Hermite polynomials (Integral)

Question

Could you please help me? How to evaluate this integral
$$\int_{-\infty }^{\infty }x e^{-x^2}H_{2n-1}(xy)dx$$
I tried to use a recurring formula like:
$$H_{2n-1}(xy)=2xyH_{2n-2}(xy)-2(2n-2)H_{2n-3}(xy)$$
So I was able to rewrite the integral as:
$$I_n=y\int_{-\infty }^{\infty }2x^2e^{-x^2}H_{2n-2}(xy)-2(2n-2)I_{n-1}$$
After that I wanted to get the recurrent formula for $I_n$ using integration by parts, but have some troubles with $erf$ function.
I think, that the solution should be easier

Answer 1

The expression can be presented in closed form. $$ H_{k}(t)=(-1)^ke^{t^2}\frac{d^k}{dt^k}e^{-t^2}$$ $$\frac{d^k}{dt^k}e^{-t^2}=\frac{d^k}{dt^k}\Bigl(\sqrt\pi\int_{-\infty}^{\infty}\frac{dq}{2\pi}e^{-\frac{q^2}{4}+iqt}\Bigr)=\sqrt\pi\int_{-\infty}^{\infty}\frac{dq}{2\pi}(iq)^{k}e^{-\frac{q^2}{4}+iqt}$$ $$I(y)=\int_{-\infty }^{\infty }x e^{-x^2}H_{2n-1}(xy)dx=\frac{1}{y^2}\int_{-\infty }^{\infty }dt\, t e^{-\frac{t^2}{y^2}}e^{t^2}(-1)^{2n-1}\frac{d^{2n-1}}{dt^{2n-1}}e^{-t^2}$$ $$=\frac{(-1)^{n+1}\sqrt\pi}{2\pi i y^2}\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }dt\,dq \,t\,e^{-\frac{1-y^2}{y^2}t^2+iqt}q^{2n-1}e^{-\frac{q^2}{4}}$$ Making full square in the power of the exponent (change $p=t-\frac{iq}{2}\frac{y^2}{1-y^2})$ $$=\frac{(-1)^{n+1}\sqrt\pi}{2\pi i y^2}\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }dp\,dq \,\Bigl(p+\frac{iq}{2}\frac{y^2}{1-y^2}\Bigr)e^{-\frac{1-y^2}{y^2}p^2}q^{2n-1}e^{-\frac{q^2}{4(1-y^2)}}$$ Integral with $p$ in the parentheses wanish (because of the parity of the function). We are left with $$I(y)=\frac{(-1)^{n+1}}{4 (1-y^2)}\sqrt{\frac{y^2}{1-y^2}}\int_{-\infty }^{\infty }dq\,q^{2n}e^{-\frac{q^2}{4(1-y^2)}}$$ Integral $\,\int_{-\infty }^{\infty }dq\,q^{2n}e^{-\frac{q^2}{4(1-y^2)}}=(-1)^n\frac{d^{n}}{da^{n}}\sqrt{\frac{\pi}{a}}\,$ at $\,a=\frac{1}{4(1-y^2)}$

Could you check and finish from here?

PS the final result is $I(y)=(-1)^{n+1}\sqrt\pi\, 2^{n-1}(2n-1)!!\,y\,(1-y^2)^{n-1}$