Editing Theta function (section)

==Integral representations==
The Jacobi theta functions have the following integral representations:

:<math>\begin{align}
\vartheta_{00} (z; \tau) &= -i\int_{i-\infty}^{i+\infty} e^{i \pi \tau u^2} \frac{\cos(2\pi uz + \pi u)}{\sin(\pi u)} \mathrm{d}u; \\[6pt]
\vartheta_{01} (z; \tau) &= -i\int_{i-\infty}^{i+\infty} e^{i \pi \tau u^2} \frac{\cos (2\pi uz)}{\sin (\pi u)} \mathrm{d}u; \\[6pt]
\vartheta_{10} (z; \tau) &= -ie^{i \pi z + \frac14 i \pi\tau} \int_{i-\infty}^{i+\infty} e^{i\pi\tau u^2} \frac{\cos(2\pi uz + \pi u + \pi \tau u)}{\sin (\pi u)} \mathrm{d}u; \\[6pt]
\vartheta_{11} (z; \tau) &= e^{i \pi z + \frac14 i \pi \tau} \int_{i-\infty}^{i+\infty} e^{i\pi\tau u^2} \frac{\cos(2\pi uz + \pi \tau u)}{\sin(\pi u)} \mathrm{d}u.
\end{align}</math>

The Theta Nullwert function <math> \theta_{3}(q) </math> as this integral identity:

:<math>\theta_{3}(q) = 1 + \frac{4q\sqrt{\ln(1/q)}}{\sqrt{\pi}} \int_{0}^{\infty} \frac{\exp[-\ln(1/q)\,x^2]\{1 - q^2\cos[2\ln(1/q)\,x]\}}{1 - 2q^2\cos[2\ln(1/q)\,x] + q^4} \,\mathrm{d}x </math>

This formula was discussed in the essay ''Square series generating function transformations'' by the mathematician Maxie Schmidt from Georgia in Atlanta.

Based on this formula following three eminent examples are given:

:<math>\biggl[\frac{2}{\pi}K\bigl(\frac{1}{2}\sqrt{2}\bigr)\biggr]^{1/2} = \theta_{3}\bigl[\exp(-\pi)\bigr] = 1 + 4\exp(-\pi) \int_{0}^{\infty} \frac{\exp(-\pi x^2)[1 - \exp(-2\pi)\cos(2\pi x)]}{1 - 2\exp(-2\pi)\cos(2\pi x) + \exp(-4\pi)} \,\mathrm{d}x </math>

:<math>\biggl[\frac{2}{\pi}K(\sqrt{2} - 1)\biggr]^{1/2} = \theta_{3}\bigl[\exp(-\sqrt{2}\,\pi)\bigr] = 1 + 4\,\sqrt[4]{2}\exp(-\sqrt{2}\,\pi) \int_{0}^{\infty} \frac{\exp(-\sqrt{2}\,\pi x^2)[1 - \exp(-2\sqrt{2}\,\pi)\cos(2\sqrt{2}\,\pi x)]}{1 - 2\exp(-2\sqrt{2}\,\pi)\cos(2\sqrt{2}\,\pi x) + \exp(-4\sqrt{2}\,\pi)} \,\mathrm{d}x </math>

:<math>\biggl\{\frac{2}{\pi}K\bigl[\sin\bigl(\frac{\pi}{12}\bigr)\bigr]\biggr\}^{1/2} = \theta_{3}\bigl[\exp(-\sqrt{3}\,\pi)\bigr] = 1 + 4\,\sqrt[4]{3}\exp(-\sqrt{3}\,\pi) \int_{0}^{\infty} \frac{\exp(-\sqrt{3}\,\pi x^2)[1 - \exp(-2\sqrt{3}\,\pi)\cos(2\sqrt{3}\,\pi x)]}{1 - 2\exp(-2\sqrt{3}\,\pi)\cos(2\sqrt{3}\,\pi x) + \exp(-4\sqrt{3}\,\pi)} \,\mathrm{d}x </math>

Furthermore, the theta examples <math> \theta_{3}(\tfrac{1}{2}) </math> and <math> \theta_{3}(\tfrac{1}{3}) </math> shall be displayed:

:<math>\theta_{3}\left(\frac{1}{2}\right) = 1+2\sum_{n = 1}^{\infty} \frac{1}{2^{n^2}} = 1 + 2\pi^{-1/2}\sqrt{\ln(2)} \int_{0}^{\infty} \frac{\exp[-\ln(2)\,x^2]\{16 - 4\cos[2\ln(2)\,x]\}}{17 - 8\cos[2\ln(2)\,x]} \,\mathrm{d}x </math>

:<math>\theta_{3}\left(\frac{1}{2}\right) = 2.128936827211877158669\ldots </math>

:<math>\theta_{3}\left(\frac{1}{3}\right) = 1+2\sum_{n = 1}^{\infty} \frac{1}{3^{n^2}} = 1 + \frac{4}{3}\pi^{-1/2}\sqrt{\ln(3)} \int_{0}^{\infty} \frac{\exp[-\ln(3)\,x^2]\{81 - 9\cos[2\ln(3)\,x]\}}{82 - 18\cos[2\ln(3)\,x]} \,\mathrm{d}x </math>

:<math>\theta_{3}\left(\frac{1}{3}\right) = 1.691459681681715341348\ldots </math>