I am interested in finding the analytic expression of the following integral: $$ I(k)=\int_0^1\frac{P_{n}(t)\sqrt{1-t^2}}{\sqrt{1-k^2t^2}}dt, $$ where $P_{n}(t)$ is an even polynomial: $P_{n}(t)=\sum_{i=0}^n a_it^{2i}$, and $k$ is a real number ($|k|<1$). From general considerations the integral can be reduced to some algebraic expression involving complete elliptic integrals of the first and second kind, $E(k)$ and $K(k)$. Is there a simple way of the reduction?

  • $\begingroup$ might in be that the $k^2$ is in the numerator? otherwise we don't get complete elliptic integrals $\endgroup$ – tired Dec 15 '16 at 13:18
  • $\begingroup$ We may set $$ J_n = \int_{0}^{1}\frac{t^{2n}\sqrt{1-t^2}}{\sqrt{1-k^2 t^2}} $$ and use integration by parts to derive a (two-terms) recurrence relation for the $\{J_n\}_{n\geq 0}$ sequence. Since $J_0$ and $J_1$ only depend on $K(k)$ and $E(k)$, hopefully every $J_n$ depends on $K(k)$ and $E(k)$ in a simple way. $\endgroup$ – Jack D'Aurizio Dec 15 '16 at 13:41
  • $\begingroup$ Why not?! For example for $n=0$ one easily finds: $I(k)=\frac{1}{k^2}E(k)+\left(1-\frac{1}{k^2}\right)K(k)$. $\endgroup$ – user Dec 15 '16 at 13:50

Let we set: $$ J_n = \int_{0}^{1}\frac{t^{2n}\sqrt{1-t^2}}{\sqrt{1-k^2 t^2}} = \int_{0}^{\pi/2}\frac{\sin^{2n}(\theta)\cos^2(\theta)}{\sqrt{1-k^2\sin^2\theta}}\,d\theta \tag{1}$$ $$ H_n = \int_{0}^{\pi/2}\frac{\sin^{2n}(\theta)\,d\theta}{\sqrt{1-k^2\sin^2\theta}}.\tag{2} $$ We have $J_n=H_n-H_{n+1}$ and: $$ \sum_{n\geq 0}H_n z^n = \int_{0}^{\pi/2}\frac{d\theta}{(1-z\sin^2\theta)\sqrt{1-k^2\sin^2\theta}}=\int_{0}^{+\infty}\frac{dt}{(1+(1-z)t^2)\sqrt{1-\frac{k^2 t^2}{1+t^2}}} $$ that is a complete elliptic integral of the third kind, $\Pi\left(z;\frac{\pi}{2},k\right)$, also denoted as $\Pi(z\,|\,k)$. Since $$ \Pi(n\,|\,m) = \sum_{k\geq 0}\sum_{j=0}^{k}\binom{2j}{j}\binom{2k}{k}\frac{m^j n^{k-j}}{4^{k+j}} \tag{3} $$ as soon as $|n|,|m|<1$, we have: $$ H_n = \sum_{\tau\geq 0}\binom{2\tau}{\tau}\binom{2n+2\tau}{n+\tau}\frac{k^{\tau}}{4^{n+2\tau}} \tag{4} $$ a not-so-horrible series whose general term behaves like $\frac{k^\tau}{\pi\sqrt{\tau(n+\tau)}}$. It is a hypergeometric series: $$ H_n = \frac{1}{4^n}\binom{2n}{n}\;\phantom{}_2 F_1\left(\frac{1}{2},n+\frac{1}{2};n+1;k\right)\tag{5} $$ that can be approximated with great accuracy through a continued fraction. The same applies to: $$ J_n = \frac{1}{4^n(2n+2)}\binom{2n}{n}\;\phantom{}_2 F_1\left(\frac{1}{2},n+\frac{1}{2};n+2;k\right).\tag{6} $$

