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I am dealing with a problem involving the complete elliptical function of the first kind, which is defined as:

$K(k)=\int_0^{\pi/2} d\theta \frac{1}{\sqrt{1-k^2\sin^2(\theta)}}=\int_0^1 dt \frac{1}{\sqrt{1-t^2}\sqrt{1-k^2 t^2}} $

for $k^2<1$. I am trying to find however how to analytically continue the function for cases where $k^2>1$, and especially when $k^2$ is complex and classify it depending on $\mathrm{Im}(k^2)$ is complex or not.

This looks like a very trivial question that must be written somewhere, but i couldn't find it explicitely not in several references or here in the forum. The closest solution i found has been in these notes:

http://www.damtp.cam.ac.uk/user/md327/fcm_2.pdf

Where they say in a last remark that the analytically continuation should be written in terms of $K(k^2) n +iK'(k')m$, where $k'$ is just the complimentary modulus and $n,m$ integers, however, right now i don't know how to prove it, and what are the $n,m$ depending on $\mathrm{Re,Im}(k^2)$

Someone can help?

mtn
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  • The integral is defined for complex $k$ (apart from values such that $k^2 >1$). So the integral itself provides an analytical continuation. I do not understand what you are asking... If you are asking for $k^2>1$, then you can take the limit for $k^2 \pm i \epsilon$ ($\epsilon\downarrow 0$). – Fabian Nov 10 '16 at 15:08
  • Sorry, maybe i didn't explain well. Actually the problem that I am try to solve find roots of an equation that involves the elliptical function and i need to find roots not only in the first Riemann surface defined by the $K(k)$ as defined above, but also in the second, third, ... Riemann surfaces. I want to know how to find systematically the shape of extra Rieman surfaces – mtn Nov 10 '16 at 15:11
  • To be even more explicit, i found in some place that the second/third Riemann surface of such elliptical function is $K+2i K'$ when $Re(k^2)>1$ and $Im(k^2)<>0$ but i don't know where does it come from. So i would like to understand where these multiple Riemann sheets emerge for this function. – mtn Nov 10 '16 at 15:22

1 Answers1

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On an abstract level, the elliptic integral $K$ is an inverse of the elliptic functions which are doubly periodic (in the complex plane) with periods $K$ and $iK'$ (this in a sense answers your question).

To be concrete, one can answer your question given the integral representation you have given (setting $m=k^2$) for convenience. We are interested to know the value of the integral for $m_\epsilon = m \pm i\epsilon$ with $m>1$ and $\epsilon>0$. We can evaluate $$\lim_{\epsilon\to0} K(m_\epsilon) = \lim_{\epsilon\to0} \int_0^{\pi/2}\!d\theta\,\frac{1}{\sqrt{1-m_\epsilon \sin^2 \theta}} = \underbrace{\int_0^{\theta^*}\!d\theta\,\frac{1}{\sqrt{1-m \sin^2 \theta}}}_{I_1}\pm i \underbrace{\int_{\theta^*}^{\pi/2}\!d\theta\,\frac{1}{\sqrt{m \sin^2 \theta-1}}}_{I_2}$$ with $\theta^*=\arcsin(1/m^{1/2})$.

Now, we treat the two integrals separately. We perform the substitution $$ m \sin^2\theta = \sin^2 \theta'$$ and obtain $$I_1 = \frac{1}{\sqrt{m}} \int_{0}^{\pi/2} \!d\theta'\,\frac{1}{\sqrt{1- m^{-1} \sin^2 \theta'}} = \frac{K(m^{-1})}{\sqrt{m}}.$$ Similarly employing the substitution $1- \sin^2 \theta =(1-m^{-1}) \sin^2 \theta'$ , we find that $$ I_2 = \frac{1}{\sqrt{m}} \int_0^{\pi/2}\!d\theta'\,\frac{1}{\sqrt{1-(1-m^{-1}) \sin^2 \theta'}} = \frac{K(1-m^{-1})}{\sqrt{m}}.$$

Thus, we have that $$ K(m) = \frac{1}{\sqrt{m}}\left[K(m^{-1}) \pm i K(1-m^{-1}) \right] . \tag{1}$$ The right hand side is the analytical continuation through the branch cut at $m>1$.

With the formula (1), you can find all the additional branches. E.g., at the point $m<1$ the right hand side becomes invalid but then using (1) you can find a the analytical continuation again.

To get a connection to your formula, you have to note that $$ \frac{K(1-m^{-1})}{\sqrt{m}} = K(1-m)= K'(m).$$ Thus, we have that (for $m>1$ as before) $$\mathop{\rm Im} K(m) = \pm K'(m)$$ and thus the branches differ by (even) multiples of $i K'(m)$. Similarly, by analytically continuing $K'(m)$ you observe that you get additional branches displaces by $K(m)$.

You find some more information e.g. here.

Fabian
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  • Thanks for your quick answer, it has been most helpful. However, i have some last question, which i still couldn't find in the reference you gave me, which is precisely the one of your last comment. How to analytically continue K'(m) to get the addittional branches. Do you know somewhere i could find it? – mtn Nov 11 '16 at 06:58
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    $K(m)= K'(1-m)$, you know the analytical continuaiton of $K$ thus you get the analytical continuation of $K'$ (for $m<0$) it hold from (1) that $K'(m)=(1-m)^{-1/2} K((1-m)^{-1}) \pm K(m)$. – Fabian Nov 12 '16 at 18:21
  • @Fabian : I'm sorry, but could You help me with one closely related question? It is possible to convert $K(k)$ for $0 < k<1$ and $\frac{1}{\sqrt{k}}K\left( \frac{1}{k}\right)$ for $1<k<\infty$ to the same form $$ \frac{1}{1+\sqrt{k}}K\left(\frac{4\sqrt{k}}{(\sqrt{k}+1)^{2}} \right) $$ by using the relation $$ K(x) = \frac{1}{1+\sqrt{x}}K\left(\frac{4\sqrt{x}}{(\sqrt{x}+1)^{2}} \right), \quad 0<x<1 $$ Therefore, $\text{Re }K(k)$ is "continuous" on all domain $0<k<\infty$ in a sense that its functional dependence is the same, despite the singularity at $k = 1$. How to explain this? – John Taylor Apr 26 '17 at 10:32
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    @JohnTaylor: The "jump" across the branch cut, just occurs in imaginary component as outlined in the answer. This is quite typical for functions that are defined (on some part of the real axis) as real functions due to Schwarz reflection principle. You observe a similar behaviour for the elementary function $z \log z$ on the (standard) branch cut along the negative real line. – Fabian Apr 26 '17 at 18:51
  • @Fabian : I invite You to answer on corresponding question https://math.stackexchange.com/questions/2252855/real-part-of-elliptic-integral , if You want, since because of its rate I see that it is interesting to people, while this comment may be out of their attention. – John Taylor Apr 26 '17 at 18:56
  • @Fabian Is there a similar formula, ( like formula (1) above ) for other elliptic integrals ? Specifically incomplete elliptic integrals ? – user91411 Jan 21 '20 at 14:49