The complete elliptic integral of the first kind $K(x)$ has a branch cut on the real axis for $|x|>0$.

One has: \begin{align} &1.) \qquad \lim_{\eta \to 0^-}K(x+i\eta)=K(x) &\text{ with } x,\eta\in \mathbb{R} \text{ and } x>1 \\ &2.) \qquad \lim_{\eta \to 0^+}K(x+i\eta)=K(x)+2i K(1-x) &\text{ with } x,\eta\in \mathbb{R} \text{ and } x>1 \end{align}

Therefore, for an evaluation on the branch cut, I have to chose an $i \eta$ prescription. My numerical tests suggest, that Mathematica has the default prescription $1.)$, but I want to be absolutely certain.

Can someone please point me to a page of Mathematica or something else, which confirms that Mathematica has a default prescription and that it is $1.)$.

On a side note: I am aware of the fact, that I could use functional identities of $K$ to map the evaluation to arguments to $|x|<1$ where everything is nice and cosy, but I would rather avoid them for some specific reasons.

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    $\begingroup$ The Wolfram Functions website describes the branch cut of the complete elliptic integral as you do. I have always assumed that the conventions are the same between the Wolfram Functions website and the Mathematica software, but I admittedly do not know for sure. $\endgroup$ May 4, 2017 at 18:25
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    $\begingroup$ Also, a "brunch cut" (line below the equations) sounds pretty tasty. :-) $\endgroup$ May 4, 2017 at 18:27
  • $\begingroup$ @MichaelSeifert Thx! $\endgroup$
    – Armin
    May 4, 2017 at 18:29
  • $\begingroup$ @MichaelSeifert But how do I know, what value it gives me if I evaluate e.g. EllipticK[2]? Does it evaluate ``K[2]+2 I K[-1]" or "K[2]"? $\endgroup$
    – Armin
    May 4, 2017 at 18:32
  • $\begingroup$ @MichaelSeifert I do understand now, what you meant and that you answered my question already... Sry, for me being so slow... $\endgroup$
    – Armin
    May 5, 2017 at 7:04

1 Answer 1


Taking the directional limits will reveal Mathematica's conventions:

Assuming[x > 1, Limit[EllipticK[x + I ε], ε -> 0, Direction -> 1]]


Assuming[x > 1, Limit[EllipticK[x + I ε], ε -> 0, Direction -> -1]]

2 I EllipticK[1 - x] + EllipticK[x]

  • $\begingroup$ many thanks. good idea! $\endgroup$
    – Armin
    May 5, 2017 at 6:12

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