Formulae in hoc propositae in eo casu, ubi modulus ipsius \(q\) unitatem aequatur, consideratione satis dignae videntur, quippe quae functiones unius variabilis pro quovis valore argumenti valore discontinuas praebeant sistant.Choice n'est pas visible ?

Primo in hunc finem formulas 1-7 integramus

\( 48) \, \int\log{k}\frac{\partial q}{q}=\)

Series quidem propositae magna ex parte pro modulo ipsius \(q\) unitati aequale non convergunt, sed integrando series convergentes inde derivari possunt; itaque primo integralia formularum 1-7 proponamus exhibemus

\begin{eqnarray*} 48) \int_0(\log k\frac{\partial q}{q}-\log 4\sqrt q)\frac{\partial q}{q}=-4\log{(1+q)}+\frac{4}{4}\log{(1+q^2)}-\frac{4}{9}\log{(1+q^3)}+\frac{4}{16}\log{(1+q^4)} \end{eqnarray*}

\begin{eqnarray*}49) \int_0- \log k'\frac{\partial q}{q}= 4 \log{\frac{1+q}{1-q}}+\frac{4}{9}\log{\frac{1+q^3}{1-q^3}}+\frac{4}{25}\log{\frac{1+q^5}{1-q^5}} \end{eqnarray*}

\begin{eqnarray*}50) \int_0\log\frac{2K}{\pi}\frac{\partial q}{q}= 4\log{(1+q)}+\frac{4}{9}\log{(1+q^3)}+\frac{4}{25}\log{(1+q^5)}\end{eqnarray*}

\begin{eqnarray*} 51)\int_0\left(\frac{2K}{\pi}-1\right)\frac{\partial q}{q}=& \mbox{[illeg.] }Suggère un changement de signe. 4\log{(1+q)}-4\log{(1-q)}+\frac{4}{3}\log{(1-q^3)}-\frac{4}{5}\log{(1-q^5)}\\ & \mbox{[illeg.] } + 2i\log{\frac{1-qi}{1+qi}}+\frac{2i}{2}\log{\frac{1-q^2i}{1+q^2i}}+\frac{2i}{3}\log{\frac{1-q^3i}{1+q^3i}}\end{eqnarray*}

\begin{eqnarray*} 52)\int_0\left(\frac{2kK}{\pi}\right)\frac{\partial q}{q}=& 4 \log{\frac{1+\sqrt q}{1-\sqrt q}}-\frac{4}{3}\log{\frac{1+\sqrt{q^3}}{1-\sqrt{q^3}}}+\frac{4}{5}\log{\frac{1+\sqrt{q^5}}{1-\sqrt{q^5}}}\\ & 4i\log{\frac{1+\sqrt qi}{1-\sqrt qi}}+\frac{4i}{3}\log{\frac{1+\sqrt q^3i}{1-\sqrt q^3i}}+\frac{4i}{5}\log{\frac{1+\sqrt q^5i}{1-\sqrt q^5i}}\end{eqnarray*}

\begin{eqnarray*} 53)\int_0\left(\frac{2 k'K}{\pi}-1\right)\frac{\partial q}{q}=& -4 \log{(1+q)}+\frac{4}{3}\log{(1+q^3)}-\frac{4}{5}\log{(1+q^5)}\\ & -\frac{2i}{2}\log{\frac{1-qi}{1+qi}}+\frac{2i}{2}\log{\frac{1-q^2i}{1+q^2i}}-\frac{2i}{3}\log{\frac{1-q^3i}{1+q^3i}}\end{eqnarray*}

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Avec les mains de Dedekind et Weber : 

Perinde obtinetur \[ 63) \, \int_0^{e^{xi}}-\log k'\frac{\frak dq}{q}=-2\log\tan\frac{x^2}{2}-\frac{2}{9}\log\tan\frac{3x^2}{2}-\frac{2}{25}\log\tan\frac{5x^2}{2}\] si \(x\) est (quantitas) / numerus surdus, non convergit, sin minus, denotando [rature] / literis \(m, n\) numeros integros inter se primes, et ponendo \(\frac{x}{2\pi}=\frac{m}{n}\) ita exhiberi potest [Page13r] \begin{align*} &+4\pi i\left(\left(\frac{x}{2\pi}\right)-\left(\frac{x}{2\pi}+\frac{1}{2}\right)\right)+\frac{4\pi i}{9}\left(\left(\frac{3x}{2\pi}\right)-\left(\frac{3x}{2\pi}+\frac{1}{2}\right)\right)+\frac{4\pi i}{25}\left(\left(\frac{5x}{2\pi}\right)-\left(\frac{5x}{2\pi}+\frac{1}{2}\right)\right)\\ &63 \, \int_0^{e^{xi}}-\log k'\frac{\frak dq}{q}=-2\log\tan\frac{x^2}{2}-\frac{2}{9}\log\tan\left(\frac{3x}{2}\right)^2-\frac{2}{25}\log\tan\left(\frac{5x}{2}\right)^2\\ &=\sum_{-\infty,+\infty}\frac{\log\tan\left(\frac{px}{2}\right)^2}{p^2}\\ \end{align*} \begin{align*} &63) \, \int_0^{e^{xi}}-\log k'\frac{\frak dq}{q}=-2\log\tan\frac{x^2}{2}-\frac{2}{9}\log\tan\left(\frac{3x^2}{2}\right)-\frac{2}{25}\log\tan\left(\frac{5x^2}{2}\right)-\ldots\\ &+4\pi i\left(\left(\frac{x}{2\pi}\right)-\left(\frac{x}{2\pi}+ \frac{1}{2}\right)\right)+\frac{4\pi i}{9}\left(\left(\frac{3x}{2\pi}\right)- \left(\frac{3x}{2\pi}+\frac{1}{2}\right)\right)+\frac{4\pi i}{25} \left(\left(\frac{5x}{2\pi}\right)-\left(\frac{5x}{2\pi}+\frac{1}{2}\right)\right)-\ldots\\ &=\sum_{-\infty,+\infty}\frac{\log\tan\left(\frac{px}{2}\right)^2}{p^2} \end{align*} \begin{align*} & 63) \, \int_0^{e^{xi}}-\log k'\frac{\mathfrak ddq}{q}=-2\log\tan\frac{x^2}{2}-\frac{2}{9}\log\tan\left(\frac{3x^2}{2}\right)-\frac{2}{25}\log\tan\left(\frac{5x^2}{2}\right)-\ldots\\ &+4\pi i\left(\left(\frac{x}{2\pi}\right)-\left(\frac{x}{2\pi}+\frac{1}{2}\right)\right)+\frac{4\pi i}{9}\left(\left(\frac{3x}{2\pi}\right)-\left(\frac{3x}{2\pi}+\frac{1}{2}\right)\right)+\frac{4\pi i}{25}\left(\left(\frac{5x}{2\pi}\right)-\left(\frac{5x}{2\pi}+\frac{1}{2}\right)\right)-\ldots\\ &=\sum_{-\infty,+\infty}\frac{\log\tan\left(\frac{px}{2}\right)^2}{p^2}+\left[4\pi i \sum_{1,\infty}^p \frac{1}{p^2}\left(\left(\frac{px}{2\pi}\right)-\left(\frac{px}{2\pi}+\frac{1}{2}\right)\right)\right] \end{align*}

