information:filtering, information:classification
Gauss curve integral

Now this formula for the squaring of an integral is set up:

This chain of equations can then be generated accordingly:

2 = ∫ 0 1 ∫ 0 ∞ 2 x exp (− x 2) exp (− x 2 y 2) d x d y = ∫ 0 1 ∫ 0 ∞ 2 x exp d x d y = {\displaystyle {\biggl }^{2}=\int _{0}^{1}\int _{0}^{\infty }2x\exp(-x^{2})\exp(-x^{2}y^{2})\,\mathrm {d} x\,\mathrm {d} y=\int _{0}^{1}\int _{0}^{\infty }2x\exp {\bigl }\,\mathrm {d} x\,\mathrm {d} y=}

= ∫ 0 1 { 1 y 2 + 1 − 1 y 2 + 1 exp } x = 0 x = ∞ d y = ∫ 0 1 1 y 2 + 1 d y = arctan (1) = π 4 {\displaystyle =\int _{0}^{1}{\biggl \{}{\frac {1}{y^{2}+1}}-{\frac {1}{y^{2}+1}}\exp {\biggl }{\biggr \}}_{x=0}^{x=\infty }\,\mathrm {d} y=\int _{0}^{1}{\frac {1}{y^{2}+1}}\,\mathrm {d} y=\arctan(1)={\frac {\pi }{4}}}

For the integral of the Gauss curve this value can be generated: