Can someone help me with the following problem?
I have the complete elliptic integral of the 3° kind defined as follow:
$$\Pi(n|m):=\int_0^{\frac{\pi}{2}}\frac{\mathrm{d}t}{(1-n\sin(t)^2)\sqrt{1-m\sin(t)^2}}$$
This integral converges for $m<1$ and $n<1$
I calculated that for $m>1$ and $n<1$ $$\Pi(n|m)=\frac{1}{\sqrt{m}}\Pi\left(\left.\frac{n}{m}\right|\frac{1}{m}\right)-\frac{i}{\sqrt{m}(1-n)}\Pi\left(\left.\frac{n(m-1)}{m(n-1)}\right|\frac{m-1}{m}\right)$$
I wanted to try to derive the formulas also for
- $m<1$ and $n>1$ (in this case this it's a real number)
- $m>1$ and $n>1$ (in this case this it's a complex number)
But I have problem to evaluate that integral since the method I used for $m>1$ doesn't work and it diverges.
Can someone give me some hint to solve it please?
Edit
I trasformed the integral into a convergent integral and I used Cauchy regularization, but I don't know how to go on:
$${\Pi(n|m)=\int_{0}^{\frac{\pi}{2}}\left[\frac{1}{\left(1-n\sin(t)^2\right)\sqrt{1-m\sin(t)^2}}+\frac{\sqrt{\frac{n}{(n-1)(n-m)}}}{2\left(t-\text{arccsc}(\sqrt{n})\right)}\right]dt-\frac{1}{2}\sqrt{\frac{n}{(n-1)(n-m)}}\ln\left(\frac{\pi}{2\text{arccsc}(\sqrt{n})}-1\right)}$$