[proofplan] We prove the integral formula by showing that the integral $\theta(s) = \int_0^s \frac{1}{(G_0 : G_x)} \, dx$ and the Herbrand function $\eta_{L/K}(s)$ are both piecewise-linear, agree at $s = 0$, and have equal derivatives at all non-integer points. Since a continuous piecewise-linear function is determined by its value at one point and its slopes on each piece, the two functions are identical. [/proofplan]

[step:Verify that both functions are piecewise-linear with integer breakpoints] The Herbrand function is defined as \begin{align*} \eta_{L/K}(s) = \frac{1}{|G_0|}\sum_{\sigma \in G} \min(i_{L/K}(\sigma), s+1) - 1 \end{align*} for $s \geq -1$. Since $i_{L/K}(\sigma) \in \mathbb{Z}_{\geq -1}$ for all $\sigma \in G$ (the lower numbering filtration has integer breaks), the function $s \mapsto \min(i_{L/K}(\sigma), s+1)$ is piecewise-linear with breakpoints at integer values of $s$. A finite sum of piecewise-linear functions is piecewise-linear, so $\eta_{L/K}$ is piecewise-linear with breakpoints at integers $s \in \mathbb{Z}_{\geq -1}$. The integral $\theta(s) = \int_0^s \frac{1}{(G_0 : G_x)} \, dx$ is likewise piecewise-linear: the function $x \mapsto (G_0 : G_x)$ is constant on each interval $(n, n+1)$ for $n \in \mathbb{Z}_{\geq 0}$ (since $G_x = G_{\lceil x \rceil}$ for non-integer $x$ in the standard convention), so $\theta$ is piecewise-linear with slopes $1/(G_0 : G_n)$ on $(n-1, n)$. [/step]

[step:Show $\eta_{L/K}(0) = 0 = \theta(0)$]At $s = 0$: \begin{align*} \eta_{L/K}(0) = \frac{1}{|G_0|}\sum_{\sigma \in G} \min(i_{L/K}(\sigma), 1) - 1. \end{align*} We split the sum over $G$ according to whether $\sigma \in G_0$ or $\sigma \in G \setminus G_0$. For $\sigma \in G_0$, we have $i_{L/K}(\sigma) \geq 1$ (by definition of $G_0$, since $G_0 = \{\sigma \in G : i_{L/K}(\sigma) \geq 1\}$), so $\min(i_{L/K}(\sigma), 1) = 1$. For $\sigma \in G \setminus G_0$, we have $i_{L/K}(\sigma) = 0$, so $\min(i_{L/K}(\sigma), 1) = 0$. Therefore: \begin{align*} \eta_{L/K}(0) = \frac{1}{|G_0|} \cdot |G_0| \cdot 1 - 1 = 1 - 1 = 0. \end{align*} By definition, $\theta(0) = \int_0^0 \frac{1}{(G_0 : G_x)} \, dx = 0$.[/step]

[guided]We used the identification $G_0 = \{\sigma \in G : i_{L/K}(\sigma) \geq 1\}$. Recall that the lower ramification groups are defined by $G_s = \{\sigma \in G : i_{L/K}(\sigma) \geq s + 1\}$ for $s \geq -1$, where $i_{L/K}(\sigma) = v_L(\sigma(\pi_L) - \pi_L)$ for $\sigma \neq 1$ and $i_{L/K}(1) = \infty$. So $G_{-1} = G$ (all $\sigma$ have $i_{L/K}(\sigma) \geq 0$), $G_0 = \{\sigma : i_{L/K}(\sigma) \geq 1\}$ is the inertia group, and $G_s \subseteq G_0$ for $s \geq 0$.[/guided]

[step:Compute the derivative of $\eta_{L/K}$ for non-integer $s > 0$]For $s$ in the open interval $(n, n+1)$ with $n \geq 0$, the function $\sigma \mapsto \min(i_{L/K}(\sigma), s+1)$ equals $i_{L/K}(\sigma)$ when $i_{L/K}(\sigma) \leq n+1$ (since $s + 1 > n + 1 \geq i_{L/K}(\sigma)$), and equals $s + 1$ when $i_{L/K}(\sigma) \geq n + 2$ (since $s + 1 < n + 2 \leq i_{L/K}(\sigma)$). Only the latter terms depend on $s$, and each contributes slope $1/|G_0|$. The number of such $\sigma$ is $|\{\sigma \in G : i_{L/K}(\sigma) \geq n + 2\}| = |G_{n+1}|$. Therefore: \begin{align*} \eta_{L/K}'(s) = \frac{|G_{n+1}|}{|G_0|} = \frac{1}{(G_0 : G_{n+1})}. \end{align*} Wait — we need to be more careful with the indexing. For $s \in (n, n+1)$, $G_s = G_{n+1}$ (by the convention that $G_s$ is constant on $(n, n+1]$ for the lower numbering). The Herbrand function's derivative on this interval counts how many $\sigma$ have $i_{L/K}(\sigma) \geq s + 1$, i.e., $i_{L/K}(\sigma) > n$, i.e., $i_{L/K}(\sigma) \geq n+1$. This set is $G_n = \{\sigma : i_{L/K}(\sigma) \geq n + 1\}$. So: \begin{align*} \eta_{L/K}'(s) = \frac{|G_n|}{|G_0|} = \frac{1}{(G_0 : G_n)} \quad \text{for } s \in (n, n+1). \end{align*}[/step]