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Test inventé :

Début de la phrase \( a+b\) suite de la phrase Début modifié de la phrase \( a+c\) suite de la phrase changée

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Avec Handshift

\[ 63) \, \int_0^{e^{xi}}-\log k'\frac{\frak dq}{q}=-2\log\tan\frac{x^2}{2}-\frac{2}{9}\log\tan\frac{3x^2}{2}-\frac{2}{25}\log\tan\frac{5x^2}{2}\] si \(x\) est (quantitas) / numerus surdus, non convergit, sin minus, denotando [rature] / literis \(m, n\) numeros integros inter se primes, et ponendo \(\frac{x}{2\pi}=\frac{m}{n}\) ita exhiberi potest [Page13r] \begin{align*} &+4\pi i\left(\left(\frac{x}{2\pi}\right)-\left(\frac{x}{2\pi}+\frac{1}{2}\right)\right)+\frac{4\pi i}{9}\left(\left(\frac{3x}{2\pi}\right)-\left(\frac{3x}{2\pi}+\frac{1}{2}\right)\right)+\frac{4\pi i}{25}\left(\left(\frac{5x}{2\pi}\right)-\left(\frac{5x}{2\pi}+\frac{1}{2}\right)\right)\\ &63 \, \int_0^{e^{xi}}-\log k'\frac{\frak dq}{q}=-2\log\tan\frac{x^2}{2}-\frac{2}{9}\log\tan\left(\frac{3x}{2}\right)^2-\frac{2}{25}\log\tan\left(\frac{5x}{2}\right)^2\\ &=\sum_{-\infty,+\infty}\frac{\log\tan\left(\frac{px}{2}\right)^2}{p^2}\\ \end{align*}

\begin{align*} &63) \, \int_0^{e^{xi}}-\log k'\frac{\frak dq}{q}=-2\log\tan\frac{x^2}{2}-\frac{2}{9}\log\tan\left(\frac{3x^2}{2}\right)-\frac{2}{25}\log\tan\left(\frac{5x^2}{2}\right)-\ldots\\ &+4\pi i\left(\left(\frac{x}{2\pi}\right)-\left(\frac{x}{2\pi}+ \frac{1}{2}\right)\right)+\frac{4\pi i}{9}\left(\left(\frac{3x}{2\pi}\right)- \left(\frac{3x}{2\pi}+\frac{1}{2}\right)\right)+\frac{4\pi i}{25} \left(\left(\frac{5x}{2\pi}\right)-\left(\frac{5x}{2\pi}+\frac{1}{2}\right)\right)-\ldots\\ &=\sum_{-\infty,+\infty}\frac{\log\tan\left(\frac{px}{2}\right)^2}{p^2} \end{align*}

\begin{align*} & 63) \, \int_0^{e^{xi}}-\log k'\frac{\mathfrak ddq}{q}=-2\log\tan\frac{x^2}{2}-\frac{2}{9}\log\tan\left(\frac{3x^2}{2}\right)-\frac{2}{25}\log\tan\left(\frac{5x^2}{2}\right)-\ldots\\ &+4\pi i\left(\left(\frac{x}{2\pi}\right)-\left(\frac{x}{2\pi}+\frac{1}{2}\right)\right)+\frac{4\pi i}{9}\left(\left(\frac{3x}{2\pi}\right)-\left(\frac{3x}{2\pi}+\frac{1}{2}\right)\right)+\frac{4\pi i}{25}\left(\left(\frac{5x}{2\pi}\right)-\left(\frac{5x}{2\pi}+\frac{1}{2}\right)\right)-\ldots\\ &=\sum_{-\infty,+\infty}\frac{\log\tan\left(\frac{px}{2}\right)^2}{p^2} [+\left[4\pi i \sum_{1,\infty}^p \frac{1}{p^2}\left(\left(\frac{px}{2\pi}\right)-\left(\frac{px}{2\pi}+\frac{1}{2}\right)\right)\right]] \end{align*}