ガウス積分(Gaussian integral)

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

$-t^2$でなく$-{\left(\frac{x+b}{c}\right)}^2$の場合($b,c\mathrm{は}\mathrm{定}\mathrm{数}$)

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