Does the existence of the integral $\int_0^\infty f(x)dx$ imply that f(x) is bounded on $[0,\infty)$ when f(x) is continuous in this same interval?

Question

Does the existence of the integral $\int_0^\infty f(x)dx$ imply that f(x) is bounded on $[0,\infty)$ when f(x) is continuous in this same interval ?

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I'm trying to use Cauchy without knowing what the next step is, stating that $\int_a^\infty f(x)dx$ exists if and only if $\lim_{t\to \infty} \int_a^xf(x)dx $ exists and approachs a particular value, which means that for all $\epsilon > 0$ there is $t_0$ such that for all $x,y > t_0$ : $$|\int_a^y f(x)dx - \int_a^x f(x)dx | < \epsilon$$ and since $$\int_a^y f(x)dx - \int_a^x f(x)dx = \int_x^y f(x)dx$$ we get that $$|\int_x^y f(x)dx|<\epsilon$$

And from here I don't really know what conclusion can be drawn. Does that mean that f(x) is indeed bounded ? It seems to me that I'm missing some counter-example to disprove the statement.

Regards

Answer 1

Does the existence of the integral $\int_0^\infty f(x)dx$ imply that f(x) is bounded on $[0,\infty)$ when f(x) is continuous in this same interval ?

No. For each interval $[n,n+1]$ define $f_n:[n,n+1]\to\mathbb{R}$ by piecewise linear extension of the following four points:

$$f_n(n)=0$$ $$f_n(n+\frac{1}{2n^3})=2n$$ $$f_n(n+\frac{1}{n^3})=0$$ $$f_n(n+1)=0$$

So it's a triangle with basis of length $1/n^3$ and height $2n$. And thus

$$\int_n^{n+1}f_n(x)dx=\frac{1}{n^2}$$

And for $n=0$ we put $f_0$ to be the constant $0$ and so $\int_0^1 f_0(x)dx=0$.

Now we glue all $f_n$:

$$f:[0,\infty)\to\mathbb{R}$$ $$f(x)=f_{\lfloor x\rfloor}(x)$$

$f$ is continuous, it is not bounded but

$$\int_0^{\infty} f(x)dx=\sum_{n=0}^\infty\int_n^{n+1}f_n(x)dx=\sum_{n=1}^\infty \frac{1}{n^2}=2$$

Answer 2

The function $\sqrt x \sin(x^2)$ is unbounded (check the points $\sqrt {\pi/2+2n\pi}).$ Letting $x=\sqrt y,$ we get

$$\int_0^\infty \sqrt x \sin(x^2)\,dx = \int_0^\infty \frac{\sin(y)}{2y^{1/4}}\,dx .$$

The integral on the right converges, conditionally, by Dirichlet's test.

Here's one that converges absolutely: $\int_0^\infty x[\cos^2x]^{x^5}\ dx.$ I discussed the proof of this here Lebesgue integrable and continuous function with divergente sequence