PDE中的积分不等式

tian27546西西

设$H_{n}(x)$是Hermite多项式,且设$k,m,n$是任意正整数,求证:
$$\int_{0}^{\infty}e^{-x^2}H_{2k+1}(x)H_{2m+1}(x)H_{2n+1}(x)\dfrac{dx}{x}\ge 0$$

hbghlyj

$H_1(x)=2x$, 要证明的是$$\int_{0}^{\infty}e^{-x^2}H_{2m+1}(x)H_{2n+1}(x)dx\ge 0$$
因为$H_{2m+1}(x),H_{2n+1}(x)$是奇函数, 等价于
$$\int_{-\infty}^{\infty}e^{-x^2}H_{2m+1}(x)H_{2n+1}(x)dx\ge 0$$
根据维基百科有公式${\displaystyle \int _{-\infty }^{\infty }H_{m}(x)H_{n}(x)\,e^{-x^{2}}\,dx={\sqrt {\pi }}\,2^{n}n!\,\delta _{nm},}$证毕.