\documentstyle [a4,11pt,axodraw]{article}
 
\begin{document}
\begin{titlepage}
\begin{center}
{\huge The FORM version of MINCER} \\ [8mm]
\end{center}
 
\begin{abstract}\noindent
The program MINCER~\cite{mince} for massless three-loop Feynman diagrams
of the propagator type has been reprogrammed in the language of
FORM~\cite{form}. The new version is thoroughly optimized for extremely 
large computations. It has been
used for some calculations that were previously not practical.
\end{abstract}
\end{titlepage}

\section{The basic algorithms}

Let us first sketch the basic algorithms that are to be used for the 
computation of massless diagrams of the propagator type. Although most 
of these formulae have been presented before in the literature we like 
to repeat them here for several reasons: it will allow us to establish a 
uniform notation and the literature contains many misprints. It will 
also make it easier to understand the later sections.

\subsection{One loop integrals}

The basic one loop diagram is given by:
\begin{center} \begin{picture}(250,90)(0,0)
\ArrowLine(100,45)(70,45)    \Text(123,77)[lb]{P} 
\ArrowLine(180,45)(150,45)   \Text(116,4)[lb]{P+Q}
\ArrowArc(125,45)(25,0,180)  \Text(80,52)[lb]{Q}  
\ArrowArcn(125,45)(25,0,180) \Text(165,52)[lb]{Q} 
\end{picture} \\ {\sl The basic one loop diagram}
\end{center}
\noindent The integral is given by~\cite{fyodor:one}
\begin{eqnarray}
\label{eq:oneloop}
\lefteqn{
   \int \frac{d^D P}{(2\pi)^D} \frac{{\cal P}_n(P)}
   {P^{2\alpha}(P-Q)^{2\beta}}  = 
} \nonumber \\
 &  \displaystyle{
    \frac{1}{(4\pi)^2}(Q^2)^{D/2-\alpha-\beta}\sum_{\sigma \ge 0}^{[n/2]}
    G(\alpha,\beta,n,\sigma)Q^{2\sigma} \left\{ \frac{1}{\sigma!}
    \left( \frac {\Box} {4} \right) ^\sigma {\cal P}_n(P)\right\}_{P=Q},
    }
\end{eqnarray}
in which
\begin{equation}
    {\cal P}_n(P) = P_{\mu_1} P_{\mu_2} \cdots P_{\mu_n}.
\end{equation}
D is the dimension of space-time and is also given by $D = 4 - 
2\epsilon$, $\Box = \partial^2/\partial P_\mu \partial P_\mu$ and $G$ 
can be expressed in terms of $\Gamma$-functions:
%
\begin{equation}
\label{eq:Gamma}
    G(\alpha,\beta,n,\sigma) = 
    (4\pi)^\epsilon\,
    \frac{\Gamma(\alpha+\beta-\sigma-D/2)
    \Gamma(D/2-\alpha+n-\sigma)\Gamma(D/2-\beta+\sigma)}
    {\Gamma(\alpha)\Gamma(\beta)\Gamma(D-\alpha-\beta+n)}.
\end{equation}

\subsection{Two loop integrals}

The two loop integrals come in three topologies. Each of these 
topologies needs its own treatment. The algorithm to deal with the 
more complicated topologies is based on integration by parts. In its 
simplest form it is called the rule of the
triangle~\cite{fyodor:one}~\cite{fyodor:two}.
\begin{center} \begin{picture}(250,90)(0,0)
\Text(85,70)[lb]{$P_1$}
\Text(155,70)[lb]{$P_2$}
\Text(122,70)[lb]{$P$}
\Line(105,60)(145,60)
\Line(145,60)(125,30)
\Line(125,30)(105,60)
\Line(75,60)(105,60)
\Line(145,60)(175,60)
\Line(125,30)(125,10)
\end{picture} \\ {\sl The triangle subgraph} \end{center}
We define the integral
\begin{equation}
    I(n,\alpha_0,\beta_1,\beta_2,\alpha_1,\alpha_2)
    = \int d^D\!P\,
    \frac{P_{\mu_1}\cdots P_{\mu_n}}
    {(P^2)^{\alpha_0}((P+p_1)^2)^{\beta_1}(p_1^2)^{\alpha_1}
    ((P+p_2)^2)^{\beta_2}(p_2^2)^{\alpha_2}},
\end{equation}
This integrals can be studied by first considering the identity:
%
\begin{equation}
\label{bla}
    \int d^D\!P\, 
    \left\{ 
       \frac {\partial} {\partial P_\mu}
       \frac {P_\mu\ \ \ \ \  {\cal P}_n(P)} 
             { (P^2)       ^{\alpha_0} 
               ((P+p_1)^2) ^{\beta_1} 
               (p_1^2)     ^{\alpha_1}
               ((P+p_2)^2) ^{\beta_2}
               (p_2^2)     ^{\alpha_2}
             } 
    \right\} = 0.
\end{equation}
%
Working out the derivative one obtains the recursion relation:
%
\begin{eqnarray}
\label{eq:tri}
\lefteqn{I(n,\alpha_0,\beta_1,\beta_2,\alpha_1,\alpha_2) = \bigl(} \nonumber \\
 && + \beta_1 ( I(n,\alpha_0 - 1,\beta_1 + 1,\beta_2,\alpha_1,\alpha_2)
     - I(n,\alpha_0,\beta_1 + 1,\beta_2,\alpha_1 - 1,\alpha_2) )
    \nonumber \\ &&
    + \beta_2 ( I(n,\alpha_0 - 1,\beta_1,\beta_2 + 1,\alpha_1,\alpha_2)
       - I(n,\alpha_0,\beta_1,\beta_2 + 1,\alpha_1,\alpha_2 - 1) )
    \nonumber \\ &&
    \bigr) / ( D + n - 2 \alpha_0 - \beta_1 - \beta_2 ).
\end{eqnarray}
By using this relation as a recursion relation one can eventually remove 
at least one of the denominators in most of the non-trivial topologies. 
It is however possible to improve upon this by `solving' the recursion. 
This gives the formula~\cite{fyodor:three}
%
\begin{eqnarray}
\label{eq:trisol}
\lefteqn{ I(n,\alpha_0,\beta_1,\beta_2,\alpha_1,\alpha_2) \Gamma(\beta_1)
          \Gamma(\beta_2) = 
        } \nonumber \\
%
\lefteqn{
   +\sum_{i=0}^{\alpha_2-1}
    \sum_{j=0}^{\alpha_0-1} \sum_{k=0}^{\alpha_0-j-1}
    (-1)^{\alpha_1+i}
    I(n,\alpha_0\!-\!j\!-\!k,\beta_1\!+\!\alpha_1\!+\!j,
        \beta_2\!+\!i\!+\!k,0,\alpha_2\!-\!i)
 }   \nonumber \\ 
 &
{\displaystyle{
    \frac{
        \Gamma(\alpha_1+i+j+k) \Gamma(\beta_1+\alpha_1+j)
        \Gamma(D+1+n-2\alpha_0-\beta_1-\beta_2-\alpha_1-i)
     }{
        \Gamma(D+1+n-2\alpha_0-\beta_1-\beta_2+j+k)
        \Gamma(\alpha_1)\ i!\ j!\ k!  
     }
}}
   \nonumber \\
 & \times \Gamma(\beta_2+k+i)
   \nonumber \\ 
\lefteqn{
    +\sum_{i=0}^{\alpha_1-1} \sum_{j=0}^{\alpha_2-1}
     \sum_{k=0}^{\alpha_0}
        (-1)^{i+j}
     I(n,0,\beta_1\!+\!i\!+\!k,\beta_2\!+\!\alpha_0\!+\!j\!-\!k,
        \alpha_1\!-\!i,\alpha_2\!-\!j)
}   \nonumber \\ 
 &
{\displaystyle{
    \frac{\alpha_0 \Gamma(\alpha_0+i+j)
          \Gamma(D+n-2\alpha_0-\beta_1-\beta_2-i-j)
          \Gamma(\beta_1+k+i)
         }{ 
           (\alpha_0-k)!\ i!\ j!\ k!\
           \Gamma(D+n-\alpha_0-\beta_1-\beta_2)
         }
}}
 \nonumber \\
& \times \Gamma(\alpha_0+\beta_2-k+j)
 \nonumber \\ 
\lefteqn{
    +\sum_{i=0}^{\alpha_1-1}
    \sum_{j=0}^{\alpha_0-1} \sum_{k=0}^{\alpha_0-j-1}
    (-1)^{i+\alpha_2}
    I(n,\alpha_0\!-\!j\!-\!k,\beta_1\!+\!i\!+\!k,
\beta_2\!+\!\alpha_2\!+\!j,\alpha_1\!-\!i,0)
} \nonumber \\ 
 &  
{\displaystyle{
    \frac{
         \Gamma(D+1+n-2\alpha_0-\beta_1-\beta_2-\alpha_2-i)
         \Gamma(i+j+k+\alpha_2)
         \Gamma(\beta_1+i+k) 
      }{
         \Gamma(D+1+n-2\alpha_0-\beta_1-\beta_2+k+j)
         \Gamma(\alpha_2) i!\ j!\ k! 
      } 
}}
 \nonumber \\
 & \times \Gamma(\beta_2+\alpha_2+j). 
\end{eqnarray}

