Hermite polynomials (Integral) 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
 A: 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 $I(y)=(-1)^{n+1}\sqrt\pi\, 2^{n-1}(2n-1)!!\,y\,(1-y^2)^{n-1}$
A: The answer relies on the "Hermite multiplication theorem" (which can be found on the wiki)
$$
H_n(xy) = \sum_{m=0}^{\left \lfloor \frac{n}{2} \right \rfloor}y^{n-2m}(y^2-1)^m {n \choose 2m}\frac{(2m)!}{m!}H_{n-2m}(x).
$$
I will change your integral a bit
$$
I_N = \int_{-\infty}^\infty xH_N(xy)\exp(-x^2)dx.
$$
Expanding (using the Hermite multiplication theorem and the fact that $x=\frac{1}{2}H_1(x)$)
$$
I_N=\sum_{m=0}^{\left \lfloor \frac{N}{2} \right \rfloor}y^{N-2m}(y^2-1)^m {N \choose 2m}\frac{(2m)!}{m!}\frac{1}{2}\int_{-\infty}^\infty H_1(x)H_{N-2m}(x)\exp(-x^2)dx.
$$
Now we can use orthogonality
$$
\int_{-\infty}^\infty H_a(x)H_b(x)\exp(-x^2)dx =2^a\sqrt{\pi}a!\delta_{a,b}
$$
to enforce $N-2m=1$, or in other words $m=\frac{N-1}{2}$.  Since $m$ is an integer, this implies $N$ must be odd (which is enforced by your notation $N=2n-1$).  Thus,
$$
I_N=\begin{cases} 0 & N \textrm{ even} \\ \sqrt{\pi}\frac{N!}{\left(\frac{N-1}{2}\right)!}y(y^2-1)^{\frac{N-1}{2}} & N \textrm{ odd} \end{cases}.
$$
Therefore, your integral is
$$
\int_{-\infty}^\infty xH_{2n-1}(xy)\exp(-x^2)dx = \sqrt{\pi}\frac{(2n-1)!}{\left(n-1\right)!}y(y^2-1)^{n-1}.
$$
You could expand this with the binomial theorem for a series in $y$ but I think this form is the most elegant.
This agrees with @svyatoslav's answer above.  To see this, note $(-1)^{n+1}=(-1)^{n-1}$ and therefore the negative sign can be taken into the $(1-y^2)^{n-1}$ term and for odd numbers, the double factorial is precisely equal to
$$
(2n-1)!! = \frac{(2n-1)!}{2^{n-1}(n-1)!}
$$
see, for example, the wiki.
A: *

*This source was used: Functions

*Evaluate:
$$I=\int_{\Bbb R}xe^{-x^2}H_{2n-1}(xy)dx=
    2\int_0^∞xe^{-x^2}\biggl((2n-1)!\sum_{k=0}^{\lfloor{n-\frac
    12}\rfloor}\frac{(-1)^k(2xy)^{2n-2k-1}}{k!(2n-2k-1)!}\biggr)dx$$

*$$2\int_0^∞xe^{-x^2}\biggl((2n-1)!\sum_{k=0}^{\lfloor{n-\frac 12}\rfloor}\frac{(-1)^k(2y)^{2n-2k-1}}{k!(2n-2k-1)!}\biggr)dx= 2(2n-1)!\sum_{k=0}^{\lfloor{n-\frac 12}\rfloor}\frac{(-1)^k(2y)^{2n-2k-1}}{k!(2n-2k-1)!} \int_0^∞e^{-x^2}x ^{2n-2k-2}dx$$

*Logically this integrand looks like the one of the gamma functions. If we expand the $x^{2n-2k-2}$ part, then:

*$$\int_0^ ∞ e^{-x^2} {(x^2)}^{n-k-\frac 12}x dx=^{u=x^2}\int_0^ ∞ e^{-u} u^{n-k-\frac 12}du=Γ\biggl(n-k+\frac 12 \biggr)$$

*Therefore our possible final answer is: $$I=2(2n-1)!\sum_{k=0}^{\lfloor{n-\frac 12}\rfloor}\frac{(-1)^k(2y)^{2n-2k-1}}{k!(2n-2k-1)!} Γ\biggl(n-k+\frac 12 \biggr),Re(k-n)<\frac 12 $$

*Proof of integral:Proof
Much easier integral using same method and substitution of summation bounds:$$\int_0^ ∞ xH_n(x) e^{-x^2} dx=\frac{\sqrt π 2^{n-2}}{Γ\biggl(\frac{3-n}{2}\biggr)}$$
Try it out and you will see it is intriguing.
As always this answer may not be correct, so please leave me feedback!I am not sure if the integral can be split apart into 0 to infinity
Or you could always just use the definition of the integral to get a series right away with a limit.
