11
$\begingroup$

Let $f(z)$ be an entire function defined by $$f(z)=\prod_{n=1}^{\infty}\bigg(1-\frac{z^{2}}{a_{n}^{2}}\bigg),\qquad z\in \mathbb C$$ where $\{a_{n}\}_{n=1}^{\infty}$ is a sequence of positive real numbers, determined so that the infinite product above defines an entire function. How can we compute the integral $$\int_{-\infty}^{\infty}|f(x)|^{2}dx$$ where $x$ is real. Or at least finding an upper bound for it (if it is finite)?

  • 2
    Do you even know of an example where the integral is finite? I can't, off the top of my head.2012-08-15
  • 0
    @HaraldHanche-Olsen: I don't know, but I think it may depend on the sequence $\{a_{n}\}$ ! So if this s the case, and the value of the integral depends on $\{a_{n}\}$, what conditions we must have?2012-08-15
  • 3
    @Harald: We have $\frac{\sin\pi z}{\pi z}=\prod_{n=1}^\infty \left(1-\frac{z^2}{n^2}\right)$.2012-08-15
  • 0
    @timur: Nice example!2012-08-15
  • 0
    So, what about evaluating the above integral? any idea!2012-08-15
  • 0
    Do you assume that $\lim\limits_{n\to\infty}a_n=\infty$?2012-08-15
  • 2
    @Norbert: Oh of course, since if not then the zeros of $f$ have a limit point, and so $f=0$, since it is entire.2012-08-15
  • 1
    More generally if $\sum_n 1/a_n^2 < \infty$, the infinite product converges uniformly on compact sets to an entire function whose zeros are $\pm a_n$.2012-08-16
  • 0
    So it looks that there is no general method to evaluate that integration?!2012-08-16
  • 3
    At least, nobody here can think of a method. Which doesn't mean there isn't one. A couple of observations: As noted by Robert Israel, the convergence of the product has to do with the convergence of a sum, i.e., it has to do with the asymptotic behaviour of the sequence $(a_n)$. But by adding just one more factor to timur's example, I can destroy the finiteness of the integral, which shows that this does not relate well to asymptotic behaviours of $(a_n)$. I think this is likely a quite hard problem.2012-08-16

1 Answers 1