LToolsFullResult is an option for
LToolsEvaluate. When set to True (default),
LToolsEvaluate will return the full result including
singularities and accompanying terms. Otherwise, only the finite part
(standard output of LoopTools) will be provided.
The full result is assembled from pieces returned by LoopTools for the \lambda^2-parameter set to -2, -1 and 0 respectively. The correct prefactor that accounts for the normalization differences between Passarino-Veltman function in FeynCalc and LoopTools is added as well.
As long as LToolsFullResult is set to True,
the value of the LToolsSetLambda option is ignored.
Disabling LToolsFullResult will most likely lead to
incorrect normalization of the results (especially if you are only
interested in the finite part). The reason for this are missing
contributions to the finite part generated from poles being multiplied
by terms proportional to \varepsilon or
\varepsilon^2.
LToolsLoadLibrary[];\text{LoopTools library loaded.}
(* ====================================================
FF 2.0, a package to evaluate one-loop integrals
written by G. J. van Oldenborgh, NIKHEF-H, Amsterdam
====================================================
for the algorithms used see preprint NIKHEF-H 89/17,
'New Algorithms for One-loop Integrals', by G.J. van
Oldenborgh and J.A.M. Vermaseren, published in
Zeitschrift fuer Physik C46(1990)425.
====================================================*)LToolsEvaluate[A0[m^2], InitialSubstitutions -> {m^2 -> 1}]\frac{1.}{\varepsilon }-0.721946
Setting LToolsFullResult to False will make
LToolsEvaluate return only the finite part since the
default value for LToolsSetLambda is 0.
However, the normalization does not agree with the FeynCalc
convention
LToolsEvaluate[A0[m^2], InitialSubstitutions -> {m^2 -> 1}, LToolsFullResult -> False]1.
Even though LToolsEvaluate includes the correct
prefactor to convert to the FeynCalc normalization, the finite
contribution generated by the 1/\varepsilon-pole is missing here.
finRes = LToolsEvaluate[A0[m^2], InitialSubstitutions -> {m^2 -> 1}, LToolsFullResult -> False, LToolsExpandInEpsilon -> False]\frac{1. \pi ^{-\varepsilon } \Gamma (1-\varepsilon )^2 \Gamma (\varepsilon +1)}{\Gamma (1-2 \varepsilon )}
By setting LToolsSetLambda->-1 we can get the
coefficient of the pole. Here it is obvious that the function is
IR-finite so that we do not need to check for the 1/\varepsilon^2-pole
poleRes = LToolsEvaluate[A0[m^2], InitialSubstitutions -> {m^2 -> 1}, LToolsFullResult -> False, LToolsExpandInEpsilon -> False, LToolsSetLambda -> -1]\frac{1. \pi ^{-\varepsilon } \Gamma (1-\varepsilon )^2 \Gamma (\varepsilon +1)}{\Gamma (1-2 \varepsilon )}
Combining both pieces and expanding in \varepsilon up to zeroth order we recover the
same result as when using the option LToolsFullResult
Series[1/Epsilon poleRes + finRes, {Epsilon, 0, 0}] // Normal\frac{1.}{\varepsilon }-0.721946