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Fubini's theorem
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=== Integrals of Complete Elliptic Integrals === The [[improper integral]] of the [[Elliptic integral#Complete elliptic integral of the first kind|Complete Elliptic Integral of first kind]] '''K''' takes the value of twice the [[Catalan constant]] accurately. The antiderivative of that K-integral belongs to the so-called ''Elliptic Polylogarithms''. The Catalan constant can only be obtained via the [[Inverse tangent integral|Arctangent Integral]], which results from the application of Fubini's theorem: <math display="block"> \int_{0}^{1} K(x) \,\mathrm{d}x ={\color{blue}\int_{0}^{1}} {\color{green}\int_ {0}^{1}} \frac{1}{\sqrt{(1 - x^2 y^2)(1 - y^2)}} \,{\color{green}\mathrm{d}y }\,{\color{blue}\mathrm{d}x} = {\color{green}\int_{0}^{1}}{\color{blue}\int_{0}^{1}} \frac{1}{\sqrt{(1 - x^2 y^2)(1 - y^2)}} \,{\color{blue}\mathrm{d}x} \,{\color{green} \mathrm{d}y}= </math> <math display="block"> = \int_{0}^{1} \frac{\arcsin(y)}{y\sqrt{1 - y^2}} \,\mathrm{d}y = {\color{RoyalBlue} \biggl\{ 2\,\mathrm{Ti}_{2} \bigl[ y\bigl(1 + \sqrt{1 - y^2}\,\bigr)^{-1} \bigr] \biggr\}_{y = 0}^{y = 1}} = 2\,\mathrm{Ti}_{2}(1) =2\beta(2) =2\,C </math> This time, the expression now in royal cyan color tone is not elementary, but it leads directly to the equally non-elementary value of the "Catalan constant" using the Arctangent Integral, also called Inverse Tangent Integral. The same procedure also works for the ''Complete Elliptic Integral of the second kind'' '''E''' in the following way: <math display="block"> \int_{0}^{1} E(x) \,\mathrm{d}x ={\color{blue}\int_{0}^{1}} {\color{green}\int_ {0}^{1}} \frac{\sqrt{1 - x^2 y^2}}{\sqrt{1 - y^2}} \,{\color{green}\mathrm{d}y} \,{\color{blue}\mathrm{d}x} = {\color{green}\int_{0}^{1}}{\color{blue}\int_{0}^{1}} \frac {\sqrt{1 - x^2 y^2}}{\sqrt{1 - y^2}} \,{\color{blue}\mathrm{d}x} \,{\color{green}\mathrm {d}y}= </math> <math display="block"> = \int_{0}^{1} \biggl[\frac{\arcsin(y)}{2y\sqrt{1 - y^2}} + \frac{1}{2}\biggr] \,\mathrm{d}y = {\color{RoyalBlue}\biggl\{ \mathrm{Ti}_{2} \bigl[ y\bigl(1 + \sqrt{1 - y^2}\,\bigr )^{-1} \bigr] + \frac{1}{2} y \biggr\}_{y = 0}^{y = 1}} = \mathrm{Ti}_{2}(1) + \frac{1}{2} =\beta(2) + \frac{1}{2} =C + \frac{1}{2} </math>
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