Editing Theta function (section)

=== Exemplary elliptic integrals ===

In the following some ''Elliptic Integral Singular Values''<ref>{{cite web|access-date=2023-04-07|title=Elliptic Integral Singular Value
|url=https://archive.lib.msu.edu/crcmath/math/math/e/e103.htm|website=msu.edu}}<!-- auto-translated by Module:CS1 translator --></ref> are derived:

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The ensuing function has the following lemniscatically elliptic antiderivative:

:<math>\frac{1}{\sqrt{x^4 + 1}} = \frac{\mathrm{d}}{\mathrm{d}x}\,\frac{1}{2} F\biggl[2\arctan(x); \frac{1}{2}\sqrt{2}\,\biggr]</math>

For the value <math>n = 2</math> this identity appears:

:<math>\frac{1}{4\sqrt{2}}\csc\bigl(\frac{\pi}{4}\bigr)\beta\bigl(\frac{1}{4}\bigr) = \int_{0}^{\infty} \frac{1}{\sqrt{x^4 + 1}} \,\mathrm{d}x = \biggl\{{\color{blue}\frac{1}{2} F\biggl[2\arctan(x); \frac{1}{2}\sqrt{2}\,\biggr]}\biggr\}_{x = 0}^{x = \infty} =</math>

:<math>= \frac{1}{2} F\bigl(\pi; \frac{1}{2}\sqrt{2}\bigr) = K\bigl(\frac{1}{2}\sqrt{2}\bigr) </math>

This result follows from that equation chain:

:<math>{\color{ForestGreen}K\bigl(\frac{1}{2}\sqrt{2}\bigr) = \frac{1}{4}\beta\bigl(\frac{1}{ 4}\bigr)}</math>
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The following function has the following equianharmonic elliptic antiderivative:

:<math>\frac{1}{\sqrt{x^6 + 1}} = \frac{\mathrm{d}}{\mathrm{d}x}\,\frac{1}{6}\sqrt [4]{27}F\biggl[2\arctan\biggl(\frac{\sqrt[4]{3}\,x}{\sqrt{x^2 + 1}}\biggr);\frac{1 }{4}(\sqrt{6} + \sqrt{2})\biggr]</math>

For the value <math>n = 4</math> that identity appears:

:<math>\frac{1}{6\sqrt[3]{2}}\csc\bigl(\frac{\pi}{6}\bigr)\beta\bigl(\frac{1}{3} \bigr) = \int_{0}^{\infty} \frac{1}{\sqrt{x^6 + 1}} \,\mathrm{d}x = \biggl\{{\color{blue}\frac{1}{6}\sqrt[4]{27}F\biggl[2\arctan\biggl(\frac{\sqrt[4]{3}\,x}{\sqrt{x^2 + 1}}\biggr);\frac{1}{4}(\sqrt{6} + \sqrt{2})\biggr]}\biggr\}_{x = 0}^{x = \infty} =</math>

:<math>= \frac{1}{6}\sqrt[4]{27} F\bigl[2\arctan(\sqrt[4]{3});\frac{1}{4}(\sqrt {6} + \sqrt{2})\bigr] = \frac{2}{9}\sqrt[4]{27} K\bigl[\frac{1}{4}(\sqrt{6} + \sqrt{2})\bigr] = \frac{2}{3}\sqrt[4]{3} K\bigl[\frac{1}{4}(\sqrt{6} - \sqrt{2}) \bigr] </math>

This result follows from that equation chain:

:<math>{\color{ForestGreen}K\bigl[\frac{1}{4}(\sqrt{6} - \sqrt{2})\bigr] = \frac{1}{2\sqrt[3 ]{2}\sqrt[4]{3}}\beta\bigl(\frac{1}{3}\bigr)}</math>
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And the following function has the following elliptic antiderivative:
:<math>\frac{1}{\sqrt{x^8 + 1}} =</math>

:<math>= \frac{\mathrm{d}}{\mathrm{d}x}\,\frac{1}{4}\sec \bigl(\frac{\pi}{8}\bigr)F\biggl\{2\arctan\biggl[\frac{2\cos(\pi/8)\,x}{\sqrt{x^4 + \sqrt{2}\,x^2 + 1} - x^2 + 1}\biggr];2\sqrt[4]{2}\sin\bigl(\frac{\pi}{8}\bigr) \biggr\} + \frac{1}{4}\sec\bigl(\frac{\pi}{8}\bigr)F\biggl\{\arcsin\biggl[\frac{2\cos(\pi/ 8)\,x}{x^2 + 1}\biggr];\tan\bigl(\frac{\pi}{8}\bigr)\biggr\}</math>

For the value <math>n = 6</math> the following identity appears:

:<math>\frac{1}{8\sqrt[4]{2}}\csc\bigl(\frac{\pi}{8}\bigr)\beta\bigl(\frac{3}{8} \bigr) = \int_{0}^{\infty} \frac{1}{\sqrt{x^8 + 1}} \,\mathrm{d}x =</math>

:<math>= \biggl\langle{\color{blue}\frac{1}{4}\sec\bigl(\frac{\pi}{8}\bigr)F\biggl\{2\arctan\biggl[\frac{2\cos(\pi/8)\,x}{\sqrt{x^4 + \sqrt{2}\,x^2 + 1} - x^2 + 1}\biggr];2\sqrt[4]{ 2}\sin\bigl(\frac{\pi}{8}\bigr)\biggr\} + \frac{1}{4}\sec\bigl(\frac{\pi}{8}\bigr)F \biggl\{\arcsin\biggl[\frac{2\cos(\pi/8)\,x}{x^2 + 1}\biggr];\tan\bigl(\frac{\pi}{8} \bigr)\biggr\}}\biggr\rangle_{x = 0}^{x = \infty} =</math>

:<math>= \frac{1}{4}\sec\bigl(\frac{\pi}{8}\bigr)F\bigl[\pi;2\sqrt[4]{2}\sin\bigl (\frac{\pi}{8}\bigr)\bigr] = \frac{1}{2}\sec\bigl(\frac{\pi}{8}\bigr)K(\sqrt{2\sqrt {2} - 2}\bigr) = 2\sin\bigl(\frac{\pi}{8}\bigr)K(\sqrt{2} - 1)</math>

This result follows from that equation chain:
:<math>{\color{ForestGreen}K(\sqrt{2} - 1) = \frac{1}{8}\sqrt[4]{2}\,(\sqrt{2} + 1)\, \beta\bigl(\frac{3}{8}\bigr)}</math>
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