# A method for calculating this integral hermite polynomials

I need proof this,

$\int_{-\infty}^{\infty}e^{-x^2}H_n^{2}(x)x^2dx=2^nn!\sqrt{\pi}(n+\frac{1}{2})$

This is the idea: Multiply $(1-t^2)^{-1/2}e^{2x^2t/(1+t)}=\underset{n=0}{\overset{\infty}\sum}\frac{H_n^{2}(x)}{2^nn!}t^n,~~~|t|<1,$ by $x^2$, integrate from $-\infty$ to $\infty$, and evaluate the integral in the left-hand side, calling the result $\varphi(t).$ Then expand $\varphi(t)$ in powers of $t$ and equate coefficients of identical powers of $t$ in both sides of the equation so obtained. I have problems in evaluating the integral, and thus expand in powers of $t$. $(H_n(x)) Hermite-Polynomials$

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Are you sure there's an $x^{2}$ in the integrand? Otherwise, the integral looks like the normalization integral for Hermite polynomials (I see you're using the physicist versions, not the probabilist): $$\int_{-\infty}^{\infty}e^{-x^{2}}H_{n}^{2}(x)\,dx=\sqrt{\pi}\, n!\,2^{n}.$$ – Adrian Keister Jul 19 '13 at 17:07
Where did you find this formula? – Mhenni Benghorbal Jul 20 '13 at 12:23
Check this related problem. – Mhenni Benghorbal Jul 20 '13 at 13:30

• Method 1: recurrence relations and symmetries of the Hermite polynomials

Hint: I would use the relations

$$H_{n+1}(x)=xH_n(x)-H'_n(x),$$

$$H'_n(x)=nH_{n-1}(x)$$

and an induction argument, as follows. Let us suppose the steps $k=1,\dots,n-1$ are true; we want to show that

$$\int e^{-x^2}H_n^2(x)x^2dx=2^nn!\sqrt{\pi}(n+\frac{1}{2}).$$

Using the above relations the l.h.s is

$$\int e^{-x^2}H_n^2(x)x^2dx=\int e^{-x^2}x^2(xH_{n-1}(x)-H'_{n-1}(x))^2dx= \int e^{-x^2}x^4H^2_{n-1}(x)dx \\ -2\int e^{-x^2}x^3H_{n-1}(x)H^{'}_{n-1}(x)dx +\int e^{-x^2}x^2(H^{'}_{n-1}(x))^2dx=\\ \int e^{-x^2}x^4H^2_{n-1}(x)dx -2(n-1)\int e^{-x^2}x^3H_{n-1}(x)H_{n-2}(x)dx +\\(n-1)^2\int e^{-x^2}x^2(H_{n-2}(x))^2dx. ~(*)$$

Let us discuss $(*)$: the first 2 terms can be reduced using integration by parts on the products

$$e^{-x^2}x^4H^2_{n-1}(x)=-(-\frac{1}{2}(2x)e^{-x^2})x^3H_{n-1}(x),$$ $$e^{-x^2}x^3H_{n-1}(x)H_{n-2}(x)=-(-\frac{1}{2}(2x)e^{-x^2})x^2H_{n-1}(x)H_{n-2}(x),$$

while the last term can be evaluated by the induction hypothesis. All we need is to remember that

$$\int e^{-x^2}x^{2q+1}H^2_{r}(x)dx=0,$$ $$\int e^{-x^2}x^{2q}H_{r}(x)H_{r-1}(x)dx=0,$$

for all $q\in\mathbb N$, $r\geq 1$ for symmetry. For the second equality we used $H_r$ odd/even $\Leftrightarrow$ $H_{r-1}$ even/odd, which follows directly from the definition of the Hermite polynomials.

• Method 2: generating function

Using the generating function suggested by the OP, one needs to integrate w.r.t $x$ on the whole real axis both sides of

$$x^2e^{-x^2}e^{-\frac{t}{1+t}x^2}\frac{1}{\sqrt{1-t^2}}= \sum_{n=1}^\infty x^2e^{-x^2}\frac{H_n^2(x)}{2^nn!}t^n,$$

and expanding w.r.t. $t$. The integral on the l.h.s. is Gaussian and can be evaluated with standard techniques reducing to the case

$$\int_{-\infty}^\infty x^2e^{-\alpha(t)x^2}dx=\frac{1}{2}\sqrt{\frac{\pi}{\alpha(t)^3}}, ~(1)$$

for $\alpha(t)>0$ and $|t|<1$.

• Generating function with $\alpha(t)=\frac{1-t}{1+t}$

In the case under exam, multiplying both sides of the generating function identity w.r.t the product $x^2 e^{-x^2}$ ($e^{-x^2}$ is the weight for the integral identities involving the Hermite polynomials) and integrating, we arrive at (the exchange between summation and integration is allowed by the properties of Hermite poly.)

$$\frac{1}{\sqrt{1-t^2}} \int_{-\infty}^\infty x^2 e^{-x^2} e^{-\frac{t}{1+t}x^2}dx= \sum_{n=1}^\infty \int_{-\infty}^\infty x^2e^{-x^2}\frac{H_n^2(x)}{2^nn!}t^n;$$

to arrive at the thesis we write (using (1))

$$g(t):=\frac{1}{\sqrt{1-t^2}}\int_{-\infty}^\infty x^2e^{-\frac{1-t}{1+t}x^2}dx=\frac{1}{\sqrt{1-t^2}}\frac{1}{2}\sqrt{\frac{\pi}{\left(\frac{1-t}{1+t}\right)^3}}=\frac{\sqrt{\pi}}{2}\frac{1+t}{(1-t)^2}$$

for $|t|<1$ and we expand w.r.t $t$ at $t=0$ the function $g(t)$, obtaining

$$g(t)=\frac{\sqrt{\pi}}{2}\left(1+3t+\frac{1}{2!}10t^2+O(t^3)\right)$$

(the higher terms are easily computed, as well). This gives the thesis. I hope it helps.

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thanks Avitus, but I need to prove to the indications given. That in order to show the greatest problems with the same method: $\frac{1}{2^nn!}\int_{-\infty}^{\infty}\frac{H_n^2(x)}{1+x^2}e^{-x^2}dx=\int_{-\‌​infty}^{\infty}\left(\frac{1-x^2}{1+x^2}\right)^n\frac{e^{-x^2}}{1+x^2}dx,~~~n=0,‌​1,2...$ – albert Jul 19 '13 at 18:02
I see: returning to your strategy, you should multiply both sides of your generating function equation by $x^2 e^{-x^2}$, am I right? The overall idea sounds promising: one needs to compute a Gaussian integral. – Avitus Jul 19 '13 at 19:44
no, in this case you should multiply both sides of your generating function equation by $\frac{1}{1+x^2}$. The problem (for my) is how to calculate the resulting integral and then expoandir in powers of $t$? – albert Jul 19 '13 at 20:55
If you do not multiply also by $e^{-x^2}$ both sides of the generating function equation, how can you arrive at integrals like $\int x^2 e^{-x^2}H_n^2 dx$? – Avitus Jul 19 '13 at 20:57
this is the problem man – albert Jul 19 '13 at 21:03