while studying a physics problem I found that asymptotically the incomplete elliptic integral of the third kind, (using the Mathematica conventions, where it is called EllipticPi),

$\Pi(n;\phi|m)=\int_0^{\sin\phi} \frac{dt}{(1-nt^2)\sqrt{(1-mt^2)(1-t^2)}},$

which seems to logarithmically diverge when $\sin^2 \phi$ approaches $1/n$, approaches the logarithm indeed in the following way:

$\lim_{\epsilon \rightarrow 0} \left[ 2(n-1)\Pi\left(n;\arcsin \sqrt{\frac{1-\epsilon}{n}}|(2-n)n\right) +\log\epsilon\right]=C(n)$

where $C(n)$ is a constant. This identity (could it be a new discovery?) I checked numerically and it might be independent of the value of $m$.

I need to know the constant $C(n)$ in terms of some known functions (or constants from integrations independent of $n$) if possible, for the value of $m$ given. Mathematica will not help. Is there any chance to do it? I have no training in this apart from undergraduate analysis and use elliptic integrals for the first time here.

(by the way, for large $n$, I can from looking at the graph see $C(n)$ being of the form $a+log(n+b)$, but for $n$ around unity this is no exact fit while being similar (divergence to minus infinity for small $n$ included).

Recent Edit:

putting the elliptic integral and the logarithm in the same integral and doing what was suggested by user8268, splitting them into two nondiverging integrals, I have now problems with the more complicated, see my comment to the answer. Can anybody tell me if this is an elliptic integral of the third kind or if Mathematica is right when it gives me a mixture of them (including ugly roots, making evaluation of the limit impossible) and if so must there be a calculation mistake or might the advice given even not be correct?

notsoimportantdetails( Background: This comes from wanting to evaluate an improper integral (§) that I physically know and numerically see to converge. In turn this integral comes frome the difference of two diverging integrals. Mathematica cannot do the improper integral of the combination but it can do the indefinite integral. But then it splits up the integrand again and the result is the elliptic integral and the logarithm, each diverging when doing the limit. Mathematica cannot do the limit and as I said astonishingly does not even know that there is a divergence for the elliptic integral. When the elliptic integral and the logarithm are put together into one integral (use $-log\epsilon=\int_0^{\sqrt{\frac{1-\epsilon}{n}}}\frac{2tn}{1-t^2n}dt$ or transform the elliptic integral to obtain (o) below) we are more or less back where we started, I suppose - well, my version of Mathematica even hangs up for the indefinite evaluation now.

Edit: Since maybe all I have done so far is making it more complicated, the original problem is to evaluate the integral (§) of $\frac{1-\sqrt{1-\frac{\text{r0} (-2 m+\text{r0})}{x (-2 m+x)}}}{\left(1-\frac{2 m}{x}\right) \sqrt{1-\frac{\text{r0} (-2 m+\text{r0})}{x (-2 m+x)}}}$ from $r0$ to infinity, where $r0>2m$. This has the obvious split into the two divergent integrals.

For your reference, the combined integral (o) I talked about in the end above reads $C(n)=\lim_{\epsilon\rightarrow 0}\int_1^\epsilon\frac{1 - n + \sqrt{(1 + n (-1 + t) - 2 t) (-1 + t) (-1 + n + t)}}{t \sqrt{(1 + n (-1 + t) - 2 t) (-1 + t) (-1 + n + t)}}$ where the integration will solve the problem by providing $C(n)$ ; $\epsilon$ can actually be replaced by $0$ right away since the limit exists.


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    $\begingroup$ still unsolved. Please have a look at my comment of the answer by user8268. Since I don't understand why to do what he suggested, I cannot check where the mistake is. $\endgroup$ – Andreas Finke Oct 13 '12 at 12:35