  • $\begingroup$ For the interested reader, a Gauss hypergeometric series also satisfies a certain ODE in $k$; this is particularly useful in the vicinity of $k=1$, where said ODE has a regular singular point leading to logarithmic behavior of the $_2F_1$ function. $\endgroup$ – Semiclassical Dec 15 '16 at 14:28
  • $\begingroup$ @Semiclassical: that is a very interesting observation. $\endgroup$ – Jack D'Aurizio Dec 15 '16 at 14:31
  • $\begingroup$ Forgot to add: As an instance of such, see the discussion in this question. $\endgroup$ – Semiclassical Dec 15 '16 at 14:35
  • $\begingroup$ @JackD'Aurizio This seems like a very round-about way to end up with a hypergeometric function, no? You could simply substitute $t=\sqrt{u}$, and use Euler's integral representation. $\endgroup$ – David H Dec 15 '16 at 14:38
  • $\begingroup$ @DavidH: sure, my "presentation" is probably very sub-optimal, my goal was just to reach $(5)$ and state "this is a good representation" :D $\endgroup$ – Jack D'Aurizio Dec 15 '16 at 14:43

As was pointed out by JackD'Aurizio, the proof is based on finding the recurrence relation connecting $J_n$. As follows from the above discussion it is better to work directly with functions $$ H_{n}(k)=\int_0^1\frac{t^{2n}\,dt}{\sqrt{1-t^2}\sqrt{1-kt^2}}.$$ We are going to prove that $$ H_{n}(k)=Q_n(k) K(k)+R_n(k) E(k), $$ where $Q_n$ and $R_n$ are some rational polynomials of $k$, $K(k)$ and $E(k)$ being the complete elliptic integrals of the first and second kind, respectively. Obviously $H_0(k)=K(k)$ and $H_1(k)=\frac{K(k)-E(k)}{k}$, so that we need to consider only the case $n\ge2$.

Introducing for simplicity function $\rho_k=(1-kt^2)^\frac{1}{2}$ we may write: $$ H_n=\int_0^1\frac{t^{2n}}{\rho_1\rho_k}dt=-\int_0^1\frac{t^{2n-1}}{\rho_k}d\rho_1=\int_0^1\rho_1 d\frac{t^{2n-1}}{\rho_k}= \int_0^1\rho_1\left[\frac{(2n-1)t^{2n-2}}{\rho_k}+\frac{kt^{2n}}{\rho_k^3}\right]dt=\int_0^1\rho_1\left[\frac{(2n-2)t^{2n-2}}{\rho_k}+\frac{t^{2n-2}}{\rho_k^3}\right]dt=(2n-2)(H_{n-1}-H_n)+\int_0^1\rho_1\frac{t^{2n-2}}{\rho_k^3}dt. $$ It follows then: $$ (2n-1)H_n-(2n-2)H_{n-1}=\int_0^1\rho_1\frac{t^{2n-2}}{\rho_k^3}dt= \frac{1}{k}\int_0^1\rho_1t^{2n-3}d\frac{1}{\rho_k}=-\frac{1}{k}\int_0^1\frac{1}{\rho_k}d(\rho_1t^{2n-3})=-\frac{1}{k}\int_0^1\left[\frac{\rho_1(2n-3)t^{2n-4}}{\rho_k}-\frac{t^{2n-2}}{\rho_1\rho_k}\right]dt=-\frac{1}{k}\left[(2n-3)(H_{n-2}-H_{n-1})-H_{n-1}\right]= \frac{(2n-2)H_{n-1}-(2n-3)H_{n-2}}{k}. $$ Finally one obtains the recurrence relation: $$ (2n-1)H_n-(2n-2)(1+k^{-1})H_{n-1}+(2n-3)k^{-1}H_{n-2}=0. $$

From this one sees immediately that $$ H_n=\frac{Q_n(k) K(k)-R_n(k) E(k)}{(2n-1)!!k^n}, $$ where $Q_n$ and $R_n$ are polynomials of degree $n-1$ with integer coefficients.

Besides, it follows that a similar relation connecting $_2F_1(\frac{1}{2},n+\frac{1}{2};n+1,k)$ with $_2F_1(\frac{1}{2},\frac{1}{2};1,k)$ and $_2F_1(\frac{1}{2},-\frac{1}{2};1,k)$ is valid.


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