I have this case: $F(\arcsin u,m)$, where $u>1$, so it ends up as $F(\frac{\pi}{2}-i\text{arccosh} u,m)$.
and I need to decompose it into real and imaginary parts. I have seen this question, but it has no answer. After some searches, looking through Abramowitz & Stegun, at 17.4.11 it describes how to do that:
$$F(\phi+i\psi,m)=F(\lambda,m)+iF(\mu,m)$$
where $\cot^2\lambda$ is the positive root of
$$x^2-\left[\cot^2\phi+m\sinh^2\psi\csc^2\phi-m_1\right]x-m_1\cot^2\phi=0\tag{1}$$
and
$$m\tan^2\mu=\tan^2\phi\cot^2\lambda-1\tag{2}$$
It turns out that the real part ends up as the complete elliptic integral of the first kind, $F(\lambda,m)=K(m)$, which suits me just fine, so all I have to do is determine the imaginary part, which means I need $\mu$, which derives from $\lambda$. For my case $\phi=\frac{\pi}{2}\Rightarrow \cot^2\phi=0,\space\csc^2\phi=1$, so $(1)$ becomes:
$$x^2-\left[m\sinh^2(\text{arccosh}u)-m_1\right]x-m_1=0$$
so, considering that $m<0.1$ (practical values are $0.01$ or less), gives the positive root (with the help of wxMaxima, I'm not versed in math):
$$\lambda_+=\frac{\sqrt{m^2(u-1)^2(u+1)^2+m_1\left[2m(1-u^2)+4\right]+m_1^2}+m(u^2-1)-m_1}{2}$$
So far, so good, but now I have problems in expressing $\mu$ from $(2)$:
$$\mu=\arctan\left(\frac{\sqrt{\tan^2\phi\cot^2\lambda-1}}{m}\right)$$
Did I do it right so far? If yes, what do I do with $\tan\phi=\tan\left(\frac{\pi}{2}\right)$? If not, how should I proceed?
Looking in the same A&S, section 17.4.8 describes the transforming of the completely imaginary argument:
$$F(i\phi,m)=iF(\theta,1-m),\space\theta=\arctan(\sinh\phi)$$
Knowing that the $\lambda$ calculated earlier is different than the $\mu=\text{arccosh}u$, I still tried plotting $F(\arctan(\sinh(\text{arccosh}u)),1-m)$ (got curious), where the argument reduces to (thanks to wxMaxima) $\arctan\sqrt{u^2-1}$. By mistake, I wrote $1+m$ in the modulus, instead of $1-m$, and this came up for $m=0.1$:
and for $m=0.01$, which is about the maximum value $m$ can take in my application:
Not only it gets closer, but for $m=0.1$ I used the hammer to tweak the modulus and the exponent to be something like $F(\arctan\sqrt{u^{2.2}-1},0.98625+m)$, and the result is this (the difference is in orange, multiplied by 100):
Is this a known "shortcut"? Is this worth following?
Best Answer
There are in fact two sets of formulas for $F(\sin^{-1}u,m)$ with $u>1$, depending on whether $u>\frac1{\sqrt m}$ or not. Both are very simple and don't have the problem of the tangent blowing up at $\pi/2$.
If $1\le u\le\frac1{\sqrt m}$, Byrd and Friedman 115.02 gives $$F(\sin^{-1}u,m)=K(m)+iF\left(\sin^{-1}\frac{\sqrt{u^2-1}}{u\sqrt{1-m}},1-m\right)$$ If $\frac1{\sqrt m}\le u$, Byrd and Friedman 115.03 gives $$F(\sin^{-1}u,m)=F\left(\sin^{-1}\frac1{u\sqrt m},m\right)+iK(1-m)$$