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)?
$L^{2}(\mathbb R)$- norm of entire function
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real-analysis
complex-analysis
convergence
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2Do you even know of an example where the integral is finite? I can't, off the top of my head. – 2012-08-15
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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
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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
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0@timur: Nice example! – 2012-08-15
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0So, what about evaluating the above integral? any idea! – 2012-08-15
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0Do you assume that $\lim\limits_{n\to\infty}a_n=\infty$? – 2012-08-15
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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
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1More 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
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0So it looks that there is no general method to evaluate that integration?! – 2012-08-16
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3At 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