# Bizarre Definite Integral

Does the following equality hold?

$$\large \int_0^1 \frac{\tan^{-1}{\left(\frac{88\sqrt{21}}{215+36x^2}\right)}}{\sqrt{1-x^2}} \, \text{d}x = \frac{\pi^2}{6}$$

The supposed equality holds to 61 decimal places in Mathematica, which fails to numerically evaluate it after anything greater than 71 digits of working precision. I am unsure of it's correctness, and I struggle to prove it's correctness.

The only progress I have in solving this is the following identity, which holds for all real $x$:

$$\tan^{-1}{\left( \frac{11+6x}{4\sqrt{21}} \right )} + \tan^{-1}{\left( \frac{11-6x}{4\sqrt{21}} \right )} \equiv \tan^{-1}{\left(\frac{88\sqrt{21}}{215+36x^2}\right)}$$

I also tried the Euler Substitution $t^2 = \frac{1-x}{1+x}$ but it looks horrible.

Addition: Is there some kind of general form to this integral?

Side thoughts: Perhaps this is transformable into the Generalised Ahmed's Integral, or something similar.

• Here is a link:math.stackexchange.com/questions/397742/… – gymbvghjkgkjkhgfkl Feb 24 '17 at 11:05
• I'm still searching for an alternative solution too. Using the change of variable of Jack Lam it's equal to $\displaystyle 2\int_0^1 \dfrac{\arctan\left(\dfrac{88\sqrt{21}(1+x^2)^2}{251x^4+358x^2+251}\right)}{1+x^2}dx$ – FDP Feb 27 '17 at 10:28
• @FDP Is it possible to apply Double Integral Exchange (DIE) or Differntiation Under The IntEgral Sign (DUTIES) in this case? – Jack Tiger Lam Feb 27 '17 at 10:58
• The double integral you get using $\displaystyle \arctan x=\int_0^1 \dfrac{x}{1+x^2t^2}dt$ looks horrible. – FDP Feb 27 '17 at 12:08
• Maybe some magic change of variable makes equal this integral to $\displaystyle \int_0^{\tfrac{\pi}{4}} \dfrac{32x}{6} dx$ – FDP Feb 27 '17 at 12:19

As pointed out in one of the comments, user @Start wearing purple demonstrated a very general approach for solving this kind of integral, see this. As an alternative approach, let me give a different argument that appeals to a specific property satisfied by OP's integral.

Step 1. (Reduction and the main claim) We begin by substituting $x = \cos(\theta/2)$. Then the integral equals

$$\frac{1}{2} \int_{0}^{\pi} \arctan\left(\frac{88\sqrt{21}}{233+18\cos\theta}\right) \, d\theta = \frac{\pi}{4} - \frac{1}{2} \int_{0}^{\pi} \arctan\left(\frac{233+18\cos\theta}{88\sqrt{21}}\right) \, d\theta.$$

So it suffices to prove that

$$\int_{0}^{\pi} \arctan\left(\frac{233+18\cos\theta}{88\sqrt{21}}\right) \, d\theta \stackrel{?}{=} \frac{\pi^2}{6}. \tag{1}$$

To evaluate this integral, let me give the punchline.

Claim. Let $0 < a <1$ and $b > 0$ satisfy $4a^2 - b^2 = \frac{4}{3}$. Then $$\int_{0}^{\pi} \arctan(a + b\cos\theta) \, d\theta = \frac{\pi^2}{6}.$$

Notice that $(a, b) = \left( \frac{233}{88\sqrt{21}}, \frac{18}{88\sqrt{21}} \right)$ satisfies the relation in the assertion of Claim. So we focus on proving this claim.

Step 2. (Definition and properties of $I$) Now define $I(a, b)$ by

$$I(a, b) = \int_{0}^{\pi} \arctan(a + b\cos\theta) \, d\theta.$$

From the substitution $\theta \mapsto \pi - \theta$, it is clear that $I(a,-b) = I(a, b)$. Then for $0 < a < 1$ and $0 < \theta < \pi$, we have

\begin{align*} &\arctan(a + b\cos\theta) + \arctan(a - b\cos\theta) \\ &\hspace{1em}= \arctan\left( \frac{2a}{1-(a^2-b^2\cos^2\theta)} \right) \\ &\hspace{2em}= \arctan\left( \frac{4a}{2-2a^2+b^2+b^2\cos(2\theta)} \right) \\ &\hspace{3em}= \frac{\pi}{2} - \arctan\left( \frac{2-2a^2+b^2}{4a} + \frac{b^2}{4a}\cos(2\theta) \right). \end{align*} Plugging this back and exploiting the symmetry of cosine, we have

$$I(a, b) = \frac{\pi^2}{4} - \frac{1}{2}I\left( \frac{2-2a^2+b^2}{4a}, \frac{b^2}{4a} \right). \tag{2}$$

Step 3. Now here comes the central observation. Let $(a, b)$ satisfy $0 < a < 1$ and $b > 0$, and define the sequence $(a_n, b_n)$ recursively by

$$(a_0, b_0) = (a, b), \qquad (a_{n+1}, b_{n+1}) = \left( \frac{2-2a_n^2+b_n^2}{4a_n}, \frac{b_n^2}{4a_n} \right).$$

Observation. Assume that $4a^2 - b^2 = \frac{4}{3}$. Then for all $n \geq 0$ we have $$\frac{1}{\sqrt{3}} \leq a_{n+1} \leq a_n, \qquad 4a_n^2 - b_n^2 = \frac{4}{3}.$$

The proof is a tedious algebra, so we skip this. Now by this observation, we have $|a_n| < 1$ for all $n$. Then a recursive application of $\text{(2)}$ gives

$$I(a, b) = \frac{\pi^2}{4}\sum_{k=0}^{n-1} \left(-\frac{1}{2}\right)^k + \left(-\frac{1}{2}\right)^n I(a_n, b_n).$$

Since $|I(a_n, b_n)| \leq \frac{\pi^2}{2}$ for all $n$, taking limist as $n\to\infty$ proves the claim.

Remark. (1) The condition $4a^2 - b^2 = \frac{4}{3}$ is crucial for our proof. For arbitrary starting point $(a, b)$, the sequence $(a_n, b_n)$ is dynamically unstable and hence the formula $\text{(2)}$ is not applicable.

(2) The claim is true for any $a > 0$ in view of the principle of analytic continuation.

(3) Again, @Start wearing purple's computation gives a more general result with a relatively economic computation: for all $a, b \in \Bbb{R}$,

$$\int_{0}^{\pi} \arctan(a + b\cos\theta) \, d\theta = \pi \arg \left(1 + ia + \sqrt{b^2 + (1+ia)^2}\right). \tag{3}$$

This follows from the formula

$$\int_{0}^{\pi} \log(1 + s \cos\theta) \, d\theta = \pi \log\left( \frac{1 + \sqrt{1-s^2}}{2} \right)$$

which is valid for any complex $s$ with $|s| < 1$. Our relation $4a^2 - b^2 = \frac{4}{3}$ ensures that the RHS of $\text{(3)}$ is always $\frac{\pi^2}{6}$, since $1 + ia + \sqrt{b^2 + (1+ia)^2} = (1+\sqrt{3}a)\left( 1 + \frac{i}{\sqrt{3}} \right)$.

• Amazing! How did you come up with the remarkable recurrence relation?! – Jack Tiger Lam Apr 26 '17 at 3:26
• @JackLam, Obviously my beginning point is the formula $\text{(2)}$. The value of $I$ is exactly $\pi^2/6$ when this formula is repeatedly applicable. That fact that $I$ is not constant for general $(a, b)$ means that this is only possibly true along the level curve $C : I(a, b) = \pi^2/6$. Then $(a_n, b_n)$ will be a sequence of points on $C$. I tried to figure out this curve by taming the dynamic instability of $(a_n)$, and luckily I succeeded. – Sangchul Lee Apr 26 '17 at 3:45