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Feb 1, 2018 at 16:35 comment added Johannes Trost I assumed that the integration is from $-\infty$ to $\infty$. To simplify notation I changed the weight function to $\frac{2}{\sqrt{\pi}}\cosh(\sqrt{2 b} x) e^{-x^2-b/2}$ and got by Gram-Schmidt the first polynomials as: $2^{-1/2}, \frac{x}{\sqrt{b+1}}, \frac{2 x^2-b-1}{2\sqrt{1+2 b}}, ...$. The higher degree polynomials are to bulky to be listed in a comment.
Jan 31, 2018 at 22:49 comment added AHusain What are the first few that you've gotten by Graham-Schmidt from the $x^n$?
Jan 31, 2018 at 0:57 history asked Jia Yiyang CC BY-SA 3.0