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The integration shown here, $$∫_{-\infty}^{+∞}x^r\mathrm{Exp}[−x^2]\mathrm{H_n}^2[x]\mathrm{d}x,$$ appears when we try to calculate the spectrum of the perturbed non-linear oscillators by using perturbation theory in quantum mechanics. Is there any direct way to perform the definite integration of the form shown above. I want the solution of the integral for $r≥4$. I hope there may exist some techniques which can be used to calculate the integration of above integral. Please i need suggestion from this forum to solve this integration. Highly appreciated!

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closed as off-topic by ZeroTheHero, Jon Custer, By Symmetry, Kyle Kanos, user191954 Feb 14 at 5:03

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  • $\begingroup$ Presuming $\{H_{n}^{}[x]\}$ are hermite polynomials and $r$ is an integer. It might be easier to work with occupation number representation. $\endgroup$ – Sunyam Feb 11 at 14:17
  • $\begingroup$ Why was this downvoted? $\endgroup$ – Steven Sagona Feb 14 at 1:05
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$\textbf{Note :}$ Below a generating function is derived for calculating the following matrix element (useful in perturbative computations) : $$O_{m,n,k}^{}=\frac{2_{}^{k}}{\sqrt{\pi}}\int_{-\infty}^{+\infty}dx_{}^{} H_{m}^{}(x) x_{}^{k} H_{n}^{}(x) e_{}^{-x_{}^{2}}.$$ where $\{H_{r}^{}(x)\}$ are hermite polynamials and $k$ is a non-negative integer.

(i) Consider the generating function of the Hermite polynomials :

$$G[z,x]=\sum_{n=0}^{\infty}\frac{z_{}^{n}}{n!}H_{n}^{}(x)=e_{}^{2 x_{}^{} z_{}^{}-z_{}^{2}}.$$

(ii) Define a generating function $Z[z_{1}^{},z_{2}^{},z_{3}^{}]$ :

$$Z[z_{1}^{},z_{2}^{},z_{3}^{}]=\frac{1}{\sqrt{\pi}}\int_{-\infty}^{+\infty}dx_{}^{}e_{}^{-x_{}^{2}}G[z_{1}^{},x]G[z_{2}^{},x]e_{}^{2x z_{3}^{}}=\frac{e_{}^{-[z_{1}^{2}+z_{2}^{2}]}}{\sqrt{\pi}}\int_{-\infty}^{+\infty}dx_{}^{}e_{}^{-x_{}^{2}+2x[z_{1}^{}+z_{2}^{}+z_{3}^{}]}$$ $$\Rightarrow Z[z_{1}^{},z_{2}^{},z_{3}^{}]=e_{}^{-[z_{1}^{2}+z_{2}^{2}]+[z_{1}^{}+z_{2}^{}+z_{3}^{}]_{}^{2}}.$$ (iii) Now notice : $$O_{m,n,k}^{}=\left(\frac{\partial}{\partial z_{1}^{}}\right)_{}^{m}\left(\frac{\partial}{\partial z_{2}^{}}\right)_{}^{n}\left(\frac{\partial}{\partial z_{3}^{}}\right)_{}^{k}Z[z_{1}^{},z_{2}^{},z_{3}^{}]\Big|_{(z_{1}^{},z_{2}^{},z_{3}^{})=(0,0,0)}^{}.$$ (iv) Hence : $$O_{m,n,k}^{}=\left(\frac{\partial}{\partial z_{1}^{}}\right)_{}^{m}\left(\frac{\partial}{\partial z_{2}^{}}\right)_{}^{n}\left(\frac{\partial}{\partial z_{3}^{}}\right)_{}^{k}e_{}^{-[z_{1}^{2}+z_{2}^{2}]+[z_{1}^{}+z_{2}^{}+z_{3}^{}]_{}^{2}}\Big|_{(z_{1}^{},z_{2}^{},z_{3}^{})=(0,0,0)}^{}$$ $$\therefore \int_{-\infty}^{+\infty}dx_{}^{} H_{m}^{}(x) x_{}^{k} H_{n}^{}(x) e_{}^{-x_{}^{2}}=\frac{\sqrt{\pi}}{2_{}^{k}}\left(\frac{\partial}{\partial z_{1}^{}}\right)_{}^{m}\left(\frac{\partial}{\partial z_{2}^{}}\right)_{}^{n}\left(\frac{\partial}{\partial z_{3}^{}}\right)_{}^{k}e_{}^{-[z_{1}^{2}+z_{2}^{2}]+[z_{1}^{}+z_{2}^{}+z_{3}^{}]_{}^{2}}\Big|_{(z_{1}^{},z_{2}^{},z_{3}^{})=(0,0,0)}^{}$$

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