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I am seeking a function $f(x)$ that satisfies this condition: $\int_{0}^{\infty }f(x)x^ndx=\sqrt{n!}$ where n is an integer. I guess that $f$ will contain $e^{-\alpha x^2}$ as one of its factors, where $\alpha$ is some fraction.

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A random idea would be to square it and then compare to the definition of the Gamma function. Of course transform variable to polar coordinates or so... –  Gerenuk Jan 6 '13 at 11:38
    
I'm a little too tired to try it right now, but you might be able to get somewhere by mimicking Example 2.1.5 in these notes. See also this question from M.SE. –  Antonio Vargas Jan 6 '13 at 11:50
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1 Answer 1

This is not a complete answer as it does not provide an explicit expression for $f(x)$.

Multiply your equation with $(-s)^n/n!$ and sum over $n$, then you obtain $$(\mathcal{L}f)(s)=\int_0^\infty f(x)e^{- sx}\,dx = \sum_{n=0}^\infty \frac{(-s)^n}{\sqrt{n!}}.$$

Inverting the Laplace transform yields $$f(x) = \frac{1}{2\pi i} \int_{-i\infty}^{i\infty} \sum_{n=0}^\infty \frac{(-s)^n}{\sqrt{n!}} e^{s x}\,ds =\frac{1}{2\pi} \int_{-\infty}^{\infty} \sum_{n=0}^\infty \frac{(-i y)^n}{\sqrt{n!}} e^{iy x}\,dy.$$

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Ah, that reminds me of en.wikipedia.org/wiki/Mellin_transform –  Gerenuk Jan 6 '13 at 14:01
    
@Gerenuk: you can also do it via the inverse of the Mellin transform. I believe that it is also not possible to invert the transform in that case... –  Fabian Jan 6 '13 at 17:57
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