TITLE: Does there exist a function such that $\int_{\mathbb{R}_+^{\star} } t^nf(t)dt=0$?
QUESTION [13 upvotes]: Let $f\in C([a,b],\mathbb{R})$ such that $\displaystyle\int_{a}^{b} t^nf(t)dt=0$ for all integer n.
We know that $f\equiv 0$. It's "easy" to prove with Weierstrass theorem or with How to prove that $\,\,f\equiv 0,$ without using Weierstrass theorem?

This theorem is wrong on $\mathbb{R^+}$, we can choose :
$$f(x)=\exp(-x^{\frac{1}{4}})\sin(x^\frac{1}{4})$$

Let 
$$
I_n=\displaystyle\int_{0}^{+\infty}t^n\exp(-\omega t)dt=\frac{n!}{\omega^{n+1}},\quad n\in \mathbb{N}, \quad \omega=exp(\frac{i\pi}{4})
$$
Proof.
$$
|t^n\exp(-\omega t)|=t^n\exp \bigr(\frac{-t \sqrt{2}}{2}\bigl)\in L^1(\mathbb{R})
$$
by integration by parts we get 
$$
I_n=\frac{n}{\omega}I_{n-1}
$$
Thus,
$$
I_n=\frac{n!}{\omega^{n+1}}
$$
Plus for $n\geq 1$, $\quad \omega^{4(n+1)}=-(1)^{n+1}$
Then, 
$$
I_{4n+3}\in \mathbb{R}
$$
Therefore, 
$$
0=\Im(I_{4n+3})=\displaystyle\int_{0}^{+\infty}x^nf(x)dx
$$


Let $f:\mathbb{R}_+ \longrightarrow \mathbb{C}$, 

I would like to prove the existence of a function $f$ such that $\int_{0}^{+\infty}t^n f(t)dt=0$,

In fact this example it's not mine (I have already read it in a book) and the question is to find $f:\mathbb{R}_+ \longrightarrow \mathbb{C}$. So I would like to know if we can proove the existence more generally or just how can I construct a such function.

Thank you in advance,

REPLY [2 votes]: Start with your example, take the imaginary part, just user every fourth step with variable $t=s^{1/4}$.  We get:
$$
F(s) = \exp(-s^{1/4})\sin(s^{1/4}),
$$
for which
$$
\int_0^\infty s^n F(s)\;ds = 0
$$
for all nonnegative integers $n$.