True value of $\int_0^{\pi/2}\frac{dx}{(1-2\sin^2 x)\sqrt{1-4\sin^2 x}}$

Question

During the generation of test values I stumbled over this question

What is the true value of the complete elliptic integral of the 3rd kind $$I=\int_0^{\pi/2}\frac{dx}{(1-2\sin^2 x)\sqrt{1-4\sin^2 x}}?$$

This is a special case of the $\Pi$ integral (see e.g. https://dlmf.nist.gov/19.2.E7) and the Cauchy principal value has to be taken.

Here are my experiments with some well-known resources:

With Wolfram/Mathematica style elliptic integrals (using parameters) the value should be $$I=\Pi(2|4)=\mathtt{EllipticPi}[2,4] \approx -0.36396057469 - 0.53912891187i$$

With Maple style integrals (using modulus) it should be $$I=\Pi(2,2)=\mathtt{EllipticPi}(2,2) \approx 1.20683575210 - 0.53912891187 i$$ (Note that the imaginary parts are the same).

In order to resolve the different results, I asked Wolfram Alpha to evaluate the integral using

int(1/((1-2*sin(x)^2)*sqrt(1-4*sin(x)^2)), x=0..Pi/2) with 30 digits

as input and got the very disturbing different (!?) results

Computation result:    1.05574680871 - 3.53800841627 i 
Decimal approximation: 1.11534422571 + 1.91703566486 i

Pari/GP 2.9.4 gives another numerical integration result

? intnum(x=0,Pi/2,1/((1-2*sin(x)^2)*sqrt(1-4*sin(x)^2)))
%1 = 1.046038054724273208736077496 + 4.797148009326248296 E27*I

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Edit (May 2018): I did not expect an answer after two month, therefore I update the question. Since I am interested in the real part only, I meanwhile use the following formulas:

For $|k|>1, \nu \ne k^2$ the following formula (derived from http://functions.wolfram.com/08.06.04.0029.01 with $x=\pi/2$): $$ \Re \Pi(\nu,k) = \frac{1}{k}\; \Pi\left(\frac{\nu}{k^2} , \frac{1}{k}\right) - \Re \left(\frac{\pi}{2} \; \sqrt{\frac{\nu}{(\nu - 1)(k^2 - \nu)}}\;\;\right) $$ (the second term is zero for $k^2 < \nu$), and for $\nu=k^2 > 1$ the result is (c.f. http://functions.wolfram.com/08.03.03.0003.01): $$ \Pi(k^2,k)\ = \frac{E(k)}{1-k^2}\;\cdot $$

Answer 1

I don't think you can get a correct numerical estimate if you don't specify that you're computing the principal value.

Using the parameter notation, $$\operatorname{v.\!p.} \int_0^{\pi/2} \frac {dx} {(1 - n \sin^2 x) \sqrt {1 - m \sin^2 x}} = \\ \frac 1 {\sqrt m} \operatorname\Pi \left( \frac n m, \frac 1 m \right) + \frac i {(n - 1) \sqrt m} \left( \operatorname K \left( \frac {m - 1} m \right) - \operatorname\Pi \left( \frac {n - 1} n, \frac {m - 1} m \right) \right), \\ m > n > 1.$$ There is no ambiguity because the principal value is well-defined and the elliptic integrals on the rhs do not depend on the analytic continuation choice.

Answer 2

To see why a p.v. integral doesn't coincide with an analytic continuation, consider $f(z) = \int_1^z dt/t, \;z \notin (-\infty, 0]$. For $z<0$, we'll have $$f(z+i0) = \lim_{\epsilon \downarrow 0} \left( \int_1^\epsilon + \int_{C_\epsilon^+} + \int_{-\epsilon}^z + \int_z^{z+i \epsilon} \right) \frac {dt} t = \\ \operatorname{v.\!p.} \int_1^z \frac {dt} t + i \pi \operatorname{Res}_{t=0} \frac 1 t = \operatorname{v.\!p.} \int_1^z \frac {dt} t + i \pi, \\ f(z-i0) = \operatorname{v.\!p.} \int_1^z \frac {dt} t - i \pi,$$ where $C_\epsilon^+$ denotes a half-circle with the radius $\epsilon$ in the upper half-plane around the origin. Whether we define $f(z)$ to be continuous from above or from below on the negative real axis, the continuation will not coincide with the p.v. integral.

Similar reasoning can be applied to a limit wrt a parameter. For the integral in question, we have $$i \pi \frac 1 {\sqrt {1-4\sin^2 \left(\frac \pi 4\right)}} \operatorname{Res}_{x=\pi/4} \frac 1 {1-2\sin^2x} = -\frac \pi 2.$$ If $\Pi(n,m)$ is defined to coincide with the integral when $ n \notin [1, \infty)$ and to be continuous from below on $(1, \infty)$, the integral will be equal to $\Pi(2,4)+\pi/2$.

