システム管理者
ガウス積分
$$I^{2}=\displaystyle \frac{1}{4}\displaystyle \int_{- \infty }^{ \infty }\displaystyle \int_{- \infty }^{ \infty }e^{-(x^{2}+y^{2})}dxdy=\displaystyle \frac{1}{4}\displaystyle \int_{0}^{2\pi }\displaystyle \int_{0}^{ \infty }e^{-r^{2 } }rdrd\theta =\displaystyle \frac{2\pi }{4}\displaystyle \int_{0}^{ \infty }e^{-r^{2 } }rdr$$
$$r^{2}=X$$とおく， $$2rdr=dX$$
$$I^{2}=\displaystyle \frac{\pi }{4}\displaystyle \int_{0}^{ \infty }e^{-X}dX=\displaystyle \frac{\pi }{4}$$, $$I=\displaystyle \frac{1}{2}\sqrt{\pi }$$
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$$\displaystyle \int_{- \infty }^{ \infty }e^{-\alpha x^{2 } }dx=\sqrt{\displaystyle \frac{\pi }{\alpha } }$$
$$\displaystyle \frac{d}{d\alpha }\displaystyle \int_{- \infty }^{ \infty }e^{-\alpha x^{2 } }dx=\displaystyle \int_{- \infty }^{ \infty }dx\displaystyle \frac{d}{d\alpha }e^{-\alpha x^{2 } }=\displaystyle \int_{- \infty }^{ \infty }dx(-x^{2})e^{-\alpha x^{2 } }$$
$$\displaystyle \frac{d}{d\alpha }\sqrt{\displaystyle \frac{\pi }{\alpha } }=\sqrt{\pi }\displaystyle \frac{d}{d\alpha }\alpha ^{-1/2}=\sqrt{\pi }\left( -1/2 \right) \alpha ^{-3/2}$$
$$\displaystyle \int_{- \infty }^{ \infty }x^{2}e^{-\alpha x^{2 } }dx=\displaystyle \frac{1}{2\alpha }\sqrt{\displaystyle \frac{\pi }{\alpha } }$$