Let $\mathrm{sn}(\theta) = \mathrm{sn}(\theta,\sqrt{m})$, $\mathrm{cn}(\theta) = \mathrm{cn}(\theta,\sqrt{m})$, $\mathrm{dn}(\theta) = \mathrm{dn}(\theta,\sqrt{m})$ be the Jacobi elliptic functions associated with $k = \sqrt{m} = \sqrt{n(2-n)}$. Introduce parametrization: $$ \theta = \int_{0}^t \frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}} \rightarrow t = \mathrm{sn}(\theta ) $$ and let $\mathrm{sn}(\theta_0)^2 = \frac{1}{n}$, we can rewrite the integral as: $$ \Pi(n;\phi|k^2)= \mathrm{sn}(\theta_0)^2 \int_0^{\mathrm{sn}^{-1}(\sin\phi)} \frac{d\theta}{\mathrm{sn}(\theta_0)^2-\mathrm{sn}(\theta)^2} $$ For given $\epsilon$, introduce $\varphi$, $\psi$ such that $\sin(\varphi) = \sqrt{\frac{1-\epsilon}{n}} = \mathrm{sn}(\psi)$.
Notice $$\begin{align} \mathrm{cn}(\theta_0) &= \sqrt{1 - \mathrm{sn}(\theta_0)^2} = \sqrt{\frac{n-1}{n}}\\ \mathrm{dn}(\theta_0) &= \sqrt{1 - k^2\mathrm{sn}(\theta_0)^2} = \sqrt{n-1}\\ \frac{d}{d\theta} \mathrm{sn}(\theta) &= \mathrm{cn}(\theta)\mathrm{dn}(\theta)\\ \end{align} $$ We see $\mathrm{cn}(\theta_0) = \mathrm{sn}(\theta_0)\mathrm{dn}(\theta_0)$ and the expression appear within the definition of $C(n)$ becomes: $$ \begin{align} &2(n-1)\Pi\left(n;\varphi|k^2\right) +\log\epsilon\\ =& 2\,\mathrm{cn}(\theta_0)^2 \int_0^{\psi} \frac{d\theta}{\mathrm{sn}(\theta_0)^2-\mathrm{sn}(\theta)^2} + \log(1- \frac{\mathrm{sn}(\psi)^2}{\mathrm{sn}(\theta_0)^2} )\\ =& \int_0^{\psi} \frac{2 d\theta}{\mathrm{sn}(\theta_0)^2-\mathrm{sn}(\theta)^2} ( \mathrm{cn}(\theta_0)^2 - \mathrm{sn}(\theta)\mathrm{cn}(\theta)\mathrm{dn}(\theta) )\\ =& \int_0^{\psi} \frac{2 d\theta}{\mathrm{sn}(\theta_0)^2-\mathrm{sn}(\theta)^2} ( \mathrm{sn}(\theta_0)\mathrm{cn}(\theta_0)\mathrm{dn}(\theta_0) - \mathrm{sn}(\theta)\mathrm{cn}(\theta)\mathrm{dn}(\theta) ) \end{align} $$ Since the numerator goes to $0$ as $\theta \to \theta_0$, the $\epsilon \to 0$ limit of the definition of $C(n)$ exists and is given by: $$ C(n) = \int_0^{\theta_0} \frac{2 d\theta}{\mathrm{sn}(\theta_0)^2-\mathrm{sn}(\theta)^2} ( \mathrm{sn}(\theta_0)\mathrm{cn}(\theta_0)\mathrm{dn}(\theta_0) - \mathrm{sn}(\theta)\mathrm{cn}(\theta)\mathrm{dn}(\theta) ) $$ I don't know how to simplify this further but at least this is an alternate approach.

  • $\begingroup$ Thanks! That is interesting and it surly looks a lot nicer. But I doubt that at this point it is more useful practically than the integral (o) at the end of my question, don't you think so? $\endgroup$ – Andreas Finke Feb 8 '13 at 11:30
  • $\begingroup$ @andreasfinke: If you just want a number, it is no more practical than the integral you have. However, the elliptic functions satisfies a bunch of addition theorems which may help you to further simplify the problem. In any event, it is easier to look up those addition theorems from a list than discover it yourself. $\endgroup$ – achille hui Feb 8 '13 at 11:44

Your $C(n)$ can be expressed as an incomplete elliptic integral of the 1st and 3rd kind. I will not make all the calculations (there would be many mistakes), just tell you a way how to do it. If I say substitute $u=1-nt^2$, your integral is $$A\int\frac{du}{uy}$$ for some constant $A$, where $$y^2=(1-au)(1-bu)(1-cu)\quad(*)$$ (for some $a,b,c$). You want to know $$ \int_0^d(\frac{1}{uy}-\frac{1}{u})du=\int_0^d\frac{1-y}{u}\frac{du}{y}=\int_0^d\frac{1-y-eu}{u}\frac{du}{y}+e\int_0^d\frac{du}{y}$$ for some $d$ ($u=0$ corresponds to $\sin^2\phi=1/n$, and I removed the logarithmic singularity) and for any $e$. The last integral is elliptic of the first kind. The next to last is elliptic of the 3rd kind in the broad sense (for any $e$), as $\frac{1-y-eu}{u}$ has two simple poles on the elliptic curve $(*)$; if you want to put it the the normal form, choose $e$ so that the line $1-y-eu=0$ is tangent to the curve $(*)$ (to get a double zero) and use new coordinates $\tilde u=u/(1-y-eu)$, $\tilde y= d\tilde u/(du/y)$.

(there is certainly a much more practical way of doing it - I'm more of a theorist :)

  • $\begingroup$ Thank you. The first transformation was what I did to arrive at the last expression in my question post. I would like to follow the next steps you gave me, but before that - is it no problem that the constant $A$ (and the one that I already have) disappeared in your following steps? Wouldn't I have $\int_0^d(\frac{B}{uy}-\frac{1}{u})du$, would it still work? $\endgroup$ – Andreas Finke Oct 3 '12 at 16:59
  • $\begingroup$ @AndreasFinke: sure, all those integrals should be multiplied by $A$ $\endgroup$ – user8268 Oct 3 '12 at 17:45
  • $\begingroup$ (I cannot notify user8268, can I?) For the second to last integral I obtained $ \int_\infty^{1/(1-e)} \frac{C d\tilde u}{\tilde u \sqrt{(n + 2 n \tilde u)^2 + (1 + \tilde u)^2 (1 + 6 \tilde u + {\tilde u}^2) - 2 n (1 + 6 \tilde u + 7 {\tilde u}^2 + 2 {\tilde u}^3)}} $ - this is according to Mathematica not just of the 3rd kind and in fact so "complicated" that even with the divergence removed I cannot evaluate the result (i.e., taking the limit to the upper integration limit, $1/(1-e)$, corresp. to $u=1$) $\endgroup$ – Andreas Finke Oct 10 '12 at 6:13

I finally after all this time found the answer by simply using the second identity on http://functions.wolfram.com/EllipticIntegrals/EllipticPi3/17/01/ which generally expresses the incomplete elliptic integral of the 3rd kind with one term of an incomplete integral of the 3rd kind, one term of an incomplete integral of the 1st kind and one logarithm, which cancels the logarithm in my limit. The result has whence indeed the form suggested by user8268.


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