The two loop topologies are given by
\begin{center} \begin{picture}(420,110)(0,0)
\ArrowLine(30,60)(0,60)           \Text(10,65)[lb]{Q}
\ArrowLine(120,60)(90,60)         \Text(105,65)[lb]{Q}
\ArrowLine(60,30)(60,90)          \Text(65,56)[lb]{$P_5$}
\ArrowArc(60,60)(30,0,90)         \Text(78,90)[lb]{$P_2$}
\ArrowArc(60,60)(30,90,180)       \Text(30,90)[lb]{$P_1$}
\ArrowArc(60,60)(30,180,270)      \Text(30,26)[lb]{$P_4$}
\ArrowArc(60,60)(30,270,360)      \Text(78,26)[lb]{$P_3$}

\ArrowLine(160,60)(130,60)        \Text(140,65)[lb]{Q}
\ArrowLine(250,60)(220,60)        \Text(235,65)[lb]{Q} 
\ArrowArc(190,60)(30,0,45)     
\ArrowArc(190,60)(30,45,135)      \Text(185,100)[lb]{$P_3$}
\ArrowArc(190,60)(30,135,180)     \Text(148,77)[lb]{$P_1$}
\ArrowArcn(190,60)(30,360,180)    \Text(185,16)[lb]{$P_2$} 
\ArrowArcn(190,102.4)(30,315,225) \Text(185,59)[lb]{$P_4$} 

\ArrowLine(290,60)(260,60)        \Text(270,65)[lb]{Q}
\ArrowLine(420,60)(390,60)        \Text(405,65)[lb]{Q}
\ArrowArc(315,60)(25,0,180)       \Text(310,95)[lb]{$P_1$}
\ArrowArcn(315,60)(25,360,180)    \Text(310,21)[lb]{$P_2$}
\ArrowArc(365,60)(25,0,180)       \Text(360,95)[lb]{$P_3$}
\ArrowArcn(365,60)(25,360,180)    \Text(360,21)[lb]{$P_4$}

\end{picture} \\ {\sl The two loop topologies T1, T2 and T3} \end{center}
The topologies T2 and T3 can be solved by applying 
equation~\ref{eq:oneloop} twice. The topology T1 is treated by applying 
the rule of the triangle. This will eliminate the denominator belonging 
to one of the lines and the resulting integrals are either of the type 
T2 or of the type T3. There is one complication that arises when the 
power of the center line in T1 is not an integer. In that case the 
application of the rule of the triangle is useless as it cannot reduce 
the power of the center line ($\alpha_0$) to zero. In that case we use different 
recursion relations:
\begin{eqnarray}
\label{eq:twop}
\lefteqn{(\beta_1-1) I(n,\alpha_0,\beta_1,\beta_2,\alpha_1,\alpha_2) Q^2
     = } \nonumber \\ &&
  +(\alpha_0+\beta_1+2\beta_2-D-1)I(n,\alpha_0,\beta_1-1,\beta_2,\alpha_1,
    \alpha_2) \nonumber \\ &&
  +(\beta_1-1)I(n,\alpha_0,\beta_1,\beta_2-1,\alpha_1,\alpha_2)
    \nonumber \\ &&
  +\alpha_0 (I(n,\alpha_0+1,\beta_1-1,\beta_2-1,\alpha_1,\alpha_2)
              \nonumber \\ && \quad
  -I(n,\alpha_0+1,\beta_1-1,\beta_2,\alpha_1,\alpha_2-1)).
\end{eqnarray}
and
\begin{eqnarray}
\label{eq:tabtwo}
\lefteqn{(\alpha_0-1+\epsilon)I(0,\alpha_0,1,1,1,1) =}
\nonumber \\ &&
   -(\alpha_0-2+2\epsilon)I(0,\alpha_0-1,1,1,1,1) Q^{-2} \\ &&
    -(3-3\epsilon-2\alpha_0)G(1,\alpha_0,0,0) G(1,\alpha_0+\epsilon,0,0)
    Q^{-2(2+\alpha_0)}. \nonumber
\end{eqnarray}
In both cases we have one integral left that cannot be solved by these 
means. It has been computed~\cite{integral311}:
\begin{eqnarray}
\label{eq:G311}
    I(0,1+\epsilon,1,1,1,1) = ( 6 \zeta_3 + 9 \zeta_4 \epsilon
    + 102 \zeta_5 \epsilon^2
    + \cdots~ ) ( 1 - 2 \epsilon )
\end{eqnarray}

