# 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

• You cannot split the integral... A necessary condition for the convergence of both $\int_{-\infty}^{+\infty} f(t)dt$ and $\int_{-\infty}^{+\infty} f(t+1)dt$ would be that $l_1=l_2=0$. Commented Oct 27, 2022 at 15:01

Hint: For finite $$a,b$$, $$\int_a^b[f(t+1)-f(t)] \,\mathrm dt = \int_{a+1}^{b+1}f(t) \,\mathrm dt-\int_{a}^{b}f(t) \,\mathrm dt=\int_b^{b+1}f(t) \,\mathrm dt-\int_a^{a+1}f(t) \,\mathrm dt.$$
• $a,b$ are not finite though... Commented Oct 28, 2022 at 5:12