Fubini's theorem
Legendre's relation

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

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

{ − 1 2 2 F } x = 0 x = 1 = 1 2 2 K (1 2 2) {\displaystyle {\biggl \{}-{\frac {1}{2}}{\sqrt {2}}\,F{\biggl }{\biggr \}}_{x=0}^{x=1}={\frac {1}{2}}{\sqrt {2}}\,K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr)}}

{ 1 2 2 F − 2 E } x = 0 x = 1 = 1 2 2 {\displaystyle {\biggl \{}{\frac {1}{2}}{\sqrt {2}}\,F{\biggl }-{\sqrt {2}}\,E{\biggl }{\biggr \}}_{x=0}^{x=1}={\frac {1}{2}}{\sqrt {2}}{\biggl }}

Inserting these two integrals into the above form gives:

= ∫ 0 1 ∫ 0 1 x 3 (y 2 + 1) (1 − x 4) (1 − x 4 y 4) d x d y = {\displaystyle {\biggl }{\biggl }=\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=}

= ∫ 0 1 { y 2 + 1 2 y 2 } x = 0 x = 1 d y = ∫ 0 1 y 2 + 1 2 y 2 a r t a n h (y 2) d y = {\displaystyle =\int _{0}^{1}{\biggl \{}{\frac {y^{2}+1}{2\,y^{2}}}{\biggl }{\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=}

= y = 0 y = 1 = arctan (1) = π 4 {\displaystyle ={\biggl }_{y=0}^{y=1}=\arctan(1)={\frac {\pi }{4}}}

For the Lemniscatic special case of Legendre's relation, this result emerges:
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