\subsection{Three loop integrals}

For three loop integrals there are 14 topologies that have to be 
considered. Some of these are rather easy but there are 5 topologies 
that need special attention. One of them is outright complicated.

The first topology to consider is the ladder topology, also called LA:
\begin{center} \begin{picture}(250,120)(0,0)
\ArrowLine(75,60)(45,60)       \Text(55,65)[lb]{$Q$}
\ArrowLine(205,60)(175,60)     \Text(190,65)[lb]{$Q$}
\ArrowLine(105,30)(145,30)     \Text(120,16)[lb]{$P_5$}
\ArrowLine(145,90)(105,90)     \Text(120,100)[lb]{$P_2$}
\ArrowLine(105,30)(105,90)     \Text(110,56)[lb]{$P_7$}
\ArrowLine(145,90)(145,30)     \Text(150,56)[lb]{$P_8$}
\ArrowArc(105,60)(30,90,180)   \Text(75,90)[lb]{$P_1$}
\ArrowArc(105,60)(30,180,270)  \Text(75,26)[lb]{$P_6$}
\ArrowArc(145,60)(30,0,90)     \Text(163,90)[lb]{$P_3$}
\ArrowArc(145,60)(30,270,360)  \Text(163,26)[lb]{$P_4$}
\end{picture} \\ {\sl The ladder or LA topology}
\end{center}
Diagrams of this type can be simplified by applying the rule of the 
triangle to either the loop with $P_7$, $P_1$ and $P_6$, or to the loop 
with $P_8$, $P_4$ and $P_3$. Assuming that we use the first triangle we 
will get then a diagram that misses either $P_7$ in the denominator 
(topology O3) or it misses $P_2$ or $P_5$ in the denominator (topology 
FA). The resulting topologies are treated later in this section.

The second basic topology is the Benz topology, also called BE:
\begin{center} \begin{picture}(250,140)(0,0)
\ArrowLine(85,70)(55,70)        \Text(65,75)[lb]{$Q$}
\ArrowLine(195,70)(165,70)      \Text(180,75)[lb]{$Q$}
\ArrowLine(125,70)(125,30)      \Text(130,46)[lb]{$P_8$}
\ArrowLine(125,70)(153.3,98.3)  \Text(137,71)[lb]{$P_7$}
\ArrowLine(96.7,98.3)(125,70)   \Text(103,71)[lb]{$P_6$}
\ArrowArc(125,70)(40,0,45)      \Text(170,93)[lb]{$P_3$}
\ArrowArc(125,70)(40,45,135)    \Text(120,120)[lb]{$P_2$}
\ArrowArc(125,70)(40,135,180)   \Text(73,93)[lb]{$P_1$}
\ArrowArc(125,70)(40,180,270)   \Text(97,22)[lb]{$P_5$}
\ArrowArc(125,70)(40,270,360)   \Text(143,22)[lb]{$P_4$}
\end{picture} \\ {\sl The Benz or BE topology}
\end{center}
Again we can apply the rule of the triangle to simplify such a diagram. 
The triangle contains the momenta  $P_2$, $P_6$ and $P_7$. After this we 
have diagrams that miss either $P_2$ (topology O1) or they miss $P_1$ or 
$P_3$ (topology BU). These topologies are treated later in this section.

The topology BU is a derivative topology. This means that it is a 
topology that can be obtained by removing one or more denominators in a 
more complicated topology with the same number of loops. It is given by:
\begin{center} \begin{picture}(250,130)(0,0)
\ArrowLine(85,60)(55,60)        \Text(65,65)[lb]{$Q$}
\ArrowLine(195,60)(165,60)      \Text(180,65)[lb]{$Q$}
\ArrowLine(85,60)(125,60)       \Text(100,47)[lb]{$P_2$}
\ArrowLine(125,60)(125,100)     \Text(130,76)[lb]{$P_4$}
\ArrowLine(125,60)(125,20)      \Text(130,36)[lb]{$P_5$}
\ArrowArc(125,60)(40,0,90)      \Text(147,100)[lb]{$P_6$}
\ArrowArc(125,60)(40,90,180)    \Text(93,100)[lb]{$P_1$}
\ArrowArc(125,60)(40,180,270)   \Text(93,15)[lb]{$P_3$}
\ArrowArc(125,60)(40,270,360)   \Text(147,15)[lb]{$P_7$}
\end{picture} \\ {\sl The topology BU}
\end{center}
Also this topology can be simplified with an application of the rule of 
the triangle. In this case the triangle is either the triangle with the 
momenta  $P_4$, $P_1$ and $P_2$ or the triangle with the momenta $P_5$, 
$P_3$ and $P_2$. The choice of the triangle depends on which triangle 
will give the fewest number of terms. Assuming that we used the rule for 
the first triangle we have diagrams that miss either $P_4$ in the 
denominator (topology O2), or they miss $P_6$ in the denominator 
(topology O4) or they miss $P_5$ in the denominator (topology O2). These 
topologies are treated later in this section.

Another derivative topology is the FA topology. It is given by
\begin{center} \begin{picture}(250,120)(0,0)
\ArrowLine(75,60)(45,60)        \Text(55,65)[lb]{$Q$}
\ArrowLine(205,60)(175,60)      \Text(190,65)[lb]{$Q$}
\Line(105,30)(145,30)
\ArrowLine(145,90)(105,90)      \Text(120,100)[lb]{$P_2$}
\ArrowLine(105,90)(125,30)      \Text(100,56)[lb]{$P_6$}
\ArrowLine(125,30)(145,90)      \Text(140,56)[lb]{$P_7$}
\ArrowArc(105,60)(30,90,180)    \Text(75,90)[lb]{$P_1$}
\CArc(105,60)(30,180,218)
\ArrowArc(105,60)(30,218,270)   \Text(75,26)[lb]{$P_5$}
\ArrowArc(145,60)(30,0,90)      \Text(163,90)[lb]{$P_3$}
\ArrowArc(145,60)(30,270,322)   \Text(163,26)[lb]{$P_4$}
\CArc(145,60)(30,322,360)
\end{picture} \\ {\sl The topology FA}
\end{center}
and it is basically the Benz topology with momentum $P_8$ missing. We 
apply again the rule of the triangle and have then either topology O1 
left or topology O2.

By far the most complicated topology is the non-planar topology, also 
called the NO topology:
\begin{center} \begin{picture}(250,120)(0,0)
\ArrowLine(75,60)(45,60)        \Text(55,65)[lb]{$Q$}
\ArrowLine(205,60)(175,60)      \Text(190,65)[lb]{$Q$}
\ArrowLine(145,90)(105,90)      \Text(120,100)[lb]{$P_2$}
\ArrowLine(105,30)(145,30)      \Text(120,16)[lb]{$P_5$}
\ArrowLine(105,90)(123,63)      \Text(100,66)[lb]{$P_7$}
\ArrowLine(127,63)(145,90)      \Text(140,66)[lb]{$P_8$}
\Line(123,63)(145,30)
\Line(105,30)(123,57)
\ArrowArc(105,60)(30,90,180)    \Text(75,90)[lb]{$P_1$}
\ArrowArc(105,60)(30,180,270)   \Text(75,26)[lb]{$P_6$}
\ArrowArc(145,60)(30,270,360)   \Text(163,26)[lb]{$P_4$}
\ArrowArc(145,60)(30,0,90)      \Text(163,90)[lb]{$P_3$}
\end{picture} \\ {\sl The non-planar or NO topology}
\end{center}
In this diagram there are no triangles, so we need additional relations. 
The first step in a reduction is to reduce the dotproducts in the 
numerator to squares of the 8 momenta $P_1$ to $P_8$, and in addition 
$Q.Q$ and either $Q.P_2$ or $P_2.P_5$. The last dotproduct causes 
problems, but it is inevitable, as there are 10 independent variables 
before integration and only 9 squares. The second step in the reduction 
removes this mixed dotproduct (or denominators which effectively 
simplifies the topology). There are two ways to do this. The first 
method was introduced in~\cite{mince}. We define the integral
\begin{equation}
N'(\alpha_1,\cdots,\alpha_8,\alpha_9) =
\int \frac{d^D p_1\, d^D p_2\, d^D p_3\ (Q.p_2)^{\alpha_9}}
    {(p_1.p_1)^{\alpha_1}\cdots(p_8.p_8)^{\alpha_8}}.
\end{equation}
and the recursion is given by
\begin{eqnarray}
N'(\alpha_1,\cdots,\alpha_9) & = & \biggl(
    +\frac{1}{2}(D-1-\alpha_1-2\alpha_2-\alpha_3+\alpha_9) {\bf 9^-}
\nonumber \\ & &
    +\frac{1}{2}\alpha_1 {\bf 1^+ 7^- 9^-}
    -\frac{1}{2}\alpha_1 {\bf 1^+ 2^- 9^-}
\nonumber \\ & &
    +\frac{1}{2}\alpha_3 {\bf 3^+ 8^- 9^-}
    -\frac{1}{2}\alpha_3 {\bf 3^+ 2^- 9^-}
\nonumber \\ & &
    +\frac{\alpha_9-1}{2} {\bf 2^- 9^- 9^-} \biggr)
    /\left(2D-1+\alpha_9-\sum_{i=1}^{8}\alpha_i \right).
\end{eqnarray}
The notation here is somewhat schematic: ${\bf 1^+ 7^- 9^-}$ means that 
in $N$ the $\alpha_1$ should be incremented by one, the $\alpha_7$ 
should be decremented by one and the $\alpha_9$ should also be lowered 
by one. This equation can be applied as many times as needed to 
eliminate all powers of $Q.p_2$. Note that there will never occur 
negative powers of $Q.p_2$ because the term that could generate such a 
negative power will have a zero coefficient.

