###### Abstract

The non-planar Feynman diagram with seven massless, scalar propagators and four on-shell legs (the crossed double box) is calculated analytically in dimensional regularization. The non-planar diagram with six propagators is also discussed.

Freiburg-THEP 99/10

hep-ph/9909506

September 1999

Non-planar massless two-loop Feynman diagrams with four on-shell legs

[3ex]
J.B. Tausk^{1}^{1}1

[3ex]

Fakultät für Physik, Albert-Ludwigs-Universität Freiburg,

Hermann-Herder-Straße 3, D-79104 Freiburg, Germany

1. Introduction

Knowledge of two-loop massless Feynman diagrams with four
massless external legs is one of the key ingredients required for
a next-to-next-to-leading order calculation of 2 jet production
rates in hadron-hadron collisions [1], as well
as for other processes, including Bhabha scattering in the high energy
limit. Among those diagrams, the most complicated are the planar and
non-planar double box diagrams, both of which contain seven propagators.
Recently, significant progress in this field has been made by Smirnov, who
has derived an analytical formula in terms of polylogarithms
for the planar double box [2] (see also
ref. [3]). More simple
planar diagrams, with less than seven propagators, are discussed in
refs. [4, 5]. Results for non-planar diagrams
have until now been missing.

The ultraviolet divergences of the planar and non-planar double box diagrams in spacetime dimensions , and were calculated in ref. [6]. In , the diagrams are ultraviolet finite, but they have a complicated structure of collinear and infrared divergences. In dimensional regularization, with , the divergences appear as poles , with . As has been demonstrated by Smirnov in the planar case, the method of Mellin-Barnes contour integrals is a very natural and convenient technique to isolate these singular contributions. The main purpose of this paper is to calculate the non-planar seven propagator diagram,

(1) |

using a similar Mellin-Barnes approach. The external momenta are lightlike () and we use Mandelstam variables defined by , and .

Starting from a Mellin-Barnes representation, which we derive in section 2, we obtain analytical formulas in terms of logarithms and polylogarithms for the seven propagator diagram , which we present in section 5, by a series of steps that are explained in sections 3 and 4. Then, in section 6, we briefly discuss the non-planar diagram with six propagators. Some final comments are made in section 7.

2. A Mellin-Barnes representation

Our main tool will be the Mellin-Barnes representation of a power of
a sum as a contour integral,

(2) |

where the integration contour separates the poles of from those of , and are complex numbers such that . The Mellin-Barnes representation of a sum of several terms, , is obtained easily by iteration of the basic formula (2). Another formula that we will need is Barnes’s first lemma (see, e.g. [7]):

(3) |

Again, the contour should separate the increasing series of poles (of ) from the decreasing ones (of ).

Introducing Feynman parameters in the standard way, the non-planar double box diagram can be written as [6]

(4) | |||||

(5) | |||||

(6) | |||||

According to the causal -prescription, a small imaginary part should be added to . We do this by giving , and positive imaginary parts. At this stage, we consider , and to be three independent variables. At the end of the calculation, after we have taken the limit where , and become real, one variable can be eliminated by using the relationship .

The next step is to represent the factor as a product of five more simple factors, at the expense of introducing four Mellin-Barnes parameters:

(7) | |||||

Now the Feynman parameter integrals can be done in terms of -functions, and we obtain the following Mellin-Barnes representation for :

(8) |

(9) | |||||

where

(10) |

In order to keep things as simple as possible, we choose the contour for each of the variables , , and to be a straight line parallel to the imaginary axis, with a real part that does not depend on any of the other variables. Eq. (9) is certainly valid if the real parts of the contours are chosen in such a way that the real parts of the arguments of are all positive. It is possible to satisfy these conditions simultaneously when . For example, if , choosing , , satifies all the conditions. Obviously, there are infinitely many other, equally suitable possibilities, but for the sake of definiteness, we will stick to these contours.

3. Analytic continuation in

Having derived the Mellin-Barnes representation (9)
near , we must now perform an analytic continuation in
to the region around . To do so, we follow the
poles of the -functions as we increase . Whenever
a pole crosses one of the integration contours, which we keep fixed,
we add its residue in the corresponding variable to the right hand
side of eq. (9). This residue term still contains poles
in its remaining variables which can cross other integration contours
as is increased further. They are treated in the same way.
We end up with a large collection of residue terms, plus the original
fourfold integral.
Some of the residue terms contain factors that diverge at .
However, these singular factors do not depend on , ,
or . The remaining integrals can simply be expanded
in under the integral sign because our contours do not go
through any poles at .
(There are never problems with convergence at the limits
).

Below, we will give some examples of the residue terms that appear. But first, let us see what happens when we substitute in the integral (9). The variables and are coupled via , and , but these -functions collapse into

(11) |

Consider the first term. The factor can be absorbed into , and then both the and the integrals can be performed using Barnes’s first lemma. Moreover, the integral gives zero:

(12) |

(Here, we are taking a contour that separates the increasing and decreasing series of poles as required by Barnes’s lemma, not a straight line). Analogously, the second term on the right hand side of eq. (11) vanishes when integrated over .

This cancellation is very nice because it means that if we only need the non-planar box up to the constant term in , once we have done the and integrals, we do not need to calculate any double integrals of the form

(13) |

Only terms in which a residue has been taken in either or , or in both, are ever needed. Such terms, whose dependence on at least one of the variables , and is trivial, are much more simple than the double integrals (13).

