scipy.integrate._cubature:_InfiniteLimitsTransform

Extended Summary

Assuming a = [a_1, ..., a_n] and b = [b_1, ..., b_n]:

If $a_i=-\infty$ and $b_i=\infty$, the i-th integration variable will use the transformation $x=\frac{1-|t|}{t}$ and $t\in (-1,1)$.

If $a_i\ne \pm \infty$ and $b_i=\infty$, the i-th integration variable will use the transformation $x=a_i+\frac{1-t}{t}$ and $t\in (0,1)$.

If $a_i=-\infty$ and $b_i\ne \pm \infty$, the i-th integration variable will use the transformation $x=b_i-\frac{1-t}{t}$ and $t\in (0,1)$.