ガウス積分(Gaussian integral)

$$\begin{array}{rcl} \displaystyle \int_{\mathbb{R}}\mathrm{e}^{-(x^2+y^2)}\mathrm{d}A &=& \displaystyle \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\mathrm{e}^{-(x^2+y^2)}\mathrm{d}x\mathrm{d}y\dotso \mathrm{d}A=\mathrm{d}x\mathrm{d}y\\ &=& \displaystyle \int_{-\infty}^{\infty}\mathrm{e}^{-x^2}\mathrm{d}x\int_{-\infty}^{\infty}\mathrm{e}^{-y^2}\mathrm{d}y\\ &=& \displaystyle \left(\int_{-\infty}^{\infty}\mathrm{e}^{-t^2}\mathrm{d}t\right)^2\\ \displaystyle \int_{\mathbb{R}}\mathrm{e}^{-(x^2+y^2)}\mathrm{d}A &=& \displaystyle \int_{0}^{2\pi}\int_{0}^{\infty}\mathrm{e}^{-r^2}r\mathrm{d}r\mathrm{d}\theta\dotso \mathrm{d}A=r\mathrm{d}r\mathrm{d}\theta\\ &=& \displaystyle 2\pi\int_{0}^{\infty}\mathrm{e}^{-r^2}r\mathrm{d}r\\ &=& \displaystyle 2\pi\int_{0}^{\infty}\mathrm{e}^{-z} r\frac{1}{2r}\mathrm{d}z \dotso z=r^2,\frac{\mathrm{d}z}{\mathrm{d}r}=2r,\mathrm{d}r=\frac{1}{2r}\mathrm{d}z\\ &=& \displaystyle \pi\int_{0}^{\infty}\mathrm{e}^{-z}\mathrm{d}z\\ &=& \displaystyle \pi\left[-\mathrm{e}^{-z}\right]_0^{\infty} = \displaystyle \pi\left[\left(-\mathrm{e}^{-\infty}\right)-\left(-\mathrm{e}^{-0}\right)\right] = \pi\left[0-\left(-1\right)\right]\\ &=& \displaystyle \pi\\ \displaystyle \left(\int_{-\infty}^{\infty}\mathrm{e}^{-t^2}\mathrm{d}t\right)^2 &=& \displaystyle \pi\\ \displaystyle \int_{-\infty}^{\infty}\mathrm{e}^{-t^2}\mathrm{d}t &=& \displaystyle \sqrt{\pi} \end{array}$$

\(-t^2\)でなく\(-\left(\frac{x+b}{c}\right)^2\)の場合(\(b,cは定数\))

$$\begin{array}{rcl} \displaystyle \int_{-\infty}^{\infty}\mathrm{e}^{-\left(\frac{x+b}{c}\right)^2}\mathrm{d}x &=& \displaystyle \int_{-\infty}^{\infty}\mathrm{e}^{-z^2}\,c\,\mathrm{d}z \,\dotso\,\frac{x+b}{c}=z,\frac{\mathrm{d}z}{\mathrm{d}x}=\frac{1}{c},\mathrm{d}x=c\mathrm{d}z\\ &=& \displaystyle c\int_{-\infty}^{\infty}\mathrm{e}^{-z^2}\mathrm{d}z\\ &=& \displaystyle c\sqrt{\pi}\\ \end{array}$$