Return to Question

The complete elliptic integral of the first K(x) kind 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.

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.

The complete elliptic integral of the first K(x) kind 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 brunch 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.

The complete elliptic integral of the first K(x) kind 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.

The complete elliptic integral of the first K(x) kind 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 brunch 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 $1.)$.

One 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.

The complete elliptic integral of the first K(x) kind 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 brunch 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.