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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.

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Why && \int_{0}^{\infty} x^nf(x)dx=0 && does not translated? – hyounghyoung Sep 5 '12 at 11:10
LaTeX is surrounded by $ signs, not & signs. – Matthew Pressland Sep 5 '12 at 11:12
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 5 '12 at 11:25
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 5 '12 at 11:54
What is the meaning of $\int_0^\infty x^n f(x)\,dx$: is it the Lebesgue integral or improper Riemann integral? – user31373 Sep 7 '12 at 21:10

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