$\int_{-\infty}^{\infty}f(x+1)-f(x)\text{d}x$ 
Let $f$ be continuous on $\mathbb R$, $\lim_{x\rightarrow \infty} f(x)=A,\ \lim_{x\rightarrow -\infty} f(x)=B$ . Calculate the integral $$\int_{-\infty}^{\infty}f(x+1)-f(x)\,\text{d}x$$

My intuition says $\frac{A+B}{2}$ (it might be wrong) but I couldn't get close to proving it. I thought of using the mean value theorem, $f(x+1)-f(x)=f'(\xi_x)$ or perhaps defining a primitive function of $f$, both of which ended up not working out.

For any $N\in\mathbb{R}^+$ we have
$$ \int_{-N}^{N}f(x+1)-f(x)\,dx = \int_{-N+1}^{N+1}f(x)\,dx - \int_{-N}^{N}f(x) = \int_{N}^{N+1}f(x)\,dx - \int_{-N}^{-N+1}f(x)\,dx $$
and $\lim_{N\to +\infty}\int_{N}^{N+1}f(x)\,dx  = A$ since $f$ is a continuous function such that $\lim_{x\to +\infty}f(x)=A$.
Similarly, $\lim_{N\to +\infty}\int_{-N}^{-N+1}f(x)\,dx = B$. It follows that $f(x+1)-f(x)$ is improperly Riemann-integrable over $\mathbb{R}$ (in the symmetric sense) and $\int_\mathbb{R} f(x+1)-f(x)\,dx = A-B$ as conjectured.
