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I'm trying to find the Noether charge for the symmetry

$x\rightarrow x+f\left(x\right)$

This transformation should leave the action invariant, so

\begin{align*} dS&=S\left(x+f\left(x\right),\dots\right)-S\left(x\right)=0\\ &=\int dt\ \mathcal{L}\left(x+f\left(x\right),\dot{x}+\dot{f},\dots,t\right)-\mathcal{L}\left(x,\dot{x},\dots,t\right) \end{align*}

Using $f\left(x+\epsilon\right)-f\left(x\right)\approx \epsilon \frac{df}{dx}$

\begin{align*} dS=\int dt\ \frac{\partial \mathcal{L}}{\partial x}f+\frac{\partial \mathcal{L}}{\partial \dot{x}}\dot{f}+\frac{\partial \mathcal{L}}{\partial \ddot{x}}\ddot{f}+\dots \end{align*} Writing the second term as a total derivative \begin{align*} dS&=\int dt\ \frac{\partial \mathcal{L}}{\partial x}f+\frac{d}{dt}\left[\frac{\partial L}{\partial \dot{x}} f\right]-f\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial\dot{x}}\right)+\frac{\partial\mathcal{L}}{\partial \ddot{x}}\ddot{f}+\dots\\ &=\frac{\partial\mathcal{L}}{\partial \dot{x}}f\bigg|_0^T+\int dt\ \frac{\partial \mathcal{L}}{\partial x}f-f\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial\dot{x}}\right)+\frac{\partial\mathcal{L}}{\partial \ddot{x}}\ddot{f}+\dots \end{align*} For the higher order terms we can do the same \begin{align*} \frac{\partial \mathcal{L}}{\partial \ddot{x}}\ddot{f}&=\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \ddot{x}}\dot{f}\right)-\dot{f}\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \ddot{x}}\right)\\ &=\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \ddot{x}}\dot{f}\right)-\boxed{\frac{d}{dt}\left(f\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial\ddot{x}}\right)\right)}+f\frac{d^2}{dt^2}\left(\frac{\partial\mathcal{L}}{d\ddot{x}}\right) \end{align*} So now the integral becomes \begin{align*} dS=\sum_{n=0}^N\frac{\partial\mathcal{L}}{\partial x^{\left(n+1\right)}}f^{(n)}\bigg|_0^T+\int dt\ f\left[\sum_{n=0}^N\left(-1\right)^n\frac{d^n}{dt^n}\left(\frac{\partial \mathcal{L}}{\partial x^{(n)}}\right)\right]-\boxed{\frac{d}{dt}\left(f\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial\ddot{x}}\right)\right)+\dots}=0\end{align*}

Where the sum under the integral represents the Euler-Lagrange equations for the unperturbed action. I am kind of expecting the boxed terms to vanish as well, leaving only the first term. Did I do it right, what steps am I missing?

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In this answer, we will just list the result without a proof. For a higher-order action

$$ S[q]~=~\int\! dt~ L(q(t), \dot{q}(t),\ddot{q}(t),\dddot{q}(t),\ldots,t) \tag{1} $$

with a vertical infinitesimal quasi-symmetry

$$ \delta q^i~=~\varepsilon Y^i(q, \dot{q},\ddot{q},\dddot{q},\ldots,t) , \tag{2} $$

the bare Noether charge is

$$Q~=~ \sum_{k\geq 1} \left(\frac{d}{dt} \right)^{k-1}\left(Y^i \sum_{m\geq k} \begin{pmatrix} m \cr k \end{pmatrix} \left(-\frac{d}{dt} \right)^{m-k}\frac{\partial L}{\partial q^{i(m)}} \right).\tag{3}$$

To unpack formula (3) for a second-order Lagrangian, see e.g. my Phys.SE answer here.

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