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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2$\begingroup$ 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 $n$. $\endgroup$– Alex BeckerCommented Dec 18, 2012 at 15:28
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$\begingroup$ Alex, thank you! Is it also obvious for integrals $\int_0^\infty x^n f(x) dx$? $\endgroup$– SashaCommented Dec 18, 2012 at 15:38
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$\begingroup$ @Sasha If we let $f$ take the value $0$ I suppose it could be done. I'll give it some thought. $\endgroup$– Alex BeckerCommented Dec 18, 2012 at 15:41
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$\begingroup$ 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. $\endgroup$– Alex BeckerCommented Dec 18, 2012 at 16:09
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$\begingroup$ @Sasha: do you mean $f\ge0$ or any $f$? $\endgroup$– Pietro MajerCommented Dec 18, 2012 at 16:33
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4 Answers
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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
$$ \mu_n=\int_0^\infty x^n \mu(dx) ,\;\;\forall n=0,1,2,\dotsc, \tag{1} $$
and
$$\int_0^\infty|\mu|(dx)<\infty.\tag{2} $$
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.
answered Dec 18, 2012 at 16:34
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$\begingroup$ Thank you! This is precisely what I need. Don't you know if there is a constructive way to restore such measure? $\endgroup$– SashaCommented Dec 19, 2012 at 12:25
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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 $$\int x(1-x)^4e^{v(x)}dx = -\frac{19}{120}$$
answered Dec 18, 2012 at 16:23
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$\begingroup$ @Gjergji, the OP didn't specify the value of the integral for $n=0$, just for positive integers $n$. $\endgroup$ Commented Dec 18, 2012 at 16:37
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$\begingroup$ But it's fixable if you replace $(1-x)^2$ with $(x^{1/2}-2x^{3/2})^2$. $\endgroup$ Commented Dec 18, 2012 at 16:42
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2$\begingroup$ 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.) $\endgroup$ Commented Dec 18, 2012 at 17:07
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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 $n=0,1,2,\dots$
answered Dec 18, 2012 at 18:07