Hermite関数をHermite多項式で展開する

Hermite多項式の一般化であるHermite関数は以下のように定義される.
\begin{align} H_{\nu}(z):=2^{\nu}\sqrt{\pi}\left(\frac 1{\Gamma\left(\frac{1-\nu}2\right)}\F11{-\frac{\nu}2}{\frac 12}{z^2}-\frac{2z}{\Gamma\left(-\frac {\nu}2\right)}\F11{\frac{1-\nu}2}{\frac 32}{z^2}\right) \end{align}
今回は, このHermite関数をHermite多項式によって展開したいと思う. まず, 合流型超幾何関数
\begin{align} M(a,b;x)&:=\F11{a}b{x}\\ U(a,b;x)&:=\frac{\Gamma(1-b)}{\Gamma(1+a-b)}M(a,b;x)+\frac{\Gamma(b-1)}{\Gamma(a)}x^{1-b}M(1+a-b,2-b;x) \end{align}
を用いると, 定義から$0< x$において,
\begin{align} H_{\nu}(x)&=2^{\nu}U\left(-\frac{\nu}2,\frac 12;x^2\right) \end{align}
と表される. $x\to\infty$において,
\begin{align} U(a,b;x)&\sim x^{-a} \end{align}
であることが良く知られているので, 積分
\begin{align} \int_0^{\infty}H_{\nu}(x)H_n(x)e^{-x^2}\,dx \end{align}
は問題なく収束する(しかし, 実数全体での積分は収束しないと思われる). Hermite多項式の母関数は
\begin{align} \sum_{0\leq n}\frac{H_n(x)}{n!}t^n&=e^{-t^2+2xt} \end{align}
となることから,
\begin{align} &\sum_{0\leq n}\frac{t^n}{n!}\int_0^{\infty}H_{\nu}(x)H_n(x)e^{-x^2}\,dx\\ &=e^{-t^2}\int_0^{\infty}H_{\nu}(x)e^{-x^2+2xt}\,dx \end{align}
となる. ここで, 前の記事 で示した合流型超幾何関数のMellin-Barnes積分表示を用いると,
\begin{align} &\int_0^{\infty}H_{\nu}(x)e^{-x^2+2xt}\,dx\\ &=2^{\nu}\int_0^{\infty}U\left(-\frac{\nu}2,\frac 12;x^2\right)e^{-x^2+2xt}\,dx\\ &=\frac{2^{\nu}}{2\pi i\Gamma\left(-\frac{\nu}2\right)\Gamma\left(\frac{1-\nu}2\right)}\int_0^{\infty}\left(\int_{-i\infty}^{i\infty}\Gamma\left(-\frac{\nu}2+s\right)\Gamma\left(\frac 12-s\right)\Gamma(-s)x^{2s}\,ds\right)e^{-x^2+2xt}\,dx\\ &=\frac{2^{\nu}}{2\pi i\Gamma\left(-\frac{\nu}2\right)\Gamma\left(\frac{1-\nu}2\right)}\int_{-i\infty}^{i\infty}\Gamma\left(-\frac{\nu}2+s\right)\Gamma\left(\frac 12-s\right)\Gamma(-s)\left(\int_0^{\infty}x^{2s}e^{-x^2+2xt}\,\,dx\right)\,ds\\ \end{align}
ここで,
\begin{align} \int_0^{\infty}x^{2s}e^{-x^2+2xt}\,\,ds&=\sum_{0\leq n}\frac{(2t)^n}{n!}\int_0^{\infty}x^{2s+n}e^{-x^2}\,dt\\ &=\frac 12\sum_{0\leq n}\frac{(2t)^n}{n!}\Gamma\left(s+\frac{n+1}2\right) \end{align}
であることから,
\begin{align} &\int_{-\infty}^{\infty}H_{\nu}(x)e^{-x^2+2xt}\,dx\\ &=\frac 12\sum_{0\leq n}\frac{(2t)^n}{n!}\frac{2^{\nu}}{2\pi i\Gamma\left(-\frac{\nu}2\right)\Gamma\left(\frac{1-\nu}2\right)}\\ &\qquad\cdot\int_{-i\infty}^{i\infty}\Gamma\left(-\frac{\nu}2+s\right)\Gamma\left(s+\frac{n+1}2\right)\Gamma\left(\frac 12-s\right)\Gamma(-s)\,ds \end{align}
となる. ここで, Barnesの第1補題 より,
\begin{align} &\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\Gamma\left(-\frac{\nu}2+s\right)\Gamma\left(s+\frac{n+1}2\right)\Gamma\left(\frac 12-s\right)\Gamma(-s)\,ds\\ &=\frac{\Gamma\left(\frac{1-\nu}2\right)\Gamma\left(-\frac{\nu}2\right)\Gamma\left(1+\frac{n}2\right)\Gamma\left(\frac{n+1}2\right)}{\Gamma\left(1+\frac{n-\nu}2\right)} \end{align}
となるから, これを代入すると,
\begin{align} &\int_{-\infty}^{\infty}H_{\nu}(x)e^{-x^2+2xt}\,dx\\ &=2^{\nu-1}\sum_{0\leq n}\frac{(2t)^n}{n!}\frac{\Gamma\left(1+\frac{n}2\right)\Gamma\left(\frac{n+1}2\right)}{\Gamma\left(1+\frac{n-\nu}2\right)}\\ &=2^{\nu-1}\sqrt{\pi}\sum_{0\leq n}\frac{t^n}{\Gamma\left(1+\frac{n-\nu}2\right)}\\ &=2^{\nu-1}\sqrt{\pi}\left(\frac 1{\Gamma\left(1-\frac{\nu}2\right)}\F11{1}{1-\frac{\nu}2}{t^2}+\frac{t}{\Gamma\left(\frac{3-\nu}2\right)}\F11{1}{\frac{3-\nu}2}{t^2}\right) \end{align}
を得る. よって, Kummerの変換公式より
\begin{align} &\sum_{0\leq n}\frac{t^n}{n!}\int_0^{\infty}H_{\nu}(x)H_n(x)e^{-x^2}\,dx\\ &=e^{-t^2}\int_0^{\infty}H_{\nu}(x)e^{-x^2+2xt}\,dx\\ &=2^{\nu-1}\sqrt{\pi}\left(\frac 1{\Gamma\left(1-\frac{\nu}2\right)}e^{-t^2}\F11{1}{1-\frac{\nu}2}{t^2}+\frac{t}{\Gamma\left(\frac{3-\nu}2\right)}e^{-t^2}\F11{1}{\frac{3-\nu}2}{t^2}\right)\\ &=2^{\nu-1}\sqrt{\pi}\left(\frac 1{\Gamma\left(1-\frac{\nu}2\right)}\F11{-\frac{\nu}2}{1-\frac{\nu}2}{-t^2}+\frac{t}{\Gamma\left(\frac{3-\nu}2\right)}\F11{\frac{1-\nu}2}{\frac{3-\nu}2}{-t^2}\right)\\ &=2^{\nu}\sqrt{\pi}\left(\frac 1{\Gamma\left(-\frac{\nu}2\right)}\sum_{0\leq n}\frac{(-1)^n}{n!(2n-\nu)}t^{2n}+\frac{1}{\Gamma\left(\frac{1-\nu}2\right)}\sum_{0\leq n}\frac{(-1)^n}{n!(2n+1-\nu)}t^{2n+1}\right) \end{align}
となる. つまり, 以下が得られた.

