Proving that $\lim_{x \to\infty}\int_{x}^{x+1} f(t) dt=0$ if $\int_{1}^{\infty}f(x)<\infty$

Question

I'm stuck with this problem:

Prove that $\lim_{x \to\infty}\int_{x}^{x+1} f(t) dt=0$ if $\int_{1}^{\infty}f(x)<\infty.$

I've tried applying the Cauchy's condition, but I think that I can't do it, because I can't choose a proper $\delta>0$.

After that, I thought that it could be true that $\lim_{x \to \infty} f(x)=0$, so it would be easy, but I think that it's not true.

Any hint will be really appreciated. Thanks.

Answer 1

$\int_x^{x+1}f(t)dt=\int_1^{x+1}f(t)dt-\int_1^xf(t)dt$. This implies that $lim_{x\rightarrow +\infty}\int_1^{x+1}f(t)dt=lim_{x\rightarrow +\infty}\int_1^{x}f(t)dt=\int_1^{+\infty}f(t)dt$

Answer 2

This does follow from Cauchy's criterion. By Cauchy's criterion, for all $\varepsilon > 0$ there exists $M \geq 1$ such that for all $A,B \geq M$ we have

$$ \left| \int_A^B f(x) \, dx \right| < \varepsilon. $$

If $x \geq M$ then by taking $A = x, B = x + 1$ we see that

$$ \left| \int_{x}^{x+1} f(x) \, dx \right| < \varepsilon $$

for all $x \geq M$.

Answer 3

$$a_n(s)=\int_{n+s}^{n+s+1}f(t)dt\quad(0\le s<1)$$ $$\int_{1}^{\infty}f(t)dt=-\int_{s}^{1}f(t)dt+\sum_{n=0}^{\infty}a_n(s)<\infty\quad\text{, for all }s$$ $$\therefore \lim_{n\to\infty}a_n(s)=0$$