誤差関数の漸近展開

誤差関数$\mathrm{erf}(x)=\frac{2}{\pi }{\int }_0^x{e}^{-t^2}dt$
相補誤差関数$\mathrm{erfc}(x)=1-\mathrm{erf}(x)$
正規相補誤差関数(scaled complementary error function)
$\mathrm{erfcx}(x)=e^{x^2}\mathrm{erfc}(x)$
漸近展開は超幾何級数を用いて
$\mathrm{erfcx}(x)=ｔ\frac{1}{x\sqrt{\pi }}{}_2{F}_0(1,\frac{1}{2};;-\frac{1}{x^2})$
となる。よって次が成り立つ。
$$\begin{array}{rl}{e}^{-\frac{1}{4}{\partial }^2}\cdot \frac{1}{x}=& \frac{1}{x}{}_2{F}_0(1,\frac{1}{2};;-\frac{1}{x^2})\\ =& \sqrt{\pi }\mathrm{erfcx}(x)\\ =& \sqrt{\pi }{e}^{x^2}(1-\mathrm{erf}(x))\end{array}$$
Q-functionといのもある。
$Q(x)=\frac{1}{2}\mathrm{erfc}(x/\sqrt{2})=\frac{1}{\sqrt{2\pi }}{e}^{-x^2/2}{e}^{-{\partial }^2/2}\cdot \frac{1}{x}=\frac{1}{\sqrt{2\pi }}{\mathcal{T}}_{+}([0,i;i,1])\cdot \frac{1}{x}$

次の$SL(2,\mathbb{C})$の式が成立する。ただし$A=1+2a^2{\sigma }^2$
$$\begin{array}{r}\left(\begin{array}{cc}A& 0\\ 0& A^{-1}\end{array}\right)={\left(\left(\begin{array}{cc}1& 0\\ -{\sigma }^{2}i& 1\end{array}\right)\left(\begin{array}{cc}1& 2i{a}^2\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ i/(2a^2)& 1\end{array}\right)\right)}^{-1}\left(\begin{array}{cc}1& 2i{a}^2/A\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ iA/(2a^2)& 1\end{array}\right)\end{array}$$
LCT を用いる次が定義式から成立する。

$$\begin{array}{r}\frac{1}{x}=\sqrt{A}{\mathcal{T}}_{+}\left(\begin{array}{cc}A& 0\\ 0& 1/A\end{array}\right)\cdot \frac{1}{x}\end{array}$$
準同型性から
$$\begin{array}{r}{\mathcal{T}}_{+}\left(\begin{array}{cc}1& 0\\ -{\sigma }^{2}i& 1\end{array}\right){\mathcal{T}}_{+}\left(\begin{array}{cc}1& 2i{a}^2\\ 0& 1\end{array}\right){\mathcal{T}}_{+}\left(\begin{array}{cc}1& 0\\ i/(2a^2)& 1\end{array}\right)\cdot \frac{1}{x}=\sqrt{A}{\mathcal{T}}_{+}\left(\begin{array}{cc}1& 2i{a}^2/A\\ 0& 1\end{array}\right){\mathcal{T}}_{+}\left(\begin{array}{cc}1& 0\\ iA/(2a^2)& 1\end{array}\right)\cdot \frac{1}{x}\end{array}$$
微分演算子表示から
$$e^{\frac{{\sigma }^2}{2}{\partial }^2}{e}^{-a^2{x}^2}{e}^{-\frac{1}{4a^2}{\partial }^2}\cdot \frac{1}{x}=\sqrt{A}{e}^{-\frac{a^2}{A}{x}^2}{e}^{-\frac{A}{4a^2}{\partial }^2}\cdot \frac{1}{x}$$

$$e^{\frac{{\sigma }^2}{2}{\partial }^2}\cdot (1-e^{-a^2{x}^2}{e}^{-\frac{1}{4a^2}{\partial }^2}\cdot \frac{1}{ax})=1-e^{-\frac{a^2}{A}{x}^2}{e}^{-\frac{A}{4a^2}{\partial }^2}\cdot \frac{\sqrt{A}}{ax}$$
$$e^{\frac{{\sigma }^2}{2}{\partial }^2}\cdot \mathrm{erf}(ax)=\mathrm{erf}(ax/\sqrt{A})$$
$$e^{\frac{{\sigma }^2}{2}{\partial }^2}\cdot \mathrm{erf}(ax+b)=\mathrm{erf}((ax+b)/\sqrt{A})$$
Weierstrass変換の積分表示から
$$\frac{1}{\sqrt{2\pi {\sigma }^2}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathrm{erf}(ax+b)\exp (-\frac{(x-\mu {)}^2}{2{\sigma }^2})dx=\mathrm{erf}\left(\frac{a\mu +b}{\sqrt{1+2a^2{\sigma }^2}}\right)$$

を得る。