0
$\begingroup$

Problem:

Let $f: [1,\infty) \to \mathbb{R}$ continuous and $\sup_{x \geq0} x^2|f(x)| \lt \infty$ holds.

Show, that the improper integral $\int_1^\infty f(x) dx$ exists and

$$\int_1^\infty f(x)dx = \int_0^1 \frac{1}{t^2}f(\frac{1}{t}) dt $$ holds.

I need to know, how to obtain the improper integral from the bounded function and out of the supremum. Is it manageable, to get $f$ into the form necessary to say that it's bounded, nothing is known about the function other than it's continuous. It would be nice if I get to see the proof of this.

$\endgroup$
1
  • 1
    $\begingroup$ Use the fact that $\int_1^{\infty} \frac 1 {x^{2}} dx=1<\infty$. $\endgroup$
    – Kavi Rama Murthy
    Jul 13, 2021 at 9:04

1 Answer 1

Reset to default
1
$\begingroup$

Using the fact the p-series converges for $p>1$, take this integral

$$\int_1^\infty\frac{1}{x^p}dx, p>1$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .