0$ away from the scatterer:
%Though $I_l = \iu[H, Q_l] = Q[\delta(x-l) - \delta(x+l)]$ commutes
%with $Q_l$, seemingly suppressing the contact terms, it has to be
%noticed that $[Q_l, \rho]\notin \ideal{2}$ in the first place.
The computation of the integrals~(\ref{3Cumul}) will make repeated use of the following expressions. The translation invariant density matrices $\rho$ and $\rho'$ have integral kernels
%
\begin{align*}
\rho(x,y) &= \frac{1}{2\pi\iu}\frac{1}{x-y-\iu 0}\cdot D(x-y)\,, \\
\rho'(x,y) &= \frac{1}{2\pi\iu}\frac{1}{y-x-\iu 0}\cdot D(x-y)
\end{align*}
%
with
%
\begin{eqnonum}
D(z) = \left(\begin{array}{ll}
\ep{\iu \mu_L z} & 0 \\ 0 & \ep{\iu \mu_R z}
\end{array}\right)\,.
\end{eqnonum}
%
Moreover,
%
\begin{eqnonum}
S\str Q S = \left(
\begin{array}{ll}
T & \mathfrak{r}'\overline{\mathfrak{t}} \\ \overline{\mathfrak{r}}'\mathfrak{t} & 1-T
\end{array}
\right)\,,\qquad[Q,S\str Q S] = \left(
\begin{array}{ll}
0 & -\mathfrak{r}'\overline{\mathfrak{t}} \\ \overline{\mathfrak{r}}'\mathfrak{t} & 0
\end{array}
\right)\,.
\end{eqnonum}
%
We compute the first integrand~(\ref{3Cumul}) by temporarily dropping the time ordering.
%
\begin{align}
\label{III}
\cumul{\widehat{I}_1\widehat{I}_2\widehat{I}_3}_\rho &= \tr(I_1 \rho' I_2 \rho' I_3 \rho) - \tr(I_1 \rho' I_3 \rho I_2 \rho)\\
&= \frac{1}{(2\pi\iu)^3}\frac{\tr\big(A(t_2-t_1)A(t_3-t_2)A(t_1-t_3)\big)-\tr\big(A(t_3-t_1)A(t_2-t_3)A(t_1-t_2)\big)}{(t_1-t_2-\iu 0)(t_2-t_3-\iu 0)(t_1-t_3-\iu 0)}\label{IIIbis}
\end{align}
%
with
%
\begin{eqwithnum}
\label{A}
A(t_i-t_j) = (S\str Q S -Q) D(t_i-t_j) = \left(\begin{array}{ll}
T & \mathfrak{r}'\overline{\mathfrak{t}} \\ \overline{\mathfrak{r}}'\mathfrak{t} & -T
\end{array}\right)D(t_i-t_j)\,,
\end{eqwithnum}
%
where we used (\ref{Current}). That results in
%
\begin{eqwithnum}
\label{IIIter}
\cumul{\widehat{I}_1\widehat{I}_2\widehat{I}_3}_\rho = \frac{-4 T^2(1-T)}{(2\pi)^3}\cdot\frac{\sin V(t_1-t_2)+\sin V(t_2-t_3)+\sin V(t_3-t_1)}{(t_1-t_2-\iu 0)(t_2-t_3-\iu 0)(t_1-t_3-\iu 0)}\,,
\end{eqwithnum}
%
where $V = \mu_L-\mu_R$. In fact, the first trace equals
%
\begin{eqnonum}
2\iu T^2(1-T)\sin V(t_3-t_1)+(\text{cyclic})
\end{eqnonum}
%
by the following argument. It may be expressed as $\sum_{i,j,k = L,R}A_{ij}A_{jk}A_{ki}$. The word $ijk=LLL$ involves a single diagonal entry of~(\ref{A}) with overall compensating phases. Its contribution, $T^3$, cancels an opposite contribution from the similar term $RRR$. For $LRR$, we find
%
\begin{eqnonum}
\mathfrak{r}'\overline{\mathfrak{t}}\,\ep{\iu\mu_R(t_2-t_1)}\cdot(-T)\ep{\iu\mu_R(t_3-t_2)} \cdot\overline{\mathfrak{r}}'\mathfrak{t}\,\ep{\iu\mu_L(t_1-t_3)} = -T^2(1-T)\ep{-\iu V(t_3-t_1)}\,,
\end{eqnonum}
%
which can be combined with $T^2(1-T)\ep{\iu V(t_3-t_1)}$ from $RLL$. The
remaining words provide the cyclic permutations. To conclude, it suffices to
note that the second trace is obtained by exchanging $2\leftrightarrow 3$,
which induces a change of sign in all the exponents.
