Define
$$
\overline\alpha(f)=\limsup_{n-m\to\infty} \frac{1}{n-m}(f(m)+\ldots+f(n-1))
$$
and
$$
\underline\alpha(f)=\liminf_{n-m\to\infty} \frac{1}{n-m}(f(m)+\ldots+f(n-1)).
$$

Claim $\max\int f\,d\mu=\overline\alpha(f)$ where the maximum is taken over translation-invariant finitely additive probability measures and $\min\int f\,d\mu=\underline\alpha(f)$.

Clearly it's sufficient to prove it for $\overline\alpha(f)$. Let $M=\overline\alpha(f)$ and let $\epsilon > 0$. By definition of $\limsup$, there exists an $N$ such that for every $n$, $(1/N)(f(n)+\ldots+f(n+N-1)) < M+\epsilon$. Hence we see
$$
\int (1/N)(f(n)+\ldots+f(n+N-1))\,d\mu(n) < M+\epsilon.
$$
By translation-invariance, this gives $\int f(n)\,d\mu(n)\le M$.

Conversely, let $n_i-m_i\to\infty$ be such that $(f(m_i)+\ldots+f(n_i-1))/(n_i-m_i)\to\overline\alpha(f)$. Define a family of linear functionals $A_i\colon l^\infty\to\mathbb R$ by $A_i(g)=(g(m_i)+\ldots+g(n_i-1))/(n_i-m_i)$. Notice that
$ \vert A_i(g) \vert \le \parallel g \parallel_{\infty}$.
Let $Y=\lbrace g \in l^{\infty} \colon \lim_{i\to\infty} A_i(g) \text{ exists} \rbrace$.

This is clearly a subspace. Define a linear operator $L$ on $Y$ by $L(g)=\lim_{i\to\infty} A_i(g)$. By the above, this has norm at most 1. Applying $L$ to a constant function, we see that $L$ has norm 1. By Hahn-Banach, notice that $L$ can be extended to a norm 1 operator on all of $l^{\infty}$. We claim that $L$ is positive in the sense that $L(h)\ge 0$ when $h\in l^\infty$ is a non-negative sequence. Supposing that $L(h) < 0$ for some non-negative $l^{\infty}$ sequence $h$ (which we can assume to be of norm 1 without loss of generality), then the sequence $k(n)=1-h(n)$ is another $l^\infty$ sequence and $\parallel k \parallel_{\infty} \le 1$. However we have that $L(k)=L(1)-L(h) > 1$. This contradicts the fact that $\parallel L\parallel_{\infty}=1$.

We need to show that $L$ is invariant, that is $L(S(h))=L(h)$ for all $h$ in $l^\infty$ where $S$ is the right shift operator. To see this, notice that $A_i((S(h)-h))\to 0$ for all $h\in l^\infty$ so that $S(h)-h\in Y$ and $L(S(h))=L(h)$.

We define a measure $\mu$ by $\mu(A)=L(1_A)$. This immediately defines a finitely-additive probability measure. By the invariance of $L$ it's shift-invariant. By definition of $L$, we have $\overline\alpha(f)=L(f)=\int f\,d\mu$.