How do solve this integral $\int_{-1}^1\frac{1}{\sqrt{1-x^2}}\arctan\frac{11-6\,x}{4\,\sqrt{21}}\mathrm dx$?

Question

I need to solve the to following integral: $$\int_{-1}^1\frac{1}{\sqrt{1-x^2}}\arctan\frac{11-6\,x}{4\,\sqrt{21}}\mathrm dx.$$

I tried this integral in Mathematica, but it was not able to solve it. An approximate numeric integration gives $1.6449340668482264364724151666460251892189...$ that is close to $\frac{\pi^2}6$. But when I tried to increase the precision above 60 decimal digits, I began to see a tiny difference, which could be interpreted either as a numerical algorithm glitch, or as $\frac{\pi^2}6$ being just an accidentally close value and not the exact answer. Indeed, $\frac{\pi^2}6$ would be a suspiciously nice result for this integral. Anyway, I need your help with this.

Answer 1

I think the most ecological approach to the problem is as follows:

1. Denote $x=\sin\varphi$ and recall that $\arctan x=\frac{1}{2i}\ln\frac{1+ix}{1-ix}$. One then obtains the integral $$\frac{1}{2i}\int_{-\pi/2}^{\pi/2}\ln\frac{4\sqrt{21}+i(11-6\sin\varphi)}{4\sqrt{21}-i(11-6\sin\varphi)}d\varphi=\frac{1}{4i}\int_{0}^{2\pi}\ln\frac{4\sqrt{21}+i(11-6\cos\varphi)}{4\sqrt{21}-i(11-6\cos\varphi)}d\varphi,\tag{1}$$ where at the last step we first used the symmetry of sine function to extend the integration to interval $[-\pi,\pi]$, and then made use of periodicity to shift the integration interval and to replace $\sin$ by $\cos$.

2. There is a well-known integral (see, for example, here) $$\int_{0}^{2\pi}\ln\left(1+r^2-2r\cos\varphi\right)d\varphi=\begin{cases} 0, &\text{for}\; |r|<1,\\ 2\pi\ln r^2, &\text{for}\; |r|>1. \end{cases}\tag{2}$$ Obviously, our integral above is a difference of two integrals of this type. Some care should be taken over multivaluedness of logarithms. This can be handled by saying that the arguments of $4\sqrt{21}\pm i(11-6\cos\varphi)$ belong to $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.

3. Now we have \begin{align} 4\sqrt{21}\pm i(11-6\cos\varphi)=A_{\pm}\left(1+r_{\pm}^2-2r_{\pm}\cos\varphi\right) \end{align} with $$r_{\pm}=\frac{11-4\sqrt7}{3}e^{\pm i\pi/3},\qquad A_{\pm}=(11+4\sqrt7)e^{\pm i\pi /6}.$$ Since $|r_{\pm}|<1$, the integral (1) reduces to $$\frac{1}{4i}\cdot2\pi\cdot \ln\frac{A_+}{A_-}=\frac{\pi^2}{6}.$$

Answer 2

Sketch of solution

Integrate by parts, substitute $x\mapsto-x$ in the integral over $[-1,0]$, then recombine:

$$\begin{align*} I &= \int_{-1}^1 \frac{1}{\sqrt{1-x^2}} \arctan\frac{11-6x}{4\sqrt{21}} \, dx \\ &= \frac\pi2 \arctan \frac{88\sqrt{21}}{251} + 24\sqrt{21} \int_{-1}^1 \frac{\arcsin x}{36x^2-132x+457} \, dx \\ &= \frac\pi2 \arctan \frac{88\sqrt{21}}{251} + 6336\sqrt{21} \int_0^1 \frac{x \arcsin x}{1296x^4+15480x^2+208849} \, dx \end{align*}$$

where the constant term is simplified with $\arctan u + \arctan v = \arctan \dfrac{u+v}{1-uv}$ $(*)$.

The remaining integral can be evaluated with complex analysis; see my answer here to another definite integral with the same basic structure,

$$\begin{align*} I(\alpha, \beta, \gamma) & := \int_0^1 \frac{x \arcsin x}{\alpha x^4+\beta x^2+\gamma} \, dx \\ &= \int_0^\infty \frac{y \arctan y}{ay^4+by^2+c} \, dy & x=\frac y{\sqrt{1+y^2}} \\ J(a,b,c) & := \frac12 \int_{-\infty}^\infty \frac{y \arctan y}{ay^4+by^2+c} \, dy \end{align*}$$

with $(a,b,c)=(\alpha+\beta+\gamma, \beta+2\gamma, \gamma)$, each parameter belonging to $\Bbb N$ such that $\beta^2-4\alpha\gamma<0$.

