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.

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    $\begingroup$ Here is a link:math.stackexchange.com/questions/397742/… $\endgroup$ Feb 24, 2017 at 11:05
  • 2
    $\begingroup$ I'm still searching for an alternative solution too. Using the change of variable of Jack Lam it's equal to $\displaystyle 2\int_0^1 \dfrac{\arctan\left(\dfrac{88\sqrt{21}(1+x^2)^2}{251x^4+358x^2+251}\right)}{1+x^2}dx$ $\endgroup$
    – FDP
    Feb 27, 2017 at 10:28
  • $\begingroup$ @FDP Is it possible to apply Double Integral Exchange (DIE) or Differntiation Under The IntEgral Sign (DUTIES) in this case? $\endgroup$ Feb 27, 2017 at 10:58
  • $\begingroup$ The double integral you get using $\displaystyle \arctan x=\int_0^1 \dfrac{x}{1+x^2t^2}dt$ looks horrible. $\endgroup$
    – FDP
    Feb 27, 2017 at 12:08
  • $\begingroup$ Maybe some magic change of variable makes equal this integral to $\displaystyle \int_0^{\tfrac{\pi}{4}} \dfrac{32x}{6} dx$ $\endgroup$
    – FDP
    Feb 27, 2017 at 12:19

3 Answers 3


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)^2} \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 $1 + ia + \sqrt{b^2 + (1+ia)^2} = (1+\sqrt{3}a)\left( 1 + \frac{i}{\sqrt{3}} \right)$.

  • 1
    $\begingroup$ Amazing! How did you come up with the remarkable recurrence relation?! $\endgroup$ Apr 26, 2017 at 3:26
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    $\begingroup$ @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. $\endgroup$ Apr 26, 2017 at 3:45
  • $\begingroup$ Suppose we let $I(a,b)=I(a,b,\theta)$ then shouldn't we have: $$I(a_n, b_n)=\frac{\pi^2}4-\frac 12I(a_{n+1},b_{n+1},2\theta)$$ or did I make an error somewhere? $\endgroup$
    – Mr Pie
    Jun 17, 2021 at 0:41
  • 1
    $\begingroup$ @MrPie, Sorry for a super late answer; I just noticed your comment. For your question, note that $$\int_{0}^{\pi}f(\cos2\theta)\,\mathrm{d}\theta=\int_{0}^{\pi}f(\cos\theta)\,\mathrm{d}\theta$$ whenever both integrals converge. $\endgroup$ Apr 9, 2023 at 2:00
  • $\begingroup$ Yes, you are correct. What you said I think I eventually noticed myself haha, a few days later from my comment after returning to your post and thinking about the problem again, but I must have not made a comment about that. $\endgroup$
    – Mr Pie
    May 4, 2023 at 9:07

Late answer, but I figured it wouldn't hurt to give my two cents for fun. We have

$$ \begin{align} I &:= \int_{0}^{1}\frac{1}{\sqrt{1-x^{2}}}\arctan\left(\frac{88\sqrt{21}}{215+36x^{2}}\right)dx \\ &= -\int_{0}^{\pi}\frac{1}{\sqrt{1-\cos^{2}\left(\frac{x}{2}\right)}}\arctan\left(\frac{88\sqrt{21}}{215+36\cos^{2}\left(\frac{x}{2}\right)}\right)d\left(\cos\left(\frac{x}{2}\right)\right) \\ &= \frac{1}{2}\int_{0}^{\pi}\arctan\left(\frac{88\sqrt{21}}{18\cos\left(x\right)+233}\right)dx \\ &= \frac{1}{2}\int_{0}^{\pi}\left(\frac{\pi}{2}-\arctan\left(\frac{233+18\cos\left(x\right)}{88\sqrt{21}}\right)\right)dx \\ &= \frac{\pi^{2}}{4}-\frac{1}{4}\int_{0}^{2\pi}\arctan\left(\frac{233+18\cos\left(x\right)}{88\sqrt{21}}\right)dx. \\ \end{align} $$ Call this integral $J$. Recall the equality $\displaystyle \arctan(\alpha)=\int_{1}^{\infty}\frac{\alpha}{\alpha^{2}+t^{2}}dt$. Using Fubini's Theorem and the complex definition of $\cos(x)$, we get $$ \begin{align} J &:= \int_{0}^{2\pi}\arctan\left(\frac{233+18\cos\left(x\right)}{88\sqrt{21}}\right)dx \\ &= \operatorname{\Large\int_{0}^{2\pi}}\operatorname{\Large\int_{1}^{\infty}}\frac{\frac{233+18\cos\left(x\right)}{88\sqrt{21}}}{\left(\frac{233+18\cos\left(x\right)}{88\sqrt{21}}\right)^{2}+t^{2}}dtdx \\ &=\operatorname{\Large\int_{1}^{\infty}}\operatorname{\Large\int_{0}^{2\pi}}\frac{\frac{233+18\cos\left(x\right)}{88\sqrt{21}}}{\left(\frac{233+18\cos\left(x\right)}{88\sqrt{21}}\right)^{2}+t^{2}}dxdt \\ &= 88\sqrt{21}\int_{1}^{\infty}\int_{0}^{2\pi}\frac{e^{ix}\left(9e^{2ix}+233e^{ix}+9\right)}{162624t^{2}e^{2ix}+4194e^{ix}+54451e^{2ix}+4194e^{3ix}+81e^{4ix}+81}dxdt \\ &= 88\sqrt{21}\int_{1}^{\infty}\oint_{C}\frac{z\left(9z^{2}+233z+9\right)}{162624t^{2}z^{2}+4194z+54451z^{2}+4194z^{3}+81z^{4}+81}\cdot\frac{1}{iz}dzdt \\ &= \frac{88\sqrt{21}}{i}\int_{1}^{\infty}\oint_{C}\frac{9z^{2}+233z+9}{81+4194z+\left(162624t^{2}+54451\right)z^{2}+4194z^{3}+81z^{4}}dzdt \end{align} $$

where we let $z = e^{ix}$ to transform the inner integral into a contour integral over $C$. This path $C$ denotes the unit circle traversed counterclockwise centered at the origin.

Next