A tricky integral - $\int_0^1 \sqrt{\frac{1}{(1-t^2)^2}-\frac{(n+1)^2t^{2n}}{(1-t^{2n+2})^2}}dt$

$$\mathbf{\mbox{Evaluate:}}\qquad \int_{0}^{1} \sqrt{\frac{1}{\left(1 - t^{2}\right)^2} - \frac{\left(n + 1\right)^{2}\,t^{2n}}{\left(\, 1 - t^{2n+2}\,\,\right)^{2}}} \,\,\mathrm{d}t$$ where $n$ is any positive integer.

Introduction: This integral came up while studying the distribution of the roots of random polynomials - and I can't crack it. It seems impervious to methods of integration I know. Neither Mathematica nor Wolfram-Alpha could find a closed form, not only for this general integral, but any special case of $n>1$.

My attempt:

For $n=1$, the integral is pretty trivial to compute - expanding the integrand gives: $$\int_0^1 \sqrt{\frac{1}{t^4-2 t^2+1}-\frac{4 t^2}{t^8-2 t^4+1}}$$ Which simplifies quite easily to: $$\int_0^1 \frac{1}{t^2+1}$$ The antiderivative of the integrand is $\tan^{-1}{t}$. Evaluating at the limits gives: $$\int_0^1 \sqrt{\frac{1}{t^4-2 t^2+1}-\frac{4 t^2}{t^8-2 t^4+1}}=\frac{\pi}{4}-0=\frac{\pi}{4}$$ However, this method does not work for $n>1$, and niether does any method I know of.

Numerical values: Listed below are the approximate numerical values for this integral. Neither Wolfram Alpha nor the Inverse Symbolic calculator were able to find closed forms for these numbers.

$$n=2 \qquad 1.01868$$ $$n=3 \qquad 1.17241$$ $$n=4 \qquad 1.28844$$ $$n=5 \qquad 1.38198$$ $$n=6 \qquad 1.46049$$

Any help on this integral would be greatly appreciated. Thank you!

• @Qwerty Are you sure you typed it in right? I've tried it multiple times and it simply spits the integral back at me. Could you link to your input? – TreFox Jul 24 '16 at 19:08
• One thing which might be useful: $$\int_0^1 \sqrt{\frac{1}{(1 - t^2)^2} - \frac{(n + 1)^2 t^{2n}}{( 1 - t^{2n+2} )^2}}dt=\int_1^\infty \sqrt{\frac{1}{(1 - t^2)^2} - \frac{(n + 1)^2 t^{2n}}{( 1 - t^{2n+2} )^2}}dt$$ – Yuriy S Jul 25 '16 at 6:21
• the integrand gets peaked quiet strongly for large $n$, so maybe you should think about kind of an asymptotic expansion! – tired Jul 25 '16 at 8:18
• Have you considered a Taylor-MacLaurin series term by term integration on the interval [0,1]? – Mathemagician1234 Jul 26 '16 at 18:09
• This is known as Kac formula. There is an asymptotic expression in the link, with the 1st term derived by Kac himself. – Wiley Jul 26 '16 at 18:49

It appears that the integral when $n=2$ can be represented in terms of elliptic integrals:
$$I(2)=\frac{\pi}{2}-\frac{1}{\sqrt{6}}\left(\Pi\left(\frac23\mid\frac13\right)-K\left(\frac13\right)\right).$$
$$K(m)=\int^{\pi/2}_{0}\frac{d\theta}{\sqrt{1-m\sin^2\theta}}$$ and $$\Pi(n\mid m)=\int^{\pi/2}_{0}\frac{d\theta}{(1-n\sin^2\theta)\sqrt{1-m\sin^2\theta}}.$$
• Well, if $I(2)$ is elliptic, I would't hold much hope for $n>2$ – Yuriy S Jul 27 '16 at 9:09
• @TreFox extend the integral to the whole real line, and integrate along a half-circle in the upper half complex plane. The integral boils down to two residues (which gives the $\pi/2$ part) and a integral along a branch cut from $(\sqrt{3}-1)i/\sqrt{2}$ to $(\sqrt{3}+1)i/\sqrt{2}$ (which gives the elliptic part). – Chen Wang Jul 27 '16 at 15:10
• @You'reInMyEye Actually, $I{(3)}$ is surprisingly elliptic, and I strongly suspect $I{(4)}$ will turn out to be elliptic as well using similar techniques as for the $n=3$ case. For the $n>4$ case, however, I have abandoned all hope. :) – David H Jul 27 '16 at 23:36