The second scheme comes from~\cite{fyodor:two}. For this we define the 
integral
\begin{equation}
N(\alpha_1,\cdots,\alpha_8,\alpha_9) =
\int \frac{d^D p_1\, d^D p_2\, d^D p_3\ (p_2.p_5)^{\alpha_9}}
    {(p_1.p_1)^{\alpha_1}\cdots(p_8.p_8)^{\alpha_8}}.
\end{equation}
First we apply the recursion
\begin{eqnarray}
\label{eq:alfa6}
N(\alpha_1,\cdots,\alpha_9) & = &
    +\frac{1}{2}({\bf 3^- 9^-} - {\bf 2^- 9^-} - {\bf 8^- 9^-})
\nonumber \\ &&
    +\frac{1}{2(\alpha_6-1)}(D-1-\alpha_1-2\alpha_2-\alpha_8
    +\alpha_9)\, {\bf 6^- 9^-}
\nonumber \\ &&
    + \frac{\alpha_1}{2(\alpha_6-1)}({\bf 1^+ 6^- 7^- 9^-}
        - {\bf 1^+ 2^- 6^- 9^-})
\nonumber \\ &&
    + \frac{\alpha_8}{2(\alpha_6-1)}({\bf 3^- 6^- 8^+ 9^-}
        - {\bf 2^- 6^- 8^+ 9^-}).
\end{eqnarray}
till either $\alpha_9 = 0$, or $\alpha_6 = 1$, or one of the other 
denominators is completely removed. In the first case we have reached 
our objective of removing the mixed dotproduct and in the last case the 
topology has been simplified. Hence we can concentrate on the second. 
case. By applying a symmetry transformation to equation~\ref{eq:alfa6} 
we can use it to make $\alpha_1$, $\alpha_3$ and $\alpha_4$ equal to 1 
(or remove $\alpha_9$ or remove one of the other denominators). Hence when 
we are done with this recursion four of the denominators have only a 
single power. The next relation is
\begin{eqnarray}
\label{eq:alfa5}
\lefteqn{ N(1,\alpha_2,1,1,\alpha_5,1,\alpha_7,\alpha_8,\alpha_9) = }
\nonumber\\&
    -\frac{D-1+\alpha_9-2\alpha_2-\alpha_7-\alpha_8}{2(\alpha_5-1)}
    {\bf 5^- 9^-}
    +\frac{\alpha_9-1}{2(\alpha_5-1)}{\bf 2^- 5^- 9^- 9^-}
\nonumber \\ &
    + \frac{\alpha_7}{2(\alpha_5-1)}({\bf 2^- 5^- 7^+ 9^-}
        - {\bf 1^- 5^- 7^+ 9^-} )
    + \frac{\alpha_8}{2(\alpha_5-1)}({\bf 2^- 5^- 8^+ 9^-}
        - {\bf 3^- 5^- 8^+ 9^-} ).
\end{eqnarray}
This recursion will make $\alpha_5 = 1$ (or again remove $\alpha_9$ or 
another denominator) and a symmetric version of it will do the same for 
$\alpha_2$. After that we need two more recursions:
\begin{eqnarray}
\label{eq:alfa8}
\lefteqn{ N(1,1,1,1,1,1,\alpha_7,\alpha_8,\alpha_9) = } 
\nonumber \\ &&
    +\frac{1}{2}(2\ {\bf 1^- 9^-}+2\ {\bf 3^- 9^-}-2\ {\bf 2^- 9^-}
        -{\bf 7^- 9^-} - Q^2\ {\bf 9^-})
\nonumber \\ &&
    +\frac{1}{2}\frac{\alpha_9-1}{\alpha_8-1}({\bf 1^- 8^- 9^- 9^-}
        -{\bf 2^- 8^- 9^- 9^-} - {\bf 7^- 8^- 9^- 9^-} )
\nonumber \\ &&
    +\frac{D-1-2 \alpha_7-\alpha_8}{2(\alpha_8-1)} {\bf 8^- 9^-}
    +\frac{1}{\alpha_8-1}({\bf 1^- 2^+ 8^- 9^-} - {\bf 2^+ 7^- 8^- 9^-}
    ), \\
\label{eq:alfa7}
\lefteqn{ N(1,1,1,1,1,1,\alpha_7,1,\alpha_9) =} 
\nonumber \\ &&
    +\frac{1}{2}(2\ {\bf 4^- 9^-} + 2\ {\bf 6^- 9^-} - 2\ {\bf 5^- 9^-}
    - {\bf 8^- 9^-} -Q^2\ {\bf 9^-})
\nonumber \\ &&
    +\frac{1}{2}\frac{\alpha_9-1}{\alpha_7-1}({\bf 6^- 7^- 9^- 9^-}
        -{\bf 5^- 7^- 9^- 9^-} - {\bf 7^- 8^- 9^- 9^-} )
\nonumber \\ &&
    +\frac{D-3-\alpha_7}{2(\alpha_7-1)} {\bf 7^- 9^-}
    +\frac{1}{\alpha_7-1}({\bf 5^+ 6^- 7^- 9^-} - {\bf 5^+ 7^- 8^- 9^-}
    ).
\end{eqnarray}
They will make $\alpha_8$ and $\alpha_7$ equal to one. This brings us to 
the last recursion
\begin{eqnarray}
\label{eq:alfa9}
N(1,1,1,1,1,1,1,1,\alpha_9) & = &
    {\bf 3^- 9^-} - {\bf 8^- 9^-} + Q^2 \frac{1+2\epsilon-\alpha_9}
    {2(\alpha_9-1-4\epsilon)}{\bf 9^-}
\nonumber \\ &&
    +\frac{Q^2}{\alpha_9-1-4\epsilon}({\bf 2^- 3^+ 9^-}
    -{\bf 3^+ 8^- 9^-}).
\end{eqnarray}
After this all integrals that are left have either a simpler topology or 
have $\alpha_9 = 0$.

The advantage of the first reduction scheme is its simplicity, while the 
second scheme is faster because at each step (except for the last) it 
does not only reduce the value of $\alpha_9$, but the sum of the other 
$\alpha$'s is also reduced and hence it leaves simpler integrals behind. 
The simplicity of the programming made that the first method was used in 
the original Mincer program~\cite{mince}. In the FORM version of Mincer 
we have selected the second method. This will be discussed in a later 
section.

After eliminating $\alpha_9$ we still have to try to remove one of the 
other $\alpha$'s. First we apply:
\begin{eqnarray}
Q^2 N(\alpha_1,\cdots,\alpha_8) & = &
    {\bf 1^-} + \frac{-2D-1+2\alpha_1+2\alpha_2+\alpha_5+\alpha_6+2\alpha_7
    +\alpha_8}{\alpha_6-1} {\bf 6^-} \nonumber \\ &&
    +\frac{\alpha_8}{\alpha_6-1} {\bf 6^- 8^+}({\bf 2^-}-{\bf 3^-})
    +\frac{\alpha_5}{\alpha_6-1} {\bf 6^- 5^+}({\bf 7^-}-{\bf 4^-}).
\end{eqnarray}
or a symmetric version of it to make $\alpha_6$, $\alpha_1$, $\alpha_3$ 
and $\alpha_4$ equal to one (or remove one of the denominators). Note 
that when $\alpha_9$ is not present the $N'$ and the $N$ integrals are 
identical. Then we use
\begin{eqnarray} 
\lefteqn{(\alpha_5-1)(2-D+2\alpha_5)
          N(1,\alpha_2,1,1,\alpha_5,1,\alpha_7,\alpha_8) = } 
\nonumber \\ &&
(-2D+4+\beta) Q^2 {\bf 5^-} ( +4-3D+2 \beta + {\bf 3^- 4^+} +
{\bf 1^- 6^+} ) 
\nonumber \\ && 
+(\alpha_5-1)( {\bf 7^-4^+} + {\bf 8^-6^+} ) 
\nonumber \\ && 
-(3-2D+\beta+\alpha_5) ( {\bf 5^- 4^+} + {\bf 5^- 6^+} ),
\end{eqnarray} 
%
with $\beta =\alpha_2+\alpha_5+\alpha_7+\alpha_8$ to make $\alpha_5$, 
$\alpha_2$, $\alpha_8$ and $\alpha_7$ equal to one. 
At this point we have either killed off at least one denominator in 
which case the most complicated topologies that are left are of the type 
FA or BU, or we are faced with the integral
\begin{equation}
\label{eq:F321}
    N(1,1,1,1,1,1,1,1) = 20 \zeta_5 + {\cal O}(\epsilon).
\end{equation}
This integral was evaluated in~\cite{fyodor:two}.
The term in $O(\epsilon)$ has been computed in \cite{kaz} but here we do 
not need it~\cite{Gscheme}.

