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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

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    $\begingroup$ 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 $l_1=l_2=0$. $\endgroup$ Commented Oct 27, 2022 at 15:01

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

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  • $\begingroup$ $a,b$ are not finite though... $\endgroup$
    – Henry Lee
    Commented Oct 28, 2022 at 5:12

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