これより, $0< x$に対し, $H_{\nu}(x)$と一致する偶関数$H^+_{\nu}$, 奇関数$H_{\nu}^-$を考えると,
\begin{align} \int_{-\infty}^{\infty}H_{\nu}^+(x)H_{2n}(x)e^{-x^2}\,dx&=2^{\nu+1}\sqrt{\pi}\frac{(-1)^n(2n)!}{\Gamma\left(-\frac{\nu}2\right)n!(2n-\nu)}\\ \int_{-\infty}^{\infty}H_{\nu}^+(x)H_{2n+1}(x)e^{-x^2}\,dx&=0\\ \int_{-\infty}^{\infty}H_{\nu}^-(x)H_{2n}(x)e^{-x^2}\,dx&=0\\ \int_{-\infty}^{\infty}H_{\nu}^-(x)H_{2n+1}(x)e^{-x^2}\,dx&=2^{\nu+1}\sqrt{\pi}\frac{(-1)^n(2n+1)!}{\Gamma\left(\frac{1-\nu}2\right)n!(2n+1-\nu)} \end{align}
となるから, Hermite多項式の直交性
\begin{align} \int_{-\infty}^{\infty}H_m(x)H_n(x)e^{-x^2}\,dx&=2^n\sqrt{\pi} n!\delta_{m,n} \end{align}
を考えると, $x\in\mathbb{R}$に対し, それぞれ
\begin{align} H_{\nu}^+(x)&=\frac{2^{\nu+1}}{\Gamma\left(-\frac{\nu}2\right)}\sum_{0\leq n}\frac{(-1)^n}{2^{2n}n!(2n-\nu)}H_{2n}(x)\\ H_{\nu}^{-}(x)&=\frac{2^{\nu+1}}{\Gamma\left(\frac{1-\nu}2\right)}\sum_{0\leq n}\frac{(-1)^n}{2^{2n+1}n!(2n+1-\nu)}H_{2n+1}(x) \end{align}
と展開できる. 特に, $0< x$において,
\begin{align} H_{\nu}(x)&=\frac{2^{\nu+1}}{\Gamma\left(-\frac{\nu}2\right)}\sum_{0\leq n}\frac{(-1)^n}{2^{2n}n!(2n-\nu)}H_{2n}(x)\\ &=\frac{2^{\nu+1}}{\Gamma\left(\frac{1-\nu}2\right)}\sum_{0\leq n}\frac{(-1)^n}{2^{2n+1}n!(2n+1-\nu)}H_{2n+1}(x) \end{align}
と展開されることが分かる.