Answer 3

To compute the principal value, we can use contour integration.

Using the substitution $z=e^{ix}$, we have $$\newcommand{\PV}{\operatorname{PV}} \begin{align} \PV\int_0^{\pi/2}\frac{\mathrm{d}x}{\left(1-2\sin^2(x)\right)\sqrt{1-4\sin^2(x)}} &=\PV\int_0^{\pi/2}\frac{\mathrm{d}x}{\cos(2x)\sqrt{2\cos(2x)-1}}\\ &=\PV\int_0^\pi\frac{\mathrm{d}x}{2\cos(x)\sqrt{2\cos(x)-1}}\\ &=\PV\frac12\int_0^{2\pi}\frac{\mathrm{d}x}{2\cos(x)\sqrt{2\cos(x)-1}}\\ &=\frac1{2i}\oint_\gamma\frac{\mathrm{d}z}{\left(z^2+1\right)\sqrt{z+1/z-1}}\\ &=\frac1{2i}\oint_\gamma\sqrt{\frac{z}{z^2-z+1}}\frac{\mathrm{d}z}{z^2+1}\tag1 \end{align} $$ where $\gamma$ is the counterclockwise circle of radius $1$ with two infinitesimal green semi-circular detours around the singularities and two infinitesimal red semi-circular detours around the branch cuts $X=e^{\pi i/3}[0,1]\cup e^{-\pi i/3}[1,\infty]$ :

Since the residues at $i$ and $-i$ are negatives, the contributions from the infinitesimal green semi-circles cancel. The integrals along the infinitesimal red semi-circles are bounded by $O\!\left(\sqrt{r}\right)$ as $r\to0$, so those integrals vanish.

$f(z)=\sqrt{\frac{z}{z^2-z+1}}$ is well-defined on $\mathbb{C}\setminus X$ if we set $f(1)=1$. On $|z|=1$, $f(z)\gt0$ for $\operatorname{Re}(z)\gt\frac12$ and $i\,f(z)\gt0$ for $\operatorname{Re}(z)\lt\frac12$.

The contour above can be contracted about the finite part of the branch cut, $e^{\pi i/3}[0,1]$:

On the right side of the branch cut, using the substitution $z=te^{\pi i/3}$ and the formula $$ \sqrt{a+bi\vphantom{b^2}}= \sqrt{\frac{a+\sqrt{a^2+b^2}}2}+i\operatorname{sgn}(b)\sqrt{\frac{-a+\sqrt{a^2+b^2}}2}\tag2 $$ we have $$ \begin{align} f(z) &=f\!\left(te^{\pi i/3}\right)\\[9pt] &=\sqrt{\frac{t}{1-t^3}\left(\frac{1-t}2+i\sqrt3\frac{1+t}2\right)}\\[3pt] &=\frac12\sqrt{\frac{t}{1-t^3}}\left(\sqrt{1-t+2\sqrt{1+t+t^2}}+i\sqrt{-1+t+2\sqrt{1+t+t^2}}\right)\\[9pt] &=g(t)\tag3 \end{align} $$ Notice that $f(z)=g(t)$ is in the first quadrant.

On the left side of the branch cut, $f(z)$ is the negative of that given above. Then $f(z)=-g(t)$ is in the third quadrant.

Furthermore, on the right side of the branch cut $$ \begin{align} \frac{\mathrm{d}z}{z^2+1} &=\frac{e^{\pi i/3}\mathrm{d}t}{t^2e^{2\pi i/3}+1}\\ &=\left(\frac{1+t^2}2+i\sqrt3\frac{1-t^2}2\right)\frac{\mathrm{d}t}{1-t^2+t^4}\\[6pt] &=h(t)\,\mathrm{d}t\tag4 \end{align} $$ on the left side of the branch cut, the direction is reversed, so the sign of $h(t)\,\mathrm{d}t$ is negated.

Thus, the integral on both sides of the branch cut is twice the integral on the right side. Therefore, $$ \begin{align} \frac1{2i}\oint_\gamma\sqrt{\frac{z}{z^2-z+1}}\frac{\mathrm{d}z}{z^2+1} &=2\cdot\frac1{2i}\int_0^1g(t)\,h(t)\,\mathrm{d}t\\ &=-i\int_0^1g(t)\,h(t)\,\mathrm{d}t\tag5 \end{align} $$ Using $(3)$, $(4)$, and $(5)$, we can compute the integral numerically: $$ 1.2068357521005973203335-0.5391289118749108088597i\tag6 $$