Editing Fubini's theorem (section)

=== Gauss curve integral ===

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

:{| class = "wikitable"
|<math>\biggl[\int_{0}^{\infty} f(x) \,\mathrm{d}x\biggr]^2 = \int_{0}^{1} \int_{0}^{\infty} 2x\,f(x) \,f(xy)\,\mathrm{d}x \,\mathrm{d}y </math>
|}

This chain of equations can then be generated accordingly:

<math display="block">\biggl[\int_{0}^{\infty} \exp(-x^2) \,\mathrm{d}x\biggr]^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[-x^2 (y^2 + 1)\bigr] \,\mathrm{d}x\,\mathrm{d}y = </math>

<math display="block">= \int_{0}^{1} \biggl\{- \frac{1}{y^2 + 1}\exp\biggl[-x^2(y^2 + 1)\biggr]\biggr\}_{x = 0}^{x = \infty} \,\mathrm{d}y = \int_{0}^{1} \frac{1}{y^2 + 1} \,\mathrm{d}y = \arctan(1) = \frac{\pi}{4} </math>

For the integral of the [[Normal distribution|Gauss curve]] this value can be generated:

: <math>\int_{0}^{\infty} \exp(-x^2) \,\mathrm{d}x = \frac{1}{2}\sqrt{\pi} </math>