We then note that~(\ref{IIIter}) is locally integrable and we drop the regularizations. Eq.~(\ref{noContact}) then follows from
%
\begin{eqwithnum}
\label{AsInt}
\int_0^{t} d^3t\,\frac{\sin V(t_1-t_2)+\sin V(t_2-t_3)+\sin V(t_3-t_1)}{(t_1-t_2)(t_2-t_3)(t_1-t_3)} = 2\pi^2 Vt + o(t)\,,\quad(t\to\infty)\,.
\end{eqwithnum}
%
Actually, the time ordering should have been reinstated, but since the
expression is permutation symmetric that is superfluous. Eq.~(\ref{AsInt}) can
be derived by reinstating the regularizations $-\iu 0$; then the integral can
be broken into three terms, the first two of which vanish in the limit of
large $t$. In fact, the first term has poles at $t_3 = t_2-\iu 0$, $t_1-\iu 0$
which do not pinch the real axis. The same applies to the second and to the $t_1$-integration. Using $\int dt_2 (t_1-t_2-\iu 0)^{-1}\cdot(t_2-t_3-\iu 0)^{-1} = 2\pi\iu(t_1-t_3-\iu 0)^{-1}$, one is left with
%
\begin{eqwithnum}
2\pi\iu\int_0^t dt_1dt_3\,\frac{\sin V(t_3-t_1)}{(t_1-t_3-\iu0)^2}
=t\cdot2\pi\iu\int dx \,\frac{\sin Vx}{(x+\iu0)^2}+ o(t) = 2\pi^2 V t + o(t)\,;
\label{AsIntBis}
\end{eqwithnum}
%
in fact, the odd part of $(x+\iu 0)^{-2}$ is
$\bigl((x+\iu 0)^{-2}-(x-\iu 0)^{-2}\bigr)/2
=\pi\iu\delta'(x)$.
We next consider the middle integral in~(\ref{3Cumul}). We find
%
\begin{align*}
\cumul{\widehat{I}_1[\widehat{Q}_2, \widehat{I}_2]}_\rho = \tr\left(I_1\rho'[Q_2, I_2]\rho\right) &= \frac{1}{(2\pi\iu)^2}\frac{1}{(t_1-t_2-\iu 0)^2}\cdot\tr\big(A(t_2-t_1)[Q, S\str Q S] D(t_1-t_2)\big) \nonumber\\
&=\frac{1}{(2\pi\iu)^2}\frac{1}{(t_1-t_2-\iu 0)^2}\cdot\sin V(t_1-t_2)\,,
\end{align*}
%
and the same result for $\cumul{[\widehat{Q}_2, \widehat{I}_2]\widehat{I}_1}_\rho$ except for $+\iu 0$ instead of $-\iu 0$. Thus, $\cumul{\T\widehat{I}_1[\widehat{Q}_2, \widehat{I}_2]}_\rho$ is odd in $t_1\leftrightarrow t_2$, and the integral vanishes.
%%
%\begin{align*}
%\int_0^{t}d^2t\, \cumul{ \T \widehat{I}_1[\widehat{Q}_2, \widehat{I}_2]} &= \frac{2\iu}{(2\pi\iu)^2}T(1-T)\int_0^{t}d^2t\,\frac{\sin V(t_1-t_2)}{t_1-t_2}\left\{\frac{\theta(t_1 - t_2)}{t_1-t_2-\iu 0} + \frac{\theta(t_2 - t_1)}{t_1-t_2+\iu 0}\right\} \\
%&= \frac{2\iu}{(2\pi\iu)^2}T(1-T)\int_0^{t}d^2t\,\frac{\sin V(t_1-t_2)}{t_1-t_2}\mathrm{P}\left(\frac{1}{t_1-t_2}\right) = 0\,.
%\end{align*}
%%
%where we used that $(x\pm\iu 0)^{-1} = \mathrm{P}(x) \mp \iu\pi\delta(x)$ and $\mathrm{P}(\cdot)$ denotes the Cauchy principal value distribution.
Finally the last integrand in~(\ref{3Cumul}), being the expectation value of a commutator, reduces to its Schwinger term
%
\begin{eqnonum}
\cumul{[\widehat{Q}_1, [\widehat{Q}_1, \widehat{I}_1]]}_\rho = \tr(\rho Q_1 \rho' [Q_1, I_1] \rho) - \tr(\rho' Q_1 \rho[Q_1, I_1]\rho')\,.
\end{eqnonum}
%
Using (\ref{commIQ}, \ref{Qt}) it equals
%
\begin{multline*}
\frac{1}{(2\pi\iu)^2}\int_{-t_1}^0dy\,\left(\frac{1}{(y+t_1+\iu 0)^2}-\frac{1}{(y+t_1-\iu 0)^2}\right)\tr(S\str Q S D(t_1+y) [Q, S\str Q S] D(-t_1-y)) \\
= \frac{-2\iu}{(2\pi\iu)^2}T(1-T)\int_{-t_1}^0dy\,\sin V(y+t_1)\left(\frac{1}{(y+t_1+\iu 0)^2}-\frac{1}{(y+t_1-\iu 0)^2}\right)\,,
\end{multline*}
%
where we noted that additional contributions from $y<-t_1$ and $y>0$ vanish because of $\tr(Q D(t_1+y) [Q, S\str Q S] D(-t_1-y))=0$. After changing the integration variable to $x=y+t_1$, both factors are odd, whence the integral equals
%
\begin{eqnonum}
\frac{1}{2}\int_{-t_1}^{t_1} dx\,\sin Vx
\left(\frac{1}{(x+\iu 0)^2}-\frac{1}{(x-\iu 0)^2}\right)
= -\pi\iu V+o(1)\,,
\end{eqnonum}
%
as in (\ref{AsIntBis}). We conclude that
%
\begin{eqwithnum}
\int_0^{t} dt_1\,\cumul{[\widehat{Q}_1, [\widehat{Q}_1, \widehat{I}_1]]}_\rho = T(1-T)\frac{Vt}{2\pi}\,,
\end{eqwithnum}
%
as claimed.
%%%%%%%%%%%
\subsection{A strictly causal scattering process}
We may trade the instantaneous scattering process in use in the previous
section with a strictly causal one, all while remaining within the model of
Section~\ref{sec3}. That is achieved by simply forfeiting a piece of length
$l>0$
of the leads in favor of the scatterer. So reinterpreted, the scattering
process lasts $2l$ and the contact terms should disappear. Nevertheless the
same result for the third cumulant will be obtained, though only the first
terms on the r.h.s. of~(\ref{3Cumul}) will contribute.
In physical terms we place the detector a distance $l>0$ away from the
scatterer; in mathematical terms we replace $Q$ by its regularization $Q_l = Q
\theta(|x|\geq l)$ and hence $I$ by $I_l = \iu[H, Q_l] = Q[\delta(x-l) -
\delta(x+l)]$. It ought to be noted that $[Q_l, \rho]\neq 0$, so that the
physical appropriateness of~(\ref{LLchi}), and hence of~(\ref{momentsbis}), could be questioned. Nevertheless, we may just view $l>0$ as a regulator before taking the limit $l\to 0$. However, even this is troublesome, at least in this form, because $[Q_l, \rho]\notin \ideal{2}$ makes $\widehat{Q}_l$ undefined. That can be remedied by smoothing the step function $\theta(|x|\geq l)$ on a length $