The leftover topologies have all a one loop two-point function subgraph 
that can be integrated directly. We still distinguish two groups. 
Diagrams that still need work at the two loop level (diagrams of type O) 
and diagrams that do not even need that work anymore (diagrams of type 
Y). The topologies of type O are

\begin{center} \begin{picture}(400,120)(0,0)
\ArrowLine(20,60)(0,60)         \Text(5,65)[lb]{$Q$}
\ArrowLine(120,60)(100,60)      \Text(110,65)[lb]{$Q$}
\ArrowArc(60,60)(40,0,90)       \Text(82,100)[lb]{$P_2$}
\ArrowArc(60,60)(40,90,180)     \Text(28,100)[lb]{$P_1$}
\ArrowArc(60,60)(40,180,270)    \Text(28,15)[lb]{$P_4$}
\ArrowArc(60,60)(40,270,360)    \Text(82,15)[lb]{$P_3$}
\ArrowLine(60,20)(60,55)        \Text(65,33.5)[lb]{$P_5$}
\Line(60,95)(60,100)
\ArrowArc(60,75)(20,270,90)     \Text(77,53.5)[lb]{$P_7$}
\ArrowArcn(60,75)(20,270,90)    \Text(35,53.5)[lb]{$P_6$}

\ArrowLine(150,60)(130,60)      \Text(135,65)[lb]{$Q$}
\ArrowLine(250,60)(230,60)      \Text(240,65)[lb]{$Q$}
\ArrowLine(180,30)(200,30)      \Text(185,16)[lb]{$P_4$}
\ArrowLine(200,30)(200,90)      \Text(205,56)[lb]{$P_5$}
\Line(180,90)(200,90)
\ArrowArc(180,60)(30,90,142)    \Text(150,90)[lb]{$P_1$}
\CArc(180,60)(30,142,180)
\ArrowArc(180,60)(30,180,270)   \Text(150,26)[lb]{$P_7$}
\ArrowArcn(150,30)(30,90,0)     \Text(168,60)[lb]{$P_6$}
\ArrowArc(200,60)(30,0,90)      \Text(218,90)[lb]{$P_2$}
\ArrowArc(200,60)(30,270,360)   \Text(218,26)[lb]{$P_3$}

\ArrowLine(280,60)(260,60)      \Text(265,65)[lb]{$Q$}
\ArrowLine(400,60)(380,60)      \Text(390,65)[lb]{$Q$}
\ArrowLine(310,30)(310,90)      \Text(315,56)[lb]{$P_5$}
\ArrowArc(310,60)(30,0,90)      \Text(328,90)[lb]{$P_2$}
\ArrowArc(310,60)(30,90,180)    \Text(280,90)[lb]{$P_1$}
\ArrowArc(310,60)(30,180,270)   \Text(280,26)[lb]{$P_4$}
\ArrowArc(310,60)(30,270,360)   \Text(330,26)[lb]{$P_3$}
\ArrowArc(360,60)(20,0,180)     \Text(357,90)[lb]{$P_6$}
\ArrowArcn(360,60)(20,360,180)  \Text(357,26)[lb]{$P_7$}
\end{picture} \\ {\sl The topologies O1, O2, O3}
\end{center}
and
\begin{center} \begin{picture}(250,140)(0,0)
\ArrowLine(85,70)(55,70)        \Text(65,75)[lb]{$Q$}
\ArrowLine(195,70)(165,70)      \Text(180,75)[lb]{$Q$}
\ArrowLine(125,70)(125,110)     \Text(130,86)[lb]{$P_5$}
\ArrowArc(125,70)(40,0,30)
\ArrowArc(125,70)(40,30,90)     \Text(141,116)[lb]{$P_2$}
\ArrowArc(125,70)(40,90,150)    \Text(100,116)[lb]{$P_1$}
\ArrowArc(125,70)(40,150,180)   \Text(73,90)[lb]{$P_6$}
\ArrowArcn(125,70)(40,360,180) \Text(120,16)[lb]{$P_7$}
\ArrowArc(125,110)(40,210,270)  \Text(100,60)[lb]{$P_4$}
\ArrowArc(125,110)(40,270,330)  \Text(141,60)[lb]{$P_3$}
\end{picture} \\ {\sl The topology O4}
\end{center}
The topologies O1-O3 have a trivial two point subintegral (with momenta 
$P_6$ and $P_7$). The topology O4 requires more care. Based on scaling 
and Lorenz invariance arguments one can treat the integral over $P_6$ 
first. This is particularly simple when there are no dotproducts in the 
numerator. If there are such dotproducts, dotproducts between momenta in 
the two loop subgraph are also trivial, dotproducts involving $P_7$ can 
be rewritten with $P_7=Q-P_6$, and $P_6 = P_1-P_4$. This leaves as only 
complicated dotproducts the ones involving $Q$ and $P_{1-5}$. Let us 
replace $Q.P_i$ by $x R(P_i)$. Then we can use\footnote{The relevant
formula (A.16) from \cite{fyodor:two} contains a misprint.}:

\begin{eqnarray}
\label{eq:o4int}
    \frac { x^n d^Dp_6 }{ {(p_6^2)} ^{x_1} {(p_7^2)} ^{x_2} } & = &
    \frac{1}{{(Q^2)}^{x_1+x_2-D/2}}
    \sum_{s=0}^{n/2} G(x_1-s,x_2,n-2 s,0) \nonumber \\ &&
    \sum_{j=0}^{n/2-s}
    \frac{\Gamma(n+2-\epsilon-2 s)\Gamma(n+1-\epsilon-2 s-j) (-1)^j}
    {\Gamma(n+2-\epsilon-s)\Gamma(n+1-\epsilon-2 s)\ s!\ j!}
\nonumber \\ && \qquad \times
    \left(\frac{Q^2\Box_R}{4}\right)^{s+j}
\end{eqnarray}
The d'Alembertian $\Box_R$ acts only on the function R. If it is defined by
$\Box_R = \partial^\mu\partial_\mu$ then $\partial^\mu R(p_i) = p_i^\mu$.
The way to evaluate these derivatives is discussed in a later section.

