Editing Fubini's theorem (section)

=== Legendre's relation ===

In this next example, the more generalized form of the equation is used again as a mold:

:{| class = "wikitable"
| <math>\biggl[\int_{0}^{1} v(x) \,\mathrm{d}x\biggr]\biggl[\int_{0}^{1} w(x) \,\mathrm{d}x\biggr]= \int_{0}^{1} \int_{0}^{1} \left(x\,v(xy) \,w(x) + x\,v(x)\,w(xy)\right) \,\mathrm{d}x \,\mathrm{d}y </math>
|}

The following integrals can be computed by using the incomplete [[Elliptic Integral]]s of the first and second kind as antiderivatives and these integrals have values that can be represented with ''Complete Elliptic Integrals'':

: <math> \int_{0}^{1} \frac{1}{\sqrt{1 - x^4}} \,\mathrm{d}x = </math>

<math display="block"> \biggl\{ - \frac{1}{2}\sqrt{2}\,F\biggl[\arccos(x);\frac{1}{2}\sqrt{2}\biggr] \biggr\}_{x = 0}^{x = 1} = \frac{1}{2}\sqrt{2}\,K\bigl(\frac{1}{2}\sqrt{2}\bigr) </math>

: <math>\int_{0}^{1} \frac{x^2}{\sqrt{1 - x^4}} \,\mathrm{d}x =</math>

<math display="block">\biggl\{\frac{1}{2}\sqrt{2}\,F\biggl[\arccos(x);\frac{1}{2}\sqrt{2}\biggr] - \sqrt{2}\,E\biggl[\arccos(x);\frac{1}{2}\sqrt{2}\biggr] \biggr\}_{x = 0}^{x = 1} = \frac{1}{2}\sqrt{2}\biggl[2\,E\bigl(\frac{1}{2}\sqrt{2}\bigr) - K\bigl(\frac{1}{2}\sqrt{2}\bigr)\biggr] </math>

Inserting these two integrals into the above form gives:

<math display="block">\biggl[\int_{0}^{1} \frac{1}{\sqrt{1 - x^4}} \,\mathrm{d}x\biggr]\biggl[\int_{0}^{1} \frac{x^2}{\sqrt{1 - x^4}} \,\mathrm{d}x\biggr] =\int_{0}^{1} \int_{0}^{1} \frac{x^3 (y^2+1)}{\sqrt{(1 - x^4)(1 - x^4 y^4)}} \,\mathrm{d}x\,\mathrm{d}y = </math>

<math display="block">= \int_{0}^{1} \biggl\{\frac{y^2 + 1}{2\,y^2}\biggl[\text{artanh}\bigl(y^2\bigr) - \text{artanh}\biggl(\frac{\sqrt{1 - x^4}\,y^2}{\sqrt{1 - x^4 y^4}}\biggr)\biggr]\biggr\}_{x = 0}^{x = 1}\,\mathrm{d}y = \int_{0}^{1} \frac{y^2 + 1}{2\,y^2}\,\mathrm{artanh}\bigl(y^2\bigr) \,\mathrm{d}y= </math>

<math display="block">= \biggl[\arctan(y) - \frac{1 - y^2}{2\,y}\,\mathrm{artanh}\bigl(y^2\bigr)\biggr]_{y = 0}^{y = 1} = \arctan(1) = \frac{\pi}{4} </math>

For the [[Lemniscate elliptic functions|Lemniscatic]] special case of [[Legendre's relation]], this result emerges:

: <math>K\bigl(\frac{1}{2}\sqrt{2}\bigr)\biggl[2\,E\bigl(\frac{1}{2}\sqrt{2}\bigr) - K\bigl(\frac{1}{2}\sqrt{2}\bigr)\biggr] = \frac{\pi}{2} </math>