# Does there exist a function such that $\int_{\mathbb{R}_+^{\star} } t^nf(t)dt=0$?

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.

• @Etienne Is exactly that except here we need $f:\mathbb{R}^+ \rightarrow \mathbb{C}$. In 'my' example $f$ value is not in $\mathbb{C}$ – user117932 Mar 3 '14 at 10:16
• Usually, one would say that if $f$ is real-valued, it is in particular complex-valued. – Etienne Mar 3 '14 at 18:18
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 $n$.