$\int_{-\infty}^{\infty}f(x+1)-f(x)\text{d}x$ 
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.

You can write the integral as: $$\sum_{n\in\mathbb Z}\left[\int_n^{n+1}f(x+1)dx-\int_n^{n+1}f(x)dx\right]=\sum_{n\in\mathbb Z} [s_{n+1}-s_n]$$ where $$s_n=\int_n^{n+1}f(x)dx$$
Then some telescoping and finding $$\lim_{n\to\infty}(s_n-s_{-n})=A-B$$
