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Probably this is a trivial question, but I am unable to find an answer: is there a function $v(x)$ such that $$ \int_{0}^\infty x^n e^{v(x)} dx =\frac{1}{n!} $$ for all positiv integer n?

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Such a function cannot be continuous, or even locally bounded. Otherwise we have some interval $(a,b)$ with $a>1$ on which $v(x)>B$, so $$\int_0^\infty x^ne^{v(x)}>\int_a^b e^B>\frac{1}{n!}$$ for sufficiently large $n$. –  Alex Becker Dec 18 '12 at 15:28
Alex, thank you! Is it also obvious for integrals $\int_0^\infty x^n f(x) dx$? –  Sasha Dec 18 '12 at 15:38
@Sasha If we let $f$ take the value $0$ I suppose it could be done. I'll give it some thought. –  Alex Becker Dec 18 '12 at 15:41
It seems merely allowing $f(x)=0$ is not sufficient. Things get messy when you allow $f$ to take negative values, so I'll need to think about that case some more. –  Alex Becker Dec 18 '12 at 16:09
@Sasha: do you mean $f\ge0$ or any $f$? –  Pietro Majer Dec 18 '12 at 16:33

4 Answers 4

up vote 7 down vote accepted

Your question is a special case of the Hamburger moment problem: given a sequence of positive numbers $(\mu_n)_{n\geq 0}$ decides if there exists a positive measure $\mu$ on $\mathbb{R}$ such that $\newcommand{\bR}{\mathbb{R}}$

$$\mu_n=\int_{\bR} x^n \mu(|dx|),\;\;\forall n=0,1,2,\dotsc. $$

There exist many necessary and sufficient conditions for the existence of such a measure. The classical text The Laplace Transform by D.V. Widder is a good place to look, especially Section 10-13 of Chapter III. Here is a link to the book.

Addendum 1. If you do not care about the positivity of the measure $\mu$ then we have the following result of R.P. Boas. (See Section 14, Chapter III of Widder's book)

For any sequence of real numbers $(\mu_n)_{n\geq 0}$ there exists a signed measure $\mu$ on $[0,\infty)$ with the following properties

$$ \mu_n=\int_0^\infty x^n \mu(dx) ,\;\;\forall n=0,1,2,\dotsc, \tag{1} $$


$$\int_0^\infty|\mu|(dx)<\infty.\tag{2} $$

Recall that any signed measure $\mu$ is the difference of two positive measures $\mu=\mu^+-\mu^-$ and the total variation mesure is $|\mu|=\mu^++\mu^-$. On the semiaxis $[0,\infty)$ a signed measure $\mu$ has the from $\mu=d\alpha$, where $\alpha$ is a function of bounded variation on compact intervals.

Addendum 2. Consider the Fourier transform of the measure $\mu$ in Addendum 1. We have $\newcommand{\ii}{\boldsymbol{i}}$

$$ \widehat{\mu}(\xi):=\int_{\bR} e^{-\ii\xi x} \mu(dx) =\sum_{n\geq 0}\mu_n\frac{(-\ii\xi)^n}{n!}.\tag{3} $$

There is an obvious problem with the above equality: the series in the right-hand side may not be convergent for all $\xi$ if the momenta $\mu_n$ grow too fast. In fact if the momenta grow fast, there exist at least two measures $\mu$, $\mu'$ satisfying both (1) and (2) above. In your case the $\mu_n$ decay very fast and my guess is that $\mu$ is unique. (The series in the right-hand side of (3) converges for any $\xi$ so it defines a continuous function $f(\xi)$ which can be viewed as the Fourier transform of $\mu$ in the sense of distributions. Now use the Fourier inversion formula to recover $\mu$.)

In any case, the space of solutions of (1) (2) is completely understood. A good place to look is Akhiezer book The Classical Moment problem or Chapter 16 in the book Unbounded selfadjoint operators on Hilbert space by Konrad Schmudgen. The story is quite rich.

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Thank you! This is precisely what I need. Don't you know if there is a constructive way to restore such measure? –  Sasha Dec 19 '12 at 12:25
See Addendum 2. –  Liviu Nicolaescu Dec 19 '12 at 13:49

Such a function would have to satisfy: $$\int(1-x)^2e^{v(x)}dx=-\frac12,$$ but the left hand side is clearly non-negative...

EDIT: Another contradiction for positive exponents is $$\int x(1-x)^4e^{v(x)}dx = -\frac{19}{120}$$

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@Gjergji, the OP didn't specify the value of the integral for $n=0$, just for positive integers $n$. –  Barry Cipra Dec 18 '12 at 16:37
But it's fixable if you replace $(1-x)^2$ with $(x^{1/2}-2x^{3/2})^2$. –  Barry Cipra Dec 18 '12 at 16:42
Or $(x-4x^2)^2$. –  Emil Jeřábek Dec 18 '12 at 16:44
A very slight modification, using $(x^{n/2}-nx^{n/2+1})^2$, gives a somewhat surprising result: The OP's equality cannot hold for any three consecutive values of $n$. (To be precise, this only works for consecutive values greater than 1; my earlier comment rules out the case starting at 1.) –  Barry Cipra Dec 18 '12 at 17:07

No. For the integral to make sense, $v$ must be at least measurable. Let $$X_k=\{x\in [1,\infty): v(x)>-k\}$$ and note that $[1,\infty)=\cup_{k\in\mathbb N} X_k$ so some set $X_k$ has positive measure by countable subadditivity. Then $$\int_0^\infty x^n e^{v(x)}dx\ge \int_{X_k} x^n e^{v(x)}dx\ge \int_{X_k} e^{-k}dx > \frac{1}{n!}$$ for sufficiently large $n$.

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Suppose you are not wedded to the interval $(0,\+\infty)$.

If $C$ is the unit circle in the complex plane, oriented in the counter-clockwise direction as usual, then $$ \frac{1}{2\pi i}\oint_C z^n \frac{e^{1/z}}{z}\;dz = \frac{1}{n!} $$ for $n=0,1,2,\dots$

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