LToolsImplicitPrefactor is an option for
LToolsEvaluate. It specifies a prefactor that does not show
up explicitly in the input expression, but is understood to appear in
front of every Passarino-Veltman function. The default value is
1.
You may want to use
LToolsImplicitPrefactor->1/(2Pi)^D when working with
1-loop amplitudes, if no explicit prefactor has been introduced from the
very beginning.
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.
====================================================*)Here the prefactor i \pi^2 arises from the conversion of \int d^D q\, 1/(q^2-m^2) to A_0(m^2)
LToolsEvaluate[FAD[{q, m}], q, InitialSubstitutions -> {m -> 5}]\frac{0.\, +246.74 i}{\varepsilon }-(0.\, +972.359 i)
LToolsEvaluate[FAD[{q, m}], q, InitialSubstitutions -> {m -> 5}, Head -> keep]\frac{i \pi ^2 \;\text{keep}(25.)}{\varepsilon }+i \pi ^2 (\text{keep}(-55.4719)-\gamma \;\text{keep}(25.)-\text{keep}(25.) \log (\pi ))
This recovers the textbook prefactor
LToolsEvaluate[FAD[{q, m}], q, InitialSubstitutions -> {m -> 5}, LToolsImplicitPrefactor -> 1/(2 Pi)^(4 - 2 Epsilon)]\frac{0.\, +0.158314 i}{\varepsilon }-(0.\, +0.0419639 i)
(PaXEvaluate[FAD[{q, m}], q, PaXImplicitPrefactor -> 1/(2 Pi)^(4 - 2 Epsilon)] /. {m -> 5, ScaleMu^2 -> 1}) // N\frac{0.\, +0.158314 i}{\varepsilon }-(0.\, +0.0419639 i)
If the input expression contains both loop and non-loop terms, only
the terms containing a PaVe-function will be multiplied by
the implicit prefactor
LToolsEvaluate[extra + FAD[{q, m}], q, InitialSubstitutions -> {m -> 2}, LToolsExpandInEpsilon -> False]\frac{(0.\, +4. i) \pi ^{2-\varepsilon } \Gamma (\varepsilon +1) \Gamma (1-\varepsilon )^2}{\varepsilon \Gamma (1-2 \varepsilon )}-\frac{(0.\, +1.54518 i) \pi ^{2-\varepsilon } \Gamma (\varepsilon +1) \Gamma (1-\varepsilon )^2}{\Gamma (1-2 \varepsilon )}+\text{extra}