Disappearing term in proof of Noether's theorem [I know this is a topic in mathematical physics, but I feel like my question is more on the mathematics than the physics. At any rate, feel free to migrate it to Physics.SE if necessary.]
In my notes, one of the implications of Noether's theorem (in the Lagrangian formalism) is stated briefly as such:

Theorem. If the Euler-Lagrange equations hold, and an infinitesimal transformation is invariant, then there exists a conserved quantity.

An infinitesimal transformation is one that takes the form $Q = q + \delta q$ and, for simplicity, we assume that it does not modify the time coordinate. An invariant transformation is one such that the Euler-Lagrange equations take the same functional form, i.e.: if the Lagrangian function is $L_q(q,\dot q,t)$ in the old coordinates and $L_Q(Q,\dot Q ,t)$ in the new, then the latter also lies in the kernel of the Euler-Lagrange operator $$\frac{\mathfrak D}{\mathfrak D q^k} = \frac{\partial}{\partial q^k} - \frac d {dt} \frac{\partial}{\partial \dot q^k}$$ expressed in the old coordinates. We have seen a lemma that states that such a transformation must be such that the new and old Lagrangians are connected by the formula
$$L_Q(Q,\dot Q,t) = L_q(Q,\dot Q,t) + \frac{d\Lambda}{dt}(Q,t) \tag1 $$
for some arbitrary regular function $\Lambda$ (actually, we have only proven that if it satisfies $(1)$, the transformation is invariant, but I trust that the other implication holds).
The demonstration goes like this: consider the infinitesimal transformation $Q = q + \delta q$ and suppose that it is invariant, so that it satisfies $(1)$. The Lagrangian is uniquely defined (its value is not dependent on the chosen chart), so $L_q(q,\dot q,t) = L_Q(Q,\dot Q,t)$ and we may revert to the old Lagrangian on the RHS of $(1)$. Substituting the formula for $Q$, we get
$$L_q(q,\dot q,t) = L_q(q+\delta q, \dot q + \delta \dot q,t) + \frac{d\Lambda}{dt}(q+\delta q,t) $$
We assume that the Lagrangian and the $\Lambda$ are sufficiently regular, so that we may expand both to first order and obtain
$$\frac{\partial L_q}{\partial q^k}\delta q^k + \frac{\partial L_q}{\partial \dot q^k}\delta \dot q^k + \frac d {dt} \left(\Lambda(q,t) + \frac{\partial \Lambda}{\partial q^k}\delta q^k \right) = 0 $$
and, by a reverse Leibniz rule,
$$\frac{\partial L_q}{\partial q^k}\delta q^k + \frac{d}{dt}\left(\frac{\partial L_q}{\partial \dot q^k}\delta q^k\right) - \frac d{dt}\left( \frac{\partial L_q}{\partial \dot q^k}\right) \delta q^k + \frac d {dt} \left(\Lambda(q,t) + \frac{\partial \Lambda}{\partial q^k}\delta q^k \right) = 0. $$
Since by hypothesis the E-L equations hold in the old variables $\forall k$, the first and third terms disappear and we are left with the total time derivative
$$\frac {d}{dt} \left(\frac{\partial L_q}{\partial \dot q^k}\delta q^k + \Lambda + \frac{\partial \Lambda}{\partial q^k}\delta q^k \right) = 0$$
which implies that the quantity in parenthesis is conserved. However, in my notes, somewhere along the way the $\Lambda$ term disappears, without justification! And indeed the correct conserved quantity is
$$\frac{\partial L_q}{\partial \dot q^k}\delta q^k + \delta \Lambda = \text{const}. $$
Why does this happen? Is $\Lambda (q,t)$ itself zero for some reason? Is its total time derivative zero? And how can either option be justified thoroughly?
 A: I don't think the last term $\frac{d\Lambda}{dt}(Q,t)$ of (1) is correct.  It needs to vanish when $\delta q=0$, but in this circumstance, it gives $\frac{d\Lambda}{dt}(q,t)$.
Let us consider an example, and denote $L[q]=L(q, \dot q,t)$.  Let $L=\dot q^2/2$, and $Q=q+vt$ (Galilean boost).  Then $L[Q]=\dot q^2+2v\dot q+v^2=L[q]+\frac{d}{dt}(2vq+v^2t)$.  Up to a constant, we have $\Lambda=2vq+v^2t$ (I will add a minus sign to $\Lambda$ here).  Since $\delta q=vt$, we can identify $v$ as the small parameter.  Observe then that $\Lambda=0$ when $v=0$.
I think a more correct way to write (1) is like this:
$$
L[Q]=L[q]+\frac{d}{dt}\left(\delta\Lambda[q]+O(\delta^2)\right).
$$
In the above example, we would have $\delta\Lambda[q]+O(\delta^2)=2vq+v^2t$, or $\delta\Lambda[q]=2vq$.  Expanding this to first order then yields:
$$
\frac{\partial L}{\partial q}\delta q+\frac{\partial L}{\partial \dot q}\delta\dot q=\frac{d}{dt}\delta\Lambda[q].
$$
And from here we can finish Noether:
$$
\delta q\left(\frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L}{\partial \dot q}\right)=\frac{d}{dt}\left(-\delta q\frac{\partial L}{\partial \dot q}+\delta\Lambda[q]\right).
$$