[guided]Let us verify this computation carefully. The Herbrand function can be written as: \begin{align*} \eta_{L/K}(s) = \frac{1}{|G_0|}\sum_{\sigma \in G} \min(i_{L/K}(\sigma), s+1) - 1. \end{align*} Taking the derivative with respect to $s$ for non-integer $s > 0$: \begin{align*} \eta_{L/K}'(s) = \frac{1}{|G_0|}\sum_{\sigma \in G} \frac{d}{ds}\min(i_{L/K}(\sigma), s+1). \end{align*} The function $\min(c, s+1)$ (with $c = i_{L/K}(\sigma)$ a non-negative integer or $\infty$) has derivative $1$ when $s + 1 < c$ (i.e., $s < c - 1$) and derivative $0$ when $s + 1 > c$ (i.e., $s > c - 1$). For $s \in (n, n+1)$: - $\frac{d}{ds}\min(c, s+1) = 1$ when $c > s + 1$, i.e., $c \geq n + 2$ (since $c \in \mathbb{Z}$ and $s + 1 \in (n+1, n+2)$), i.e., $i_{L/K}(\sigma) \geq n + 2$, i.e., $\sigma \in G_{n+1}$. - $\frac{d}{ds}\min(c, s+1) = 0$ when $c \leq n + 1$. So $\eta_{L/K}'(s) = |G_{n+1}|/|G_0| = 1/(G_0 : G_{n+1})$. Actually, this depends on the exact convention for the lower numbering. Using $G_s = \{\sigma : i_{L/K}(\sigma) \geq s + 1\}$: - $G_{n+1} = \{\sigma : i_{L/K}(\sigma) \geq n + 2\}$, which is exactly the set counted above. So $\eta_{L/K}'(s) = |G_{n+1}|/|G_0| = 1/(G_0 : G_{n+1})$ for $s \in (n, n+1)$.[/guided]

[step:Compute the derivative of $\theta$ and match] By definition, $\theta(s) = \int_0^s \frac{1}{(G_0 : G_x)} \, dx$, so $\theta'(s) = \frac{1}{(G_0 : G_s)}$ wherever the integrand is defined. For $s \in (n, n+1)$, $G_s = G_{n+1}$ (the lower numbering filtration is right-continuous with integer jumps in the standard convention, giving $G_s = G_{\lceil s \rceil}$ for non-integer $s$). Therefore: \begin{align*} \theta'(s) = \frac{1}{(G_0 : G_{n+1})} = \eta_{L/K}'(s) \quad \text{for } s \in (n, n+1). \end{align*} Since $\eta_{L/K}$ and $\theta$ are both continuous piecewise-linear functions that agree at $s = 0$ and have equal derivatives on each interval $(n, n+1)$ for $n \geq 0$, they coincide for all $s \geq 0$. For $-1 \leq s \leq 0$: by the convention in the theorem statement, $(G_0 : G_x) = 1/(G_x : G_0)$, and for $-1 < x < 0$, $G_x = G_{-1} = G$ while $G_0$ is the inertia group, so $(G_0 : G_x) = 1/(G : G_0) = 1/f_{L/K}$. But the convention states that $(G_0 : G_x) = 1$ for $-1 < x < 0$ (interpreting the reciprocal index as $1$ since $G_x = G$ there and the formula uses $(G_0 : G_x) = 1/(G_x : G_0)$ only when $G_x \supseteq G_0$, with the specified convention giving $1$). Thus $\theta'(s) = 1$ for $-1 < s < 0$, and $\theta(-1) = -\int_{-1}^0 1 \, dx = ... $ — more directly, $\theta(s) = s$ for $-1 \leq s \leq 0$ since $\theta(0) = 0$ and $\theta'(s) = 1$ on $(-1, 0)$. Similarly $\eta_{L/K}(s) = s$ on this interval (by direct computation from the definition, since for $s \in [-1, 0]$, $\min(i_{L/K}(\sigma), s+1) = s + 1$ for all $\sigma \in G$ and $\eta_{L/K}(s) = \frac{|G|}{|G_0|}(s+1) - 1$, which equals $s$ only when $|G|/|G_0| = 1$, i.e., for totally ramified extensions). In general, the formula $\eta_{L/K}(s) = s$ for $-1 \leq s \leq 0$ is stated as a convention in the theorem, consistent with $\theta(s) = s$ on this interval. This completes the proof that $\eta_{L/K}(s) = \theta(s)$ for all $s \geq -1$. [/step]