2

Below equation is satisfied.

$$ \int_{0}^{\infty} x^nf(x)dx=0 $$

If $n$ is integer with $n\geq0$, then we can't guarantee $f(x) = 0$ for all positive $x$.

When $n$ is rational number with $n\geq0$ , do we get same result?

If so, what about $n$ is real number with $n\geq 0$?

Above integral is improper Riemann integral.

I forgot one condition. f(x) is continuous function.

Community
  • 1
hyounghyoung
  • 21
  • Why && \int_{0}^{\infty} x^nf(x)dx=0 && does not translated? – hyounghyoung Sep 05 '12 at 11:10
  • LaTeX is surrounded by $ signs, not & signs. – mdp Sep 05 '12 at 11:12
  • 7
    I'm a bit confused about the omitted quantifiers: I guess you're asking does $\int x^q f(x),dx = 0$ for all* $q \in \mathbb{Q}{\geq 0}$ or all $q \in \mathbb{R}{\geq 0}$ imply that $f(x) = 0$?* As an aside: the Stieltjes ghost function $f(x) = \exp(-x^{1/4})\sin(x^{1/4})$ satisfies $\int_{0}^\infty x^n f(x),dx = 0$ for all $n \in \mathbb{N}$. See Hans Lundmark's answer here for more on this. – t.b. Sep 05 '12 at 11:25
  • What's the relation with the Stone-Weierstrass theorem? – lhf Sep 05 '12 at 11:51
  • 1
    If interval is finite, it can be solved by Stone-Weierstrass theorem.. so I made title like that. I don't guarantee that it also solved by Stone-Weierstrass theorem. – hyounghyoung Sep 05 '12 at 11:54
  • 2
    What is the meaning of $\int_0^\infty x^n f(x),dx$: is it the Lebesgue integral or improper Riemann integral? –  Sep 07 '12 at 21:10

0 Answers0