The Y topologies are given by
\begin{center} \begin{picture}(400,240)(0,0)
\ArrowLine(20,180)(0,180)          \Text(0,185)[lb]{$Q$}
\ArrowLine(120,180)(100,180)       \Text(112,185)[lb]{$Q$}
\ArrowArc(60,180)(40,0,45)
\ArrowArc(60,180)(40,45,135)       \Text(55,230)[lb]{$P_3$}
\ArrowArc(60,180)(40,135,180)      \Text(10,200)[lb]{$P_1$}
\ArrowArcn(60,180)(40,360,315)
\ArrowArcn(60,180)(40,315,225)     \Text(55,125)[lb]{$P_6$}
\ArrowArcn(60,180)(40,225,180)     \Text(10,160)[lb]{$P_2$}
\ArrowArcn(60,236.57)(40,315,225)  \Text(48,185)[lb]{$P_4$}
\ArrowArc(60,123.43)(40,45,135)    \Text(62,172)[lb]{$P_5$}

\ArrowLine(160,180)(140,180)          \Text(140,185)[lb]{$Q$}
\ArrowLine(260,180)(240,180)          \Text(252,185)[lb]{$Q$}
\ArrowArc(200,180)(40,0,15)
\ArrowArc(200,180)(40,15,75)          \Text(232,217)[lb]{$P_5$}
\ArrowArc(200,180)(40,75,105)         \Text(195,230)[lb]{$P_1$}
\ArrowArc(200,180)(40,105,165)        \Text(158,217)[lb]{$P_3$}
\ArrowArc(200,180)(40,165,180)
\ArrowArcn(200,180)(40,360,180)       \Text(195,126)[lb]{$P_2$}
\ArrowArcn(248.99,228.99)(40,255,195) \Text(209,187)[lb]{$P_6$}
\ArrowArcn(151.01,228.99)(40,345,285) \Text(183,187)[lb]{$P_4$}

\ArrowLine(300,180)(280,180)                \Text(280,185)[lb]{$Q$}
\ArrowLine(400,180)(380,180)                \Text(392,185)[lb]{$Q$}
\ArrowArc(340,180)(40,0,20)
\ArrowArc(340,180)(40,20,45)
\ArrowArc(340,180)(40,45,135)               \Text(335,230)[lb]{$P_5$}
\ArrowArc(340,180)(40,135,160)              \Text(295,210)[lb]{$P_3$}
\ArrowArc(340,180)(40,160,180)              \Text(288,195)[lb]{$P_1$}
\ArrowArcn(340,180)(40,360,180)             \Text(335,126)[lb]{$P_2$}
\ArrowArcn(340,236.57)(40,315,225)          \Text(335,205)[lb]{$P_6$}
%\ArrowArcn(340,236.57)(55.39,318.15,221.85) \Text(335,165)[lb]{$P_4$}
\ArrowArcn(340,236.57)(55.39,312,228) \Text(335,165)[lb]{$P_4$}

\ArrowLine(50,60)(30,60)            \Text(30,65)[lb]{$Q$}
\ArrowLine(190,60)(170,60)          \Text(182,65)[lb]{$Q$}
\ArrowArc(150,60)(20,0,180)         \Text(145,90)[lb]{$P_5$}
\ArrowArcn(150,60)(20,360,180)      \Text(145,26)[lb]{$P_6$}
\ArrowArc(90,60)(40,0,45)
\ArrowArc(90,60)(40,45,135)         \Text(85,110)[lb]{$P_3$}
\ArrowArc(90,60)(40,135,180)        \Text(40,80)[lb]{$P_1$}
\ArrowArcn(90,60)(40,360,180)       \Text(85,6)[lb]{$P_2$}
\ArrowArcn(90,116.57)(40,315,225)   \Text(85,60)[lb]{$P_4$}

\ArrowLine(230,60)(210,60)          \Text(210,65)[lb]{$Q$}
\ArrowLine(370,60)(350,60)          \Text(362,65)[lb]{$Q$}
\ArrowArc(250,60)(20,0,180)         \Text(245,90)[lb]{$P_1$}
\ArrowArcn(250,60)(20,360,180)      \Text(245,26)[lb]{$P_2$}
\ArrowArc(290,60)(20,0,180)         \Text(285,90)[lb]{$P_3$}
\ArrowArcn(290,60)(20,360,180)      \Text(285,26)[lb]{$P_4$}
\ArrowArc(330,60)(20,0,180)         \Text(325,90)[lb]{$P_5$}
\ArrowArcn(330,60)(20,360,180)      \Text(325,26)[lb]{$P_6$}
\end{picture} \\ {\sl The topologies Y1-Y5}
\end{center}
In the case of these topologies one can directly apply
equation~\ref{eq:oneloop} leaving a diagram of topology T2 or T3.

\subsection{Harmonic projections}

A harmonic tensor of order n in momentum P is defined as a totally 
symmetric tensor with n indices that is traceless and in which the 
indices are either coupled to Kronecker delta's or the momentum P. Such 
harmonic tensors are Lorenz invariants. Hence, if we have such a tensor 
of which the indices are contracted with momenta we are integrating 
over, after the integration we will still have such a tensor, even 
though the indices are now contracted with other momenta. We will use 
this property frequently. It has actually already been used in 
equations~\ref{eq:oneloop} and \ref{eq:o4int}. We need it however in a 
very visible way when the moments of structure functions in lepton 
hadron scattering are studied. In that case we have an effective 
four-point function and two different external momenta Q and P. Such 
integrals cannot be dealt with directly. When we make an expansion in 
terms of P around zero we effectively eliminate one of the external 
momenta at the cost of much more complicated integrals. Because the 
moments of structure functions actually correspond to powers of 
$P.Q/Q.Q$ we can eliminate all terms that contain $P.P$ and the 
extraction of the n-th power of $P.Q/Q.Q$ corresponds to a harmonic 
projection. 
\begin{equation}
    P^{\mu_1}\cdots P^{\mu_n} T^{\mu_1\cdots \mu_n}
    = (P.Q/Q.Q)^n H^{\mu_1\cdots \mu_n} T^{\mu_1\cdots \mu_n}
\end{equation}
in which the left hand side represents our integral and H is the 
harmonic projection operator. This operator is again a harmonic tensor 
of degree n in Q. It has some rather nasty properties. The number of 
terms in it is a very rapidly increasing function of n:
\begin{center}
\begin{tabular}{|r|r|} \hline n & terms \\ \hline
2  & 2 \\
3  & 4 \\
4  & 10 \\
5  & 26 \\
6  & 76 \\
7  & 232 \\
8  & 764 \\
9  & 2620 \\
10 & 9496 \\
11 & 35696 \\
12 & 140152 \\
\hline
\end{tabular} \\
{\sl Number of terms in harmonic tensor of order n}
\end{center}
When calculating longitudinal crosssections one has to make a 
longitudinal projection that involves two extra powers of P and hence 
the computation of a n=6 moment involves a harmonic projection with 8 
indices. For a n=10 moment the harmonic projection operator has 9496 
terms for the transverse projection and 140152 terms for the 
longitudinal projection! It should be notted that at the moment of 
projection we have already made a rather deep expansion in P, so there 
are already many terms. This presents us with quite a problem.

The first solution that we have used~\cite{2loopmoments} was to avoid 
the harmonic projection completely. This was possible because it 
involved a 2 loop computation. When there are only integer powers of the 
denominators in the topology T1 we can keep a `foreign' momentum like P. 
Even though this makes the integrals more complicated it seems to be the 
most economic method. Unfortunately, for three loop integrals this 
method cannot be used. The non-planar recursions cannot deal with an 
extra momentum, and neither can the two loop recursions when there is a 
denominator with a non-integer power. Hence we need a better solution.

Let us first look at what a harmonic tensor looks like:
\begin{equation}
    H^{\mu_1\mu_2}(Q) = h^2_0 Q^{\mu_1} Q^{\mu_2} + h^2_2 
        \delta^{\mu_1\mu_2} Q.Q
\end{equation}
The value of $h^2_2$ can be expressed in terms of $h^2_0$ with the 
condition that the tensor should be traceless. The normalization is 
given by the condition that
\begin{equation}
    H^{\mu_1\mu_2}(Q) Q^{\mu_1} Q^{\mu_2} = (Q.Q)^2
\end{equation}
which is equivalent to the statement that $(P.Q)^n$ should be invariant 
under the whole procedure sketched above. If we define 
$h^n_j$ as the coefficient of all terms in the n-th order tensor that 
have $j/2$ Kronecker delta's then the normalization condition is that
\begin{equation}
    \sum_{j=0, j even}^{n} h^n_j = 1
\end{equation}
The formula for the $h^n_j$ is given by
\begin{equation}
    h^n_j = (-1)^j 2^{n-j} \frac{\Gamma(D-2)
    \Gamma(n+D/2-1-j)}{\Gamma(n+D-2)\Gamma(D/2-1)}
\end{equation}
which, as one can see, was already incorporated in eq~\ref{eq:o4int}.
The FORM code for the tensor becomes now:
\begin{verbatim}
  id  P.Q = Q.Q;
  ToTensor,nosquare,T,P;
  id  T(??) = sum_(j,0,nargs_(..),2,Q.Q^(j/2)*
              acc(hfac(nargs_(..),j))*distrib_(1,j,dd,T,..));
  tovector,T,Q;
  id  dd(??) = dd_(..);
\end{verbatim}
Let us look at these statements in detail as most are quite powerful. 
The first statement reflects the normalization property that
\begin{equation}
    H^{\mu_1 \cdots \mu_n} Q^{\mu_n} = H^{\mu_1 \cdots \mu_{n-1}} Q.Q
\end{equation}
The second statement is rather new in FORM (the statements 2,4 and 5 can 
be used since version 2.2) and collects all occurrences of the vector P 
in a single tensor T. in older versions of FORM one would use
\begin{verbatim}
    id  P = T(?);
    repeat id T(?)*T(??) = T(.,..);
\end{verbatim}
The fourth statement is exactly the opposite: the indices in the tensor 
T are connected to (multiple) occurrences of the vector Q.
The third statement sums over the number of indices that will end up in 
Kronecker delta's and then the distrib\_ function splits the list of 
indices in all different distributions of the indices for the delta's, 
and those that will end up in the vectors Q. The function dd\_ 
distributes the indices in its arguments over the Kronecker delta's in 
all different ways. Its power is that when indices are contracted with 
identical vectors this function takes this into account with the 
appropriate factor. It does not generate the same term twice.
The factor $h^n_j$ is called hfac and it has been tabulated as 
powerseries in $\epsilon$ to save 
time. The use of the function acc has already been explained.

The above program by itself is no guarantee for speed. Although the 
functions distrib\_ and dd\_ can take care of many symmetries a straight 
forward application would still be very impratical for the higher 
moments. We need another observation: We a diagram of a topology like NO 
will be integrated there are 9 different momenta involved. This means 
that the tensor T can be contracted with 8 different momenta (we 
eliminated the contractions with Q). There are however only 3 
independent momenta of those 8, so if we rewrite the momenta to three 
such momenta, before we try the harmonic projection there will be many 
identical momenta in T and the number of terms generated will be 
relatively modest. This is even more so for two loop integrals in which 
there are only two independent mementa (apart from Q). This way it 
turned out to be possible to redo the two-loop calculation in hardly 
more time than was needed for the calculation in which no harmonic 
projection was used, even for the n=10 moment in the longitudinal 
projection. 

 
\begin{thebibliography}{99}
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P-0330 (Moscow, 1984);  S.G. Gorishny, S.A. Larin, L.R. Surguladze and 
F.V. Tkachov,Comp. Phys. Comm. 55 (1989) 381.
%
\bibitem{form} J.A.M. Vermaseren, Symbolic Manipulation with FORM,
published by CAN (Computer Algebra Nederland), Kruislaan 413, 1098 SJ 
Amsterdam, 1991, ISBN 90-74116-01-9. An
earlier version of FORM can be obtained for free by means of anonymous
ftp from nikhefh.nikhef.nl. Version 1 is however insufficient to run
the program described in this publication.
%
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%
\bibitem{kino} See for instance S.Laporta and E.Remiddi,
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\bibitem{collins} J.C.Collins, "Renormalization", Cambridge Univ. Press,
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\end{thebibliography}

\end{document}

\begin{center} \begin{picture}(250,120)(0,0)
\end{picture} \\ {\sl The or LA topology}
\end{center}

