Jan
19
awarded Supporter
Jan
19
comment real-value solution
Thanks for your answer! It is exactly what I have been looking for. However let me note that the transformation should be $z= \sqrt{\lambda\rho}\frac{x-1}{x+1}$. If $z = \sqrt{\lambda\rho}\frac{x+1}{x- 1}$, then $dz = -2\sqrt{\lambda \rho}\frac{1}{(x-1)^2}dx$ and one obtains the integral with sign $"-"$.
Jan
18
awarded Editor
Jan
18
revised real-value solution
"the elliptic integral of the first kind" instead of "the Jacobi elliptic function"
Jan
18
awarded Student
Jan
18
asked real-value solution