---

Evaluating $J(a,b,c)$ and $I(\alpha,\beta,\gamma)$

We introduce the function

$$f(z) = \frac{z \arctan z}{az^4+bz^2+c} = -\frac i2 \frac z{az^4+bz^2+c} \left(\log\left\lvert\frac{i-z}{i+z}\right\rvert + i \arg\left(\frac{i-z}{i+z}\right)\right),$$

to be integrated along the contour $C$ below, bypassing the branch cut taken along the positive imaginary axis:

$f(z)$ has two simple poles in the upper-half plane $\Im z>0$,

$$\omega_\pm = \pm \sqrt{\frac{-b \pm \sqrt{b^2-4ac}}{2a}}$$

where $\omega_\pm$ is the pole with $\pm$ real part. The residues are

$$\begin{align*} \underset{\omega_+}{\operatorname{Res}} f(z) &= \color{red}+ \frac1{2\sqrt{b^2-4ac}} \arctan \omega_+ \\ \underset{\omega_-}{\operatorname{Res}} f(z) &= \color{red}- \frac1{2\sqrt{b^2-4ac}} \arctan \omega_- \end{align*}$$

so by the residue theorem,

$$\begin{align*} \oint_C f(z) \, dz &= i2\pi \sum_{\omega_\pm} \operatorname{Res} f(z) \\ &= \frac{i\pi}{\sqrt{b^2-4ac}} \left(\arctan \omega_+ - \arctan \omega_-\right) \\ &= \frac{i\pi}{\sqrt{b^2-4ac}} \arctan \frac{\omega_+ - \omega_-}{1 + \omega_+\omega_-} \\ &= \frac{i\pi}{\sqrt{b^2-4ac}} \arctan \frac{\sqrt{-b-\sqrt{b^2-4ac}} + \sqrt{-b+\sqrt{b^2-4ac}}}{\sqrt2 \left(\sqrt a - \sqrt c\right)} \end{align*}$$

In the limit, the integrals along the paths to either side of the cut together contribute

$$\begin{align*} \int_{\lambda_1\cup\lambda_2} f(z) \, dz &\to \pi \int_1^\infty \frac{t}{at^4 - bt^2 + c} \, dt \\ &= \frac\pi{\sqrt{b^2-4ac}} \left(\operatorname{artanh}\frac{b-2c}{\sqrt{b^2-4ac}} - \operatorname{artanh}\frac b{\sqrt{b^2-4ac}}\right) \\ &= \frac\pi{\sqrt{b^2-4ac}} \operatorname{artanh} \frac{\sqrt{b^2-4ac}}{2a-b} \\ &= \frac{i\pi}{\sqrt{b^2-4ac}} \arctan \frac{\sqrt{-b^2+4ac}}{2a-b} \end{align*}$$

using the hyperbolic version of $(*)$, $\operatorname{artanh}u+\operatorname{artanh}v=\operatorname{artanh}\dfrac{u+v}{1\color{red}+uv}$, as well as $\operatorname{artanh}(iz)=i\arctan z$.

Hence

$$\begin{align*} J(a,b,c) &= \frac{i\pi}{2\sqrt{b^2-4ac}} \left(\arctan \frac{\sqrt{-b-\sqrt{b^2-4ac}} + \sqrt{-b+\sqrt{b^2-4ac}}}{\sqrt2 \left(\sqrt a - \sqrt c\right)} - \arctan \frac{\sqrt{-b^2+4ac}}{2a-b}\right) \\ &= \frac{i\pi}{2\sqrt{b^2-4ac}} \arctan \frac{\frac{\sqrt{-b-\sqrt{b^2-4ac}}+\sqrt{-b+\sqrt{b^2-4ac}}}{\sqrt2\left(\sqrt a-\sqrt c\right)} - \frac{\sqrt{-b^2+4ac}}{2a-b}}{1 + \frac{\sqrt{-b-\sqrt{b^2-4ac}}+\sqrt{-b+\sqrt{b^2-4ac}}}{\sqrt2\left(\sqrt a-\sqrt c\right)}\cdot\frac{\sqrt{-b^2+4ac}}{2a-b}} \end{align*}$$

$$\implies \bbox[#faa, 2pt, border:2pt solid black]{I(\alpha,\beta,\gamma) = \frac{i\pi}{2\sqrt{\beta^2-4\alpha\gamma}} \arctan \frac{\frac{\sqrt{-\beta-2\gamma-\sqrt{\beta^2-4\alpha\gamma}}+\sqrt{-\beta-2\gamma+\sqrt{\beta^2-4\alpha\gamma}}}{\sqrt2\left(\sqrt{\alpha+\beta+\gamma}-\sqrt\gamma\right)} - \frac{\sqrt{-\beta^2+4\alpha\gamma}}{2\alpha+\beta}}{1 + \frac{\sqrt{-\beta-2\gamma-\sqrt{\beta^2-4\alpha\gamma}}+\sqrt{-\beta-2\gamma+\sqrt{\beta^2-4\alpha\gamma}}}{\sqrt2\left(\sqrt{\alpha+\beta+\gamma}-\sqrt \gamma\right)}\cdot\frac{\sqrt{-\beta^2+4\alpha\gamma}}{2\alpha+\beta}}}$$

(Is there a simpler expression for this?)

$$\begin{align*} \implies I(1296, 15480, 208849) &= \frac\pi{12672\sqrt{21}} \arctan\frac{88352\sqrt{21} - 225625\sqrt3}{424871} \end{align*}$$

$$\begin{align*} \implies I &= \frac\pi2 \left(\arctan \frac{88\sqrt{21}}{251} + \arctan \frac{88352\sqrt{21}-225625\sqrt3}{424871}\right) \\ &\stackrel{!}= \frac\pi2 \arctan \sqrt3 = \boxed{\frac{\pi^2}6} \end{align*}$$