6.2.2.2 step 2

Using quantities calculated in step \(1\), the algorithm now searches for a non-negative integer \(d\) using

\begin{align*} d&=\frac {1}{2} \left ( e_{\infty } - \sum _{c\in \Gamma } e_c \right ) \end{align*}

Where in the above \(e_c \in E_c\), \(e_\infty \in E_\infty \) found in step \(1\). If non-negative \(d\) is found, then

\begin{align*} \theta &= \frac {1}{2} \sum _{c\in \Gamma } \frac {e_c}{x-c} \end{align*}

If no non-negative integer \(d\) could be found, then no Liouvillian solution exists using this case. Case three is tried next if it is available.