2
$\begingroup$

My question concerns the complete elliptic integral of the first kind, specifically in its Legendre form, defined below:

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

This function is well defined for $|m|<1$, but must be analytically continued to treat larger $m$. The proper way to do this continuation has already been presented here:

https://math.stackexchange.com/a/2008409/469366.

However, I do not understand the first equation of this answer. Namely, if one defines $m_\epsilon = m \pm i\epsilon$ with $m>1$ and $\epsilon>0$, how can one perform the expansion:

$$\lim_{\epsilon\rightarrow0}K(m_\epsilon) = \lim_{\epsilon\rightarrow0} \int_0^{\pi/2} \frac{d\theta}{\sqrt{1-m_\epsilon\sin^2\theta}} = \int_0^{\theta^*} \frac{d\theta}{\sqrt{1-m\sin^2\theta}} \pm i \int_{\theta^*}^{\pi/2} \frac{d\theta}{\sqrt{m\sin^2\theta-1}},$$

where the $\pm$ in this expansion corresponds to that in $m_\epsilon$, and the value $\theta^* = \arcsin(1/\sqrt{m})$ is the input for which the denominator becomes zero.

I understand that in the second integral, the denominator would have a negative real part, and so we can reverse the order and pull out a $-i$ in front of the whole integral, but the introduction of a $\pm$ due to the $\pm$ inside $m_\epsilon$ is something I have not been able to figure out.

I feel that this may have something to do with the branch cuts present in the elliptic integrals, but I have only encountered those cuts when $m$ is taken as the independent variable, not when dealing with the integral over $\theta$, itself.

If anyone can shed light on how to perform this expansion, I would greatly appreciate it. To give you an idea of my background: I have a mathematics bachelors, but I only encountered complex analysis in physics graduate school, so I am familiar but not highly advanced in this subject.

$\endgroup$
  • $\begingroup$ we can choose the cut such as it goes from $\arcsin(1/\sqrt{m})$ to $+\infty$. in the case of positive $\epsilon$ we are walking just above it and in the case of negative $\epsilon$ just below it $\endgroup$ – tired Aug 3 '17 at 21:30
0
$\begingroup$

I apologize for putting this as an answer instead of another comment, tired, but I am new to this site and don't have enough reputation to comment.

I think I need a little more clarification on your idea. My confusion is that the branch cut should be in the complex plane of $\theta$, so it isn't immediately clear to me why adding a small imaginary part to $m$ will shift the input ($\theta$) just above or below the branch cut.

I'm also not sure the structure of the integrand as a function of $\theta$ has only one branch cut from $\theta^*$ to $\infty$. This might be an unimportant distinction, but there should be infinite singularities owing to $\sin^2\theta$ being periodic over intervals of $\pi$. This might not matter since our interval is only $\theta = 0...\pi/2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.