-The exact for depends on the helicity and direction (forward/backwards) for the quarks. Currently all different contractions of incoming and outgoing states are defined in \lstinline!joi!, \lstinline!jio! and \lstinline!joo!.
+The exact form depends on the helicity and direction (forward/backwards) for the quarks.
\subsubsection{Gluon}
In \HEJ the currents for gluons and quarks are the same, up to a colour factor $K_g/C_F$, where
The Integrals as such are provided by \QCDloop{} (see wrapper functions \lstinline!B0DD! and \lstinline!C0DD! in \texttt{src/Hjets.cc}).
In the code we are sticking to the convention of~\cite{DelDuca:2003ba}, thus instead of the $T_{1/2}$ we implement (in the functions \lstinline!A1! and \lstinline!A2!)
with $t_1 = (p_a-p_1)^2$, $t_2 = (p_a-p_1-p_H)^2$ and $R = 8 \pi^2$\todo{Code has $R=8\pi^2 i$}. Like other special currents, eq.~\eqref{eq:jH_same_helicity}
is implemented in \texttt{current\_generator/include/currents.frm}.
The currents with a helicity flip is given through
and implemented again in \texttt{current\_generator/include/currents.frm}.\todo{sign mismatch in line 5 and negative-helicity polarisation vectors}
If we instead choose the gluon momentum in the $+$ direction, so that
$p_a^- = p_{a\perp} = 0$, the corresponding currents are obtained by
replacing $p_1^- \to p_1^+, p_a^- \to p_a^+,
\frac{p_{1\perp}}{|p_{1\perp}|} \to -1$ in the second line of
eq.~\eqref{eq:jH_same_helicity} and eq.~\eqref{eq:jH_helicity_flip}..
The form factors $H_1,H_2,H_4,H_5, H_{10}, H_{12}$ are given in~\cite{DelDuca:2003ba}, and are implemented as \lstinline!H1DD,H2DD! etc. in \texttt{src/Hjets.cc}. They reduce down to fundamental QCD integrals, which are again provided by \QCDloop.
\subsubsection{Peripheral Higgs emission - Infinite top mass}
\label{sec:jH_eff}
To get the result with infinite top mass we could either take the limit $m_t\to \infty$ in~\eqref{eq:jH_helicity_flip} and~\eqref{eq:jH_same_helicity}, or use the \textit{impact factors} as given in~\cite{DelDuca:2003ba}. Both methods are equivalent, and lead to the same result. For the first one would find
\begin{align}
\lim_{m_t\to\infty} m_t^2 H_1 &= i \frac{1}{24 \pi^2}\\