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?

  • 2
    $\begingroup$ 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? :-) $\endgroup$ – a concerned citizen Jul 8 '18 at 19:41
  • $\begingroup$ 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! $\endgroup$ – m3tro Jul 9 '18 at 14:46
  • $\begingroup$ In case you're still interested, I managed to half-deduce/half-hammer a result. It might apply to your case, as well. $\endgroup$ – a concerned citizen Jul 9 '18 at 22:13

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