# Separate incomplete elliptic integral into real and imaginary parts

I am working in a problem that involves Incomplete Elliptic Integrals of the First and Second kind of the form $F(\sin^{-1}x~|~m)$ and $E(\sin^{-1}x~|~m)$ where the parameters $m$, $x$ are real numbers in the range $m>1$ and $1/\sqrt{m} \le x \le 1$.

($x$ and $m$ are related to the commonly used argument $\phi$ and modulus $m$ by $x \equiv \sin \phi$ and $m \equiv k^2$)

As can be seen by plotting them, in this range the real part of the integrals is independent of $x$ while the imaginary part isn't. As an example, see this plot for F and this other for E for a value $m=5$.

What I would like to do is to separate the real and imaginary part of this integrals, at least in this particular range. In other words, finding the real valued functions $f_{re} (m)$, $g_{re} (m)$, $f_{im} (x,m)$ and $g_{im} (x,m)$ that satisfy:

$$F(\sin^{-1}x~|~m) \equiv f_{re} (m) + \text{i} f_{im} (x,m)$$ $$E(\sin^{-1}x~|~m) \equiv g_{re} (m) + \text{i} g_{im} (x,m)$$ in the range $m>1$ and $1/\sqrt{m} \le x \le 1$.

By using the Reciprocal Modulus Transformations (see DLMF section 19.7) and taking the limit $x\rightarrow 1/\sqrt{m}$, I have found the real parts to be:

$$f_{re}(m) \equiv \frac{1}{\sqrt{m}} K\left(\frac{1}{m}\right)$$ $$g_{re}(m) \equiv \sqrt{m} \left[ E \left( \frac{1}{m} \right) - K \left( \frac{1}{m} \right) \right] + \frac{1}{\sqrt{m}} K\left(\frac{1}{m}\right)$$

However, the imaginary parts $f_{im} (x,m)$, $g_{im} (x,m)$ escape me. I reckon there should be a way of expressing them in terms of incomplete elliptic integrals with parameters in the real valued range.

If I use the reciprocal modulus transformations I will bring the parameter inside the range $0<m<1$ but the argument will now be complex as I will have $x>1$. I have looked everywhere in the literature but I can't seem to find any identity that solves the problem. I could perhaps do something if there was a way of expressing elliptic integrals of complex argument as a combination of elliptic integrals of real argument and imaginary pure argument, but I don't know how it can be done.

Does someone have any insight on how those imaginary parts could be found?

• Even if this is a zombie comment, I have a similar problem (the question is also linked) where I managed to make use of Abramowitz & Stegun up to a certain point. Maybe it helps you, or, if you found out the answer, maybe you can hint me? :-) Commented Jul 8, 2018 at 19:41
• Hey there! Thanks for the heads up. I never managed to find the answer (although I did move forward in my own specific problem because I only needed to look at some limit values of x and m). Good luck with your question! Commented Jul 9, 2018 at 14:46
• In case you're still interested, I managed to half-deduce/half-hammer a result. It might apply to your case, as well. Commented Jul 9, 2018 at 22:13
• @aconcernedcitizen I have solved this question. Commented Nov 21, 2020 at 16:15

The reciprocal modulus transformation is Byrd and Friedman 114.01: $$F(\sin^{-1}x,m)=\frac1{\sqrt m}F(\sin^{-1}x\sqrt m,1/m)$$ $$E(\sin^{-1}x,m)=\frac1{\sqrt m}(mE(\sin^{-1}x\sqrt m,1/m)+(1-m)F(\sin^{-1}x\sqrt m,1/m))$$ Now the parameter is in $$[0,1]$$ but the sine-amplitude is a real number in $$[1,\sqrt m]$$. Thus B&F 115.02 applies: $$F(\sin^{-1}x\sqrt m,1/m)=K(1/m)-iF(A,1-1/m)$$ $$E(\sin^{-1}x\sqrt m,1/m)=E(1/m)-i\left(F(A,1-1/m)-E(A,1-1/m)+\frac{(1-1/m)\sin A\cos A}{\sqrt{1-(1-1/m)\sin^2A}}\right)$$ where $$A=\sin^{-1}\frac{\sqrt{mx^2-1}}{x\sqrt{m-1}}$$ Note that I have flipped the signs of the imaginary parts from B&F to match the values of $$E(\cdot)$$ and $$F(\cdot)$$ as calculated by Mathematica and mpmath. Finally we get $$F(\sin^{-1}x,m)=\frac1{\sqrt m}(K(1/m)-iF(A,1-1/m))$$
$$E(\sin^{-1}x,m)=\frac1{\sqrt m}\left[mE(1/m)+(1-m)K(1/m)\\ -i\left(F(A,1-1/m)-mE(A,1-1/m)+\frac{(m-1)\sin A\cos A}{\sqrt{1-(1-1/m)\sin^2A}}\right)\right]$$