特殊な形の積分

$\displaystyle \int_{-\infty}^{\infty}e^{-ax^2}dx = \sqrt{\frac{\pi}{a}}$ (5.13)

$\displaystyle \int_{0}^{\infty}e^{-ax^2}dx = \frac{1}{2} \sqrt{\frac{\pi}{a}}$ (5.14)

$\displaystyle \int_{-\infty}^{\infty}e^{-ax^2}x^2dx = \frac{1}{2a}\sqrt{\frac{\pi}{a}}$ (5.15)

$\displaystyle \int_{0}^{\infty}e^{-ax^2}x^2dx = \frac{1}{4a}\sqrt{\frac{\pi}{a}}$ (5.16)

$\displaystyle \int_{-\infty}^{\infty}e^{-ax^2}x^{s}dx = \frac{1}{2}\left(1+(-1)^s\right)  a^{-\frac{1}{2}(1+s)}  \Gamma\left(\frac{1+s}{2}\right)$ (5.17)

$\displaystyle \int_{0}^{\infty}e^{-ax^2}x^{s}dx = \frac{1}{2} a^{-\frac{1}{2}(1+s)}  \Gamma\left(\frac{1+s}{2}\right)$ (5.18)

$\displaystyle \int_{\alpha}^{\beta} (x-\alpha)^m(\beta-x)^n dx = \frac{m!  n!}{(m+n+1)!}(\beta-\alpha)^{m+n+1}$ (5.19)

物理のかぎプロジェクト / 平成19年1月14日