\begin{align*} & 48) \int_0(\log k-\log 4\sqrt q)\frac{\cancel{\mathfrak d} {d}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)}{-\ldots}\\ & 49) \int_0- \log k'\frac{\cancel{\mathfrak d}{d}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}}\\ & 50) \int_0\log\frac{2K}{\pi}\frac{\cancel{\mathfrak d}{d}q}{q}= 4\log{(1+q)}+\frac{4}{9}\log{(1+q^3)}+\frac{4}{25}\log{(1+q^5)} \end{align*}

Perinde obtinetur \begin{align*} & 63) \, \int_0^{e^{xi}}-\log k'\frac{ \cancel{\mathfrak d} {d} q}{q}=-2\log\tg~\frac{x^2}{2}-\frac{2}{9}\log\tg~\left(\frac{3x^2}{2}\right)-\frac{2}{25}\log\tg~\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\tg~\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*}