We now return to the task of collecting the various residue contributions. The first poles to cross contours are at and , coming from and , respectively. (The poles of and are harmless because they depend on with the opposite sign, and therefore move away from the contours when increases). To account for the residues of these two poles, we replace , where

(14) | |||||

is a similar term, where the residue in is taken, and

(15) | |||||

Let us concentrate on . Note that in eliminating , we have changed
the -dependence of the remaining -functions.
As a consequence, the first poles of
,
and
cross the , and contours, respectively. These crossings
are taken into account by replacing
.
In a number of these terms, there are still more poles that cross
contours before finally reaches . They are dealt with
by making the following further replacements:
,
,
,
,
,
and
.
In each term, the subscripts indicate the -functions
of which the residue of the first pole has been taken. An exception
to this is , where the prime means that we have taken
the residue of the second pole of (i.e., where the
argument equals -1) instead of the first pole. The last term,
, gets a relative minus sign because the pole in
crosses the contour backwards.^{2}^{2}2
Eventually, cancels against a descendant of .

Due to the symmetry of the problem in and , there is a one to one correspondence between the descendants of and those of , and their contributions to the final result are the same. The analysis of runs along similar lines. Here, some poles are second order, with residues involving derivatives of -functions.

4. Evaluating the ’s

In all, we find 39 contributions. (This number depends on the
contours one uses. Had we chosen different ones,
there would have been other poles crossing contours and we would have
obtained a different, but equivalent, collection of residue terms).
The simplest contributions are those where a residue in four variables has
been taken, e.g.

(16) |

Note that in this particular case, the expansion starts off with a pole.

In several terms, there is one integral left. For instance, , which we get by taking residues at , and , is given by

(17) |

To calculate the integral, we expand the integrand up to the second order in . Then, closing the contour to the left or right and summing the residues of the poles inside it, we obtain harmonic series which can be expressed in terms of , and .

An example in which the final integral still involves powers of Mandelstam variables is

(18) |

where we have already set . This integral can also be calculated by closing the contour, and the result can be expressed in terms of polylogarithms [8] and Nielsen’s generalized polylogarithms [9, 10], which are defined by

(19) |

The cases , and which occur in this paper can all be written as combinations of , and .

In the term , we perform the first integration by Barnes’s lemma (in a slightly modified form, see [2]):

(20) | |||||

The integral can be then be done by closing the contour, as in the previous example. Three other ’s are related to by and/or .

Finally, there are four terms^{3}^{3}3We are leaving out
and , which contain a factor of in their
denominator and are therefore already of order . in which
both the and the integrals still survive, namely ,
, and . However, by the mechanism explained above
(see eq. (S0.Ex7)), these terms cancel against each other at
after integration over and .

5. Result for the seven propagator diagram

Adding together all the contributions discussed in the previous section,
we obtain an expression for , up to the constant term in
, in terms of (poly)logarithms of (ratios of) , and ,
which are, at this point, still independent variables. In particular,
the result is also valid when , and are all negative. This
region is, of course, unphysical, but it is nevertheless interesting to
consider because it provides us with an opportunity to check our result
against a calculation by Binoth and Heinrich [11]. They
have obtained results for the negative , , region by extracting
the divergences from the Feynman parametric integral (4)
and then performing a multi-dimensional numerical integration.

At the symmetric point, , we find

(21) | |||||

The coefficients in (21) agree with the results obtained by the numerical method [11]. We also find good agreement at the asymmetric point . In both cases the precision of the numerical method ranges from about 1 per mille for the -term to 1 per cent for the constant term.

Since the complete formula for general , and is very long and not relevant for physical applications anyway, we do not present it here. Instead, we specialize to the physically relevant case where . We also use transformation formulas for the arguments of the polylogarithms in order to avoid ending up on a cut when we make , , and real. Depending on which legs correspond to incoming and outgoing particles, there are three different kinematical regions to consider: (i), where ; (ii), where ; and (iii), where . In all regions, we have

(22) |

We will use the abbreviations , , .

In region (i), with , and , is given by

(23) | |||||

and can be obtained by interchanging in the expression for .

In region (ii), with , and , and are given by

(24) | |||||

(25) | |||||

The formulas for and in region (iii) can be obtained easily from the ones for region (ii) by using the fact the is symmetric under . This follows from the symmetry between the legs of the non-planar double box carrying momenta and .

6. Six propagator diagram

The non-planar diagram with six propagators,

(26) |

is far less difficult to calculate than . Perhaps the easiest way is by considering a scale transformation,

(27) |

Differentiating both sides of eq. (27) with respect to at gives an identity which can be used to reduce to a sum of three planar diagrams:

(28) |

In the diagrams on the right hand side of eq. (28),
the propagator on the diagonal is squared. Such boxes with a diagonal
propagator are calculated in ref. [4]. They are,
in fact, rather similar to one-loop massless box diagrams [12].
The pole at is a reflection of the linear infrared
divergence^{4}^{4}4It would diverge as if we used a
mass to regularize it. in (26) coming from the
region where the loop momenta and are both soft. Another curious
feature of eq. (28) is that although the three terms on the
right hand side contain third order poles at , the leading
singularity of the sum is only .

The six propagator diagram is completely symmetric under permutations of its external momenta. Therefore, it is sufficient to consider only region (i), where and . We obtain

(29) | |||||