# 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? Jul 24, 2016 at 19:08
• One thing which might be useful: $$\int_0^1 \sqrt{\frac{1}{(1 - t^2)^2} - \frac{(n + 1)^2 t^{2n}}{( 1 - t^{2n+2} )^2}}dt=\int_1^\infty \sqrt{\frac{1}{(1 - t^2)^2} - \frac{(n + 1)^2 t^{2n}}{( 1 - t^{2n+2} )^2}}dt$$ Jul 25, 2016 at 6:21
• the integrand gets peaked quiet strongly for large $n$, so maybe you should think about kind of an asymptotic expansion! Jul 25, 2016 at 8:18
• Have you considered a Taylor-MacLaurin series term by term integration on the interval [0,1]? Jul 26, 2016 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. Jul 26, 2016 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 $n>2$ Jul 27, 2016 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). Jul 27, 2016 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. :) Jul 27, 2016 at 23:36