An inverse definite integral problem

Question

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.

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