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For $p,q=1,3,5,\cdots $, I would like to evaluate certain integrals involving the Hermite polynomials. Recall that the $n$-th Hermite polynomial is defined as $$H_n(x)=(-1)^ne^{x^2}\frac{d^n}{dx^n}e^{-x^2}.$$

Now the integrals I like to evaluate are the following: $$I_1(p,q)=\int_0^{\infty}\frac{d}{dx}(H_pe^{\frac{-x^2}{2}})H_qe^{\frac{-x^2}{2}}dx$$

$$I_2(p,q)=\int_0^{\infty}\frac{d}{dx}(H_pe^{\frac{-x^2}{2}})xH_qe^{\frac{-x^2}{2}}dx$$ and

$$I_3(p,q)=\int_0^{\infty}(1+mx)(H_pe^{\frac{-x^2}{2}})H_q dx$$ where $m$ is a constant. I tried to use matlab, it gives answers for certain small values of $p$ and $q$. I believe that there is a closed form of the solution in terms of $p$ and $q$.

jack
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  • The first two integrals can be calculated by applying orthogonality of Hermite polynomials and the recurrence relation they satisfy $$ 2xH_{n} = H_{n+1}+2nH_{n-1}$$ There is a slight misprint in your formula: $(-1)^n$ in place of $(-1)^x.$ In the last integral I guess $e^{-x^2/2}$ is missing. – Ryszard Szwarc Jul 15 '22 at 23:56
  • Also $H_n'=2n H_{n-1}.$ – Ryszard Szwarc Jul 16 '22 at 00:26
  • Thanks for pointing out the typo. To use the orthogonality relation the range should be $(-\infty, \infty)$. In our case the range is just $(0,\infty)$. Could you please elaborate a bit ? Does the 1st integral have a closed formula in terms of $p$ and $q$ ? – jack Jul 16 '22 at 16:10
  • Sorry, I have overlooked that only the positive half line is considered. On this interval $H_p$ and $H_q$ are orthogonal if $p+q$ is even, $p\neq q.$ This is due to the fact that the product $H_pH_q$ is even function. Concerning the first integral, applying the derivative the integrals of the products $H_{p\pm 1}H_q$ show up. I have no idea at the moment how to deal with this case. The key point would be evaluation of $\int_0^\infty H_pe^{-x^2},dx$ – Ryszard Szwarc Jul 16 '22 at 19:19
  • Perhaps this https://math.stackexchange.com/questions/3259582/integration-of-hermite-polynomials could be helpful. – Ryszard Szwarc Jul 16 '22 at 19:28

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