Difference integral of continuous function Let $f:\mathbb R\rightarrow \mathbb R$ be a continuous function such that
$$\lim_{t\rightarrow -\infty}f(t)=l_1,\hspace{0.4cm} \lim_{t\rightarrow +\infty}f(t)=l_2$$
Evaluate
$$\int_{-\infty}^{+\infty}\left[f(t+1)-f(t)\right]dt$$
I thought of variable change : We have
$$\int_{-\infty}^{+\infty}f(t+1)dt-\int_{-\infty}^{+\infty}f(t)dt$$
By substituting $u=t+1$ in the first integral, we obtain
$$\int_{-\infty}^{+\infty}f(u)du-\int_{-\infty}^{+\infty}f(t)dt$$
My mind says it's the same integral so it must be equal to $0$.
But we still have no idea about the convergence of the integrals.
I tried also to parametrize the integral : Let
$$I(\alpha)=\int_{-\infty}^{+\infty}t^\alpha f(t)dt$$
Which in the end will lead to $I(0)-I(0)=0$
But here I still hesitate whether I defined $I$ as a diverging function

Hint: For finite $a,b$,
$$\int_a^b[f(t+1)-f(t)] \,\mathrm dt = \int_{a+1}^{b+1}f(t) \,\mathrm dt-\int_{a}^{b}f(t) \,\mathrm dt=\int_b^{b+1}f(t) \,\mathrm dt-\int_a^{a+1}f(t) \,\mathrm dt.$$
