Need help with an exercise (Improper integral).

Question

The exercise:

if $ f(x)$ is continuous in $[a,\infty]$ and $\lim_{x\to\infty} \int_{x}^{2x} f(t)dt = 0$ then $\int_{a}^\infty f(t)dt$ converges.

I belive the statement is true and can be proven using Cauchy's test for integral convergence.

Thanks.

Answer 1

It doesn't seem to be true, since

$$\lim_{x\to+\infty}\int_x^{2x}f(t)dt$$

$$ \implies $$

$$\forall \epsilon>0\;\exists A>0\;:$$

$$\forall x\geq A\;\; |\int_x^{2x}f(t)dt|<\epsilon$$

But not

$$\forall \epsilon>0\; \exists A>0\;:$$

$$\forall y>x\geq A\;\;\; | \int_x^yf(t)dt|<\epsilon$$

Take for example $y=2^Nx$ or $e^Nx$.

Answer 2

The statement is incorrect.

The function $f(x) = \frac{1}{x \ln(x)}$ is continuous on $[2, \infty]$ (including $\infty$ with $f(\infty) = 0$). The antiderivative of $f$ is $\ln(\ln(x)) + C$, so we have $$\int_x^{2x} f(t) \, dt = \ln(\ln(2x)) - \ln(\ln(x)) = \ln \left(1 + \frac{\ln(2)}{\ln(x)}\right) \xrightarrow{x \to \infty} 0.$$

Yet $$\int_2^\infty f(t) \, dt = \lim \limits_{x \to \infty} \ln(\ln(x)) - \ln(\ln(2)) = \infty.$$