2つのHermite関数の積の積分

上の展開とHermite多項式の直交性を用いることで, 以下のような計算ができる.
\begin{align} &\int_0^{\infty}H_{\mu}(x)H_{\nu}(x)e^{-x^2}\,dx\\ &=\frac{2^{\mu+1}}{\Gamma\left(-\frac{\mu}2\right)}\frac{2^{\nu+1}}{\Gamma\left(-\frac{\nu}2\right)}\sum_{0\leq m,n}\frac{(-1)^m}{2^{2m}m!(2m-\mu)}\frac{(-1)^n}{2^{2n}n!(2n-\nu)}\int_0^{\infty}H_{2m}(x)H_{2n}(x)e^{-x^2}\,dx\\ &=\frac{2^{\mu+\nu+1}\sqrt{\pi}}{\Gamma\left(-\frac{\mu}2\right)\Gamma\left(-\frac{\nu}2\right)}\sum_{0\leq n}\frac{(2n)!}{2^{2n}n!^2(2n-\mu)(2n-\nu)}\\ &=\frac{2^{\mu+\nu}\sqrt{\pi}}{\Gamma\left(-\frac{\mu}2\right)\Gamma\left(-\frac{\nu}2\right)(\mu-\nu)}\sum_{0\leq n}\frac{\left(\frac 12\right)_n}{n!}\left(\frac 1{n-\frac \mu2}-\frac 1{n-\frac \nu2}\right)\\ &=\frac{2^{\mu+\nu}\sqrt{\pi}}{\Gamma\left(-\frac{\mu}2\right)\Gamma\left(-\frac{\nu}2\right)(\mu-\nu)}\sum_{0\leq n}\frac{\left(\frac 12\right)_n}{n!}\int_0^1(t^{n-\frac{\mu}2-1}-t^{n-\frac{\nu}2-1})\,dt\\ &=\frac{2^{\mu+\nu}\sqrt{\pi}}{\Gamma\left(-\frac{\mu}2\right)\Gamma\left(-\frac{\nu}2\right)(\mu-\nu)}\int_0^1(t^{-\frac{\mu}2-1}-t^{-\frac{\nu}2-1})(1-t)^{-\frac 12}\,dt\\ &=\frac{2^{\mu+\nu}\sqrt{\pi}}{\Gamma\left(-\frac{\mu}2\right)\Gamma\left(-\frac{\nu}2\right)(\mu-\nu)}\left(\frac{\Gamma\left(-\frac{\mu}2\right)\sqrt{\pi}}{\Gamma\left(\frac{1-\mu}2\right)}-\frac{\Gamma\left(-\frac{\nu}2\right)\sqrt{\pi}}{\Gamma\left(\frac{1-\nu}2\right)}\right)\\ &=\frac{2^{\mu+\nu}\pi}{\mu-\nu}\left(\frac{1}{\Gamma\left(\frac{1-\mu}2\right)\Gamma\left(-\frac{\nu}2\right)}-\frac{1}{\Gamma\left(\frac{1-\nu}2\right)\Gamma\left(-\frac{\mu}2\right)}\right) \end{align}
となる. つまり以下を得る.

特に, $\mu\to\nu$とすると,
\begin{align} \int_0^{\infty}H_{\nu}(x)^2e^{-x^2}\,dx&=\frac{2^{2\nu-1}\pi}{\Gamma\left(\frac{1-\nu}2\right)\Gamma\left(-\frac{\nu}2\right)}\left(\psi\left(\frac{1-\nu}2\right)-\psi\left(-\frac{\nu}2\right)\right)\\ &=\frac{2^{\nu-2}\sqrt{\pi}}{\Gamma(-\nu)}\left(\psi\left(\frac{1-\nu}2\right)-\psi\left(-\frac{\nu}2\right)\right) \end{align}
つまり以下を得る.