ガウス積分
\(\int e^{-ax^2} dx\)
\[ \begin{align} \int ^{\infty} _{-\infty} e^{-ax^2} dx &= \sqrt{\frac{\pi}{a}} \\ \int ^{\infty} _{0} e^{-ax^2} dx &= \frac{1}{2}\sqrt{\frac{\pi}{a}} \end{align} \]
- 証明
\(I = \int ^{\infty} _{-\infty} e^{-ax^2} dx\)と置く, \[ \begin{align} I^2 &= \left(\int ^{\infty} _{-\infty} e^{-ax^2} dx\right)^2 \\ &= \left(\int ^{\infty} _{-\infty} e^{-ax^2} dx\right)\left(\int ^{\infty} _{-\infty} e^{-ay^2} dy\right) \\ &= \int ^{\infty} _{-\infty} \int ^{\infty} _{-\infty} e^{-a(x^2+y^2)} dy dx \end{align} \] ここで,\(x = r\cos\theta, y = r\sin\theta\)と変数変換する. \[ \begin{align} \int ^{\infty} _{-\infty} \int ^{\infty} _{-\infty} e^{-a(x^2+y^2)} dy dx &= \int ^{\infty} _{0} \int ^{2\pi} _{0} e^{-ar^2}rdrd\theta \\ &= 2\pi \int ^{\infty} _{0} e^{-ar^2} rdr \\ &= 2\pi \left[ -\frac{e^{-ar^2}}{2a} \right]^{\infty} _{0} \\ &= \frac{\pi}{a} \end{align} \] から\(I^2=\frac{\pi}{a}\)より, \[ \int ^{\infty} _{-\infty} e^{-ax^2} dx = \sqrt{\frac{\pi}{a}} \] \[ \begin{align} \int ^{\infty} _{-\infty} e^{-ax^2} dx &= \int ^{\infty} _{0} e^{-ax^2} dx + \int ^{0} _{-\infty} e^{-ax^2} dx \end{align} \] 右辺第2項は\(x=-y\)と変数変換すると, \[ \int ^{0} _{-\infty} e^{-ax^2} dx = \int^{0} _{\infty} e^{-ay^2} -1 \cdot dy = \int^{\infty} _{0} e^{-ay^2} dy \] から, \[ \int ^{\infty} _{-\infty} e ^{-ax^2} dx = 2 \int ^{\infty} _{0} e ^{-ax^2} dx \] となり, \[ \int ^{\infty} _{0} e^{-ax^2} dx = \frac{1}{2} \sqrt{\frac{\pi}{a}} \]
\(\int xe^{-ax^2} dx\)
\[ \begin{align} \int ^{\infty} _{-\infty} xe^{-ax^2} dx &= 0 \\ \int ^{\infty} _{0} xe^{-ax^2} dx &= \frac{1}{2a} \end{align} \]
- 証明
\[ \int ^{\infty} _{-\infty} xe^{-ax^2} dx = \left[ -\frac{e^{-ax^2}}{2a} \right] ^{\infty} _{-\infty} = 0 \] \[ \int ^{\infty} _{0} xe^{-ax^2} dx = \left[ -\frac{e^{-ax^2}}{2a} \right] ^{\infty} _{0} = \frac{1}{2a} \]
\(\int x^2e^{-ax^2} dx\)
\[ \begin{align} \int ^{\infty} _{-\infty} x^2 e^{-ax^2} dx &= \frac{1}{4}\sqrt{\frac{\pi}{a^3}} \\ \int ^{\infty} _{0} x^2 e^{-ax^2} dx &= \frac{1}{2}\sqrt{\frac{\pi}{a^3}} \end{align} \]
- 証明 \[ I_0(a) = \int ^{\infty} _{-\infty} e^{-ax^2} dx \\ I_2(a) = \int ^{\infty} _{-\infty} x^2 e^{-ax^2} dx \] とおく, \[ I_2(a) = -\frac{1}{da} \int ^{\infty} _{-\infty} e^{-ax^2} dx = -\frac{1}{da} I_0(a) = \int ^{\infty} _{-\infty} x^2 e^{-ax^2} dx \\ \] から, \[ I_2(a) = -\frac{1}{da} \sqrt{\frac{\pi}{a}} = \frac{1}{2}\sqrt{\frac{\pi}{a^3}} \]
\[ \begin{align} \int ^{\infty} _{-\infty} x^2 e^{-ax^2} dx &= \int ^{\infty} _{0} x^2 e^{-ax^2} dx + \int ^{0} _{-\infty} x^2 e^{-ax^2} dx \\ &= 2 \int ^{\infty} _{0} x^2 e^{-ax^2} dx \\ \end{align} \] から, \[ \int ^{\infty} _{0} x^2 e^{-ax^2} dx = \frac{1}{2} \int ^{\infty} _{-\infty} x^2 e^{-ax^2} dx = \frac{1}{4}\sqrt{\frac{\pi}{a^3}} \]