Editing Elliptic integral (section)

=== Special identity for the lemniscatic case ===
For the lemniscatic case, the elliptic modulus or specific eccentricity ε is equal to half the square root of two. Legendre's identity for the lemniscatic case can be proved as follows:

According to the [[Chain rule]] these derivatives hold:

: <math>\frac{\mathrm{d}}{\mathrm{d}y} \,K\bigl(\frac{1}{2}\sqrt{2}\bigr) - F\biggl[\arccos (xy);\frac{1}{2}\sqrt{2}\biggr] = \frac{\sqrt{2}\,x}{\sqrt{1 - x^4 y^4}}</math>
: <math>\frac{\mathrm{d}}{\mathrm{d}y} \,2E\bigl(\frac{1}{2}\sqrt{2}\bigr) - K\bigl(\frac {1}{2}\sqrt{2}\bigr) - 2E\biggl[\arccos(xy);\frac{1}{2}\sqrt{2}\biggr] + F\biggl[\arccos(xy );\frac{1}{2}\sqrt{2}\biggr] = \frac{\sqrt{2}\,x^3 y^2}{\sqrt{1 - x^4 y^4}} </math>

By using the [[Fundamental theorem of calculus]] these formulas can be generated:

: <math>K\bigl(\frac{1}{2}\sqrt{2}\bigr) - F\biggl[\arccos (x);\frac{1}{2}\sqrt{2}\biggr] = \int_{0}^{1} \frac{\sqrt{2}\,x}{\sqrt{1 - x^4 y^4}} \,\mathrm{d}y </math>
: <math>2E\bigl(\frac{1}{2}\sqrt{2}\bigr) - K\bigl(\frac {1}{2}\sqrt{2}\bigr) - 2E\biggl[\arccos(x);\frac{1}{2}\sqrt{2}\biggr] + F\biggl[\arccos(x);\frac{1}{2}\sqrt{2}\biggr] = \int_{0}^{1} \frac{\sqrt{2}\,x^3 y^2}{\sqrt{1 - x^4 y^4}} \,\mathrm{d}y </math>

The [[Linear combination]] of the two now mentioned integrals leads to the following formula:

: <math>
\frac{\sqrt{2}}{\sqrt{1 - x^4}} \biggl\{2E\bigl(\frac{1}{2}\sqrt{2}\bigr) - K\bigl(\frac{1}{2}\sqrt{2}\bigr) - 2E\biggl[\arccos(x);\frac{1}{2}\sqrt{2}\biggr] + F\biggl[\arccos( x);\frac{1}{2}\sqrt{2}\biggr]\biggr\} \,+
</math>
: <math>
+ \,\frac{\sqrt{2} \,x^2}{\sqrt{1 - x^4}} \biggl\{K\bigl(\frac{1}{2}\sqrt{2}\bigr) - F\biggl[\arccos(x);\frac{1}{2}\sqrt{2}\biggr]\biggr\} = \int_{0}^{1} \frac{2\,x ^3 (y^2 + 1)}{\sqrt{(1 - x^4)(1 - x^4\,y^4)}} \,\mathrm{d}y
</math>

By forming the original antiderivative related to x from the function now shown using the [[Product rule]] this formula results:

: <math> \biggl\{K\bigl(\frac{1}{2}\sqrt{2}\bigr) - F\biggl[\arccos(x);\frac{1}{2}\sqrt{ 2}\biggr]\biggr\}\biggl\{2E\bigl(\frac{1}{2}\sqrt{2}\bigr) - K\bigl(\frac{1}{2}\sqrt{2 }\bigr) - 2E\biggl[\arccos(x);\frac{1}{2}\sqrt{2}\biggr] + F\biggl[\arccos(x);\frac{1}{2} \sqrt{2}\biggr]\biggr\} =
</math>
: <math> = \int_{0}^{1} \frac{1}{y^2}(y^2 + 1)\biggl[\text{artanh}(y^2) - \text{artanh} \bigl(\frac{\sqrt{1 - x^4}\,y^2}{\sqrt{1 - x^4 y^4}}\bigr)\biggr] \mathrm{d}y </math>

If the value <math>x = 1</math> is inserted in this integral identity, then the following identity 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] = \int_{0}^{1} \frac{1}{y^2}(y^2 + 1) \,\text{artanh}(y^2) \,\mathrm{d}y = </math>

: <math> = \biggl[2\arctan(y) - \frac{1}{y}(1 - y^2)\,\text{artanh}(y^2)\biggr]_{y = 0}^{y = 1} = 2\arctan(1) = \frac{\pi}{2} </math>

This is how this lemniscatic excerpt from Legendre's identity appears:

: <math>2E\bigl(\frac{1}{2}\sqrt{2}\bigr)K\bigl(\frac{1}{2}\sqrt{2}\bigr) - K\bigl(\frac{1}{2}\sqrt{2}\bigr)^2 = \frac{\pi}{2}</math>