PRML 第1章 1.18(標準)

---

$$
\begin{array}{ll}
(1.126) & I = {\displaystyle \int_{-\infty}^\infty} \exp \left( -\dfrac{1}{2\sigma^2}x^2 \right) dx = (2\pi\sigma^2)^{1/2}\\[1.5em]
(1.141) & \Gamma(x) \equiv {\displaystyle \int_0^\infty} u^{x-1} e^{-u} du\\[1.5em]
(1.142) & {\displaystyle \prod_{i=1}^D \int_{-\infty}^\infty} e^{-x_i^2} dx_i = S_D {\displaystyle \int_0^\infty} e^{-r^2} r^{D-1} dr
\end{array}
$$

---

解答

　(1.126)で$${2\sigma^2 = 1}$$とし，(1.142)で$${u = r^2}$$と変数変換すると，

$$
\begin{array}{ll}
& {\displaystyle \prod_{i=1}^D \int_{-\infty}^\infty} e^{-x_i^2} dx_i = S_D {\displaystyle \int_0^\infty} e^{-r^2} r^{D-1} dr\\[1.5em]
\Longleftrightarrow & {\displaystyle \prod_{i=1}^D} \pi^{1/2} = S_D {\displaystyle \int_0^\infty} e^{-u} u^{(D-1)/2} \dfrac{u^{-1/2}}{2} du\\[1.5em]
\Longleftrightarrow & \pi^{D/2} = \dfrac12 S_D {\displaystyle \int_0^\infty} e^{-u} u^{D/2-1} du
\end{array}
$$

となる．よって，(1.141)より，

$$
\begin{array}{l}
S_D = \dfrac{2\pi^{D/2}}{\Gamma(D/2)}
\end{array}
$$

が成り立つ．また，

$$
\begin{array}{l}
V_D = {\displaystyle \int_0^1} S_D~r^{D-1} dr = S_D \left[ \dfrac{r^D}{D} \right]_0^1 = \dfrac{S_D}{D}
\end{array}
$$

となる．

　以上により，

$$
\begin{array}{ll}
S_2 = \dfrac{2\pi}{\Gamma(1)} = 2\pi,
& V_2 = \dfrac{S_2}{2} = \pi\\[1.5em]
S_3 = \dfrac{2\pi^\frac32}{\Gamma(3/2)} = 4\pi,
& V_3 = \dfrac{S_3}{3} = \dfrac43 \pi
\end{array}
$$

となり，既知の結果が得られる．