= Solution
The observed statistic is the \b[conditional mean among passages completed before $T=t_{\rm obs}$], not a mean in which unfinished trajectories are assigned duration $T$. Its normalization is
$$
Z_T=\int_0^Tp(t)\,dt=\sum_lA_l(1-e^{-\nu_lT}).
$$
For each contribution,
$$
\int_0^Tt\nu_le^{-\nu_lt}dt=\left[-te^{-\nu_lt}\right]_0^T+\frac{1-e^{-\nu_lT}}{\nu_l}=\frac{1-e^{-\nu_lT}(1+\nu_lT)}{\nu_l}.
$$
Dividing the weighted sum of these first moments by $Z_T$ gives
$$
\boxed{\mathcal T(T)=\frac{\sum_l(A_l/\nu_l)[1-e^{-\nu_lT}(1+\nu_lT)]}{\sum_lA_l(1-e^{-\nu_lT})}.}
$$
The denominator records the fraction of trajectories actually observed to finish.
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