k}}} \right),\,\,\,\,n = 2,3,4, \ldots \\ \end{gathered} $$

The above table of integrals was programmed in computer code [280] in the FORM language [166, 167]. The code makes it possible to automate operations of integration, reduction of similar terms and simplification of arguments of polylogarithmic functions. The table of integrals was applied in calculating radiative corrections to some processes of particle interactions, see, e.g., [45, 93, 110, 254, 281]. The presented set of integrals allows the convolution problems to be solved for a rather wide class of problems arising in calculation of the higher-order radiative corrections of perturbation theory in both QED and QCD. The larger part of the above integrals can be found in other literature sources, including automatic codes such as MATHEMATICA [282]. However, the consolidation of a necessary set of mathematical tools aimed at solving problems of a certain class seems to be completely appropriate and useful. Additionally, in the above expressions, the identical transformations were systematically used for the functions whose asymptotics as $z \to 1$ can be easily seen.

Main Properties of Polylogarithms

In the derivation of analytical expressions for contributions of the logarithmic corrections under study, special mathematical functions arise systematically; in particular, polylogarithms and the Riemann zeta function. Following the notation of [279], see also [283, 284], we write the Nielsen’s polylogarithm in general terms as

$$\begin{gathered} {{{\text{S}}}_{{n,m}}}\left( z \right) = \frac{{{{{( - 1)}}^{{n + m - 1}}}}}{{(n - 1)!m!}}\int\limits_0^1 {{\text{d}}x} \frac{{l{{n}^{{n - 1}}}(x)l{{n}^{m}}(1 - xz)}}{x}, \\ n = 1,2,3 \ldots ,\,\,\,\,m = 1,2,3 \ldots \\ \end{gathered} $$

((B.1))

In particular, we define a dilogarithm, considered by Euler in 1768 and often called Spence’s function, as follows:

$${\text{L}}{{{\text{i}}}_{2}}\left( z \right) \equiv {{{\text{S}}}_{{1,1}}}\left( z \right) = - \int\limits_0^1 {{\text{d}}x} \frac{{\ln(1 - xz)}}{x}.$$

((B.2))

A typical integral, leading to the occurrence of Spence’s function, is

$$\begin{gathered} \int {{\text{d}}x} \frac{{\ln(a + bx)}}{{c + ex}} = \frac{1}{e}\left[ {\ln\left( {\frac{{ae - bc}}{e}} \right)} \right.\ln(c + ex) \\ \left. { - \,\,{\text{L}}{{{\text{i}}}_{2}}\left( { - b\frac{{c + ex}}{{ae - bc}}} \right)} \right]. \\ \end{gathered} $$

((B.3))

Additionally, polylogarithms of higher orders also were encountered in this work:

$$\begin{gathered} {{{\text{S}}}_{{1,2}}}\left( z \right) = \frac{1}{2}\int\limits_0^1 {{\text{d}}x} \frac{{{{\ln}^{2}}(1 - xz)}}{x}, \\ {\text{L}}{{{\text{i}}}_{3}}\left( z \right) \equiv {{{\text{S}}}_{{2,1}}}\left( z \right) = \int\limits_0^1 {{\text{d}}x} \frac{{\ln(x)\ln(1 - xz)}}{x} \\ = \int\limits_0^z {{\text{d}}x} \frac{{{\text{L}}{{{\text{i}}}_{2}}\left( x \right)}}{x},\,\,\,\,{\text{L}}{{{\text{i}}}_{4}}\left( z \right) \equiv {{{\text{S}}}_{{3,1}}}\left( z \right) \\ = - \frac{1}{2}\int\limits_0^1 {{\text{d}}x} \frac{{{{\ln}^{2}}(x)\ln(1 - xz)}}{x}, \\ {{{\text{S}}}_{{1,3}}}\left( z \right) = - \frac{1}{6}\int\limits_0^1 {{\text{d}}x} \frac{{{{\ln}^{3}}(1 - xz)}}{x}, \\ {{{\text{S}}}_{{2,2}}}\left( z \right) = - \frac{1}{2}\int\limits_0^1 {{\text{d}}x} \frac{{\ln(x){{\ln}^{2}}(1 - xz)}}{x}. \\ \end{gathered} $$

((B.4))

It is useful to give several particular values of the Riemann $\zeta$-function:

$$\begin{gathered} \zeta (n) = \sum\limits_{k = 1}^\infty {\frac{1}{{{{k}^{n}}}}} ,\,\,\,\,\zeta (2) = \frac{{{{\pi }^{2}}}}{6}, \\ \zeta (3) \approx 1.20205690315959, \\ \zeta (4) = \frac{{{{\pi }^{4}}}}{{90}},\,\,\,\,\zeta (5) \approx 1.03692775514337. \\ \end{gathered} $$

((B.5))

To transform polylogarithms from the argument $z$ to functions of argument $(1 - z),$ the following relations were used:

$$\begin{gathered} {\text{L}}{{{\text{i}}}_{2}}\left( z \right) = - {\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right) - l\ln{z}\ln(1 - z) + \zeta (2), \\ {{{\text{S}}}_{{1,2}}}\left( z \right) = - {\text{L}}{{{\text{i}}}_{3}}\left( {1 - z} \right) + {\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)\ln(1 - z) \\ + \,\,\frac{1}{2}{{\ln}^{2}}(1 - z)\ln{z} + \zeta (3),\,\,\,{\text{L}}{{{\text{i}}}_{3}}\left( z \right) = - {{{\text{S}}}_{{1,2}}}\left( {1 - z} \right) \\ {\text{--}}\,{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)\ln{z} - \,\,\frac{1}{2}\ln(1 - z){{\ln}^{2}}z + \zeta (2)\ln{z} + \zeta (3), \\ \end{gathered} $$

$$\begin{gathered} {{{\text{S}}}_{{1,3}}}\left( z \right) = - {\text{L}}{{{\text{i}}}_{4}}\left( {1 - z} \right) + {\text{L}}{{{\text{i}}}_{3}}\left( {1 - z} \right)\ln(1 - z) \\ - \,\,\frac{1}{2}{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right){{\ln}^{2}}(1 - z) \\ - \frac{1}{6}{{\ln}^{3}}(1 - z)\ln{z} + \zeta (4), \\ \end{gathered} $$

((B.6))

$$\begin{gathered} {{{\text{S}}}_{{2,2}}}\left( z \right) = - {{{\text{S}}}_{{2,2}}}\left( {1 - z} \right) - {\text{L}}{{{\text{i}}}_{3}}\left( {1 - z} \right)\ln{z} \\ + \,\,{{{\text{S}}}_{{1,2}}}\left( {1 - z} \right)\ln(1 - z) + {\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right)\ln(1 - z)\ln{z} \\ + \,\,\frac{1}{4}{{\ln}^{2}}(1 - z){{\ln}^{2}}z + \zeta (3)\ln{z} + \frac{1}{4}\zeta (4), \\ {\text{L}}{{{\text{i}}}_{4}}\left( z \right) = - {{{\text{S}}}_{{1,3}}}\left( {1 - z} \right) - {{{\text{S}}}_{{1,2}}}\left( {1 - z} \right)\ln{z} \\ - \,\,\frac{1}{2}{\text{L}}{{{\text{i}}}_{2}}\left( {1 - z} \right){{\ln}^{2}}z - \frac{1}{6}\ln(1 - z){{\ln}^{3}}z \\ + \,\,\frac{1}{2}\zeta (2){{\ln}^{2}}z + \zeta (3)\ln{z} + \zeta (4). \\ \end{gathered} $$

The change from more complicated arguments, arising in practice, to the simplest ones $(z$ and $1 - z)$ was also performed by methods of identical transformations, sometimes using certain additional relations between polylogarithms of more complicated arguments, e.g.,

$$\begin{gathered} {\text{L}}{{{\text{i}}}_{2}}\left( { - \frac{z}{{1 - z}}} \right) = - {\text{L}}{{{\text{i}}}_{2}}\left( z \right) - \frac{1}{2}{{\ln}^{2}}(1 - z), \\ {\text{L}}{{{\text{i}}}_{2}}({{z}^{2}}) = 2\left[ {{\text{L}}{{{\text{i}}}_{2}}\left( z \right) + {\text{L}}{{{\text{i}}}_{2}}\left( { - z} \right)} \right] \\ \end{gathered} $$

((B.7))

etc. We note that in some of the presented relations, imaginary parts are discarded, which, if necessary, can be restored using the analytic properties of relevant functions [284].

Some explicit expressions for polylogarithms of constant arguments are useful:

$$\begin{gathered} {{{\text{S}}}_{{n,m}}}\left( 0 \right) = 0,\,\,\,\,n = 1,2,3 \ldots ,\,\,\,\,m = 1,2,3 \ldots \\ {\text{L}}{{{\text{i}}}_{n}}\left( 1 \right) = \zeta (n),\,\,\,\,n = 2,3,4, \ldots \\ {{{\text{S}}}_{{2,2}}}\left( 1 \right) = \frac{1}{4}\zeta (4),\,\,\,\,{{{\text{S}}}_{{1,n}}}\left( 1 \right) = \zeta (n + 1),\,\,\,\,n = 1,2,3 \ldots \\ {\text{ReL}}{{{\text{i}}}_{2}}\left( 2 \right) = \frac{3}{2}\zeta (2),\,\,\,\,{\text{L}}{{{\text{i}}}_{2}}\left( { - 1} \right) = - \frac{1}{2}\zeta (2), \\ {\text{L}}{{{\text{i}}}_{2}}\left( {\frac{1}{2}} \right) = \frac{1}{2}\zeta (2) - \frac{1}{2}{{\ln}^{2}}2, \\ {\text{ReL}}{{{\text{i}}}_{3}}\left( 2 \right) = \frac{7}{8}\zeta (3) + \frac{3}{2}\ln{2}\zeta (2), \\ {\text{ReL}}{{{\text{i}}}_{3}}\left( { - 1} \right) = - \frac{3}{4}\zeta (3),\,\,\,\,{\text{Re}}{{{\text{S}}}_{{1,2}}}\left( 2 \right) = \frac{7}{4}\zeta (3), \\ {\text{L}}{{{\text{i}}}_{3}}\left( {\frac{1}{2}} \right) = \frac{7}{8}\zeta (3) - \frac{1}{2}\zeta (2)\ln{2} + \frac{1}{6}{{\ln}^{3}}2, \\ {{{\text{S}}}_{{1,2}}}\left( { - 1} \right) = \frac{1}{8}\zeta (3),\,\,\,\,{{{\text{S}}}_{{1,2}}}\left( {\frac{1}{2}} \right) = \frac{1}{8}\zeta (3) - \frac{1}{6}{{\ln}^{3}}2. \\ \end{gathered} $$

((B.8))

To expand the solutions to evolution equations in a series obtained in the approximation of soft bremsstrahlung, it is convenient to use the following presentation for the $\Gamma$-function:

$$\frac{1}{{\Gamma (z)}} = z{{{\text{e}}}^{{Cz}}}\prod\limits_{n = 1}^\infty {\left( {1 + \frac{z}{n}} \right){{{\text{e}}}^{{{{ - z} \mathord{\left/ {\vphantom {{ - z} n}} \right. \kern-0em} n}}}}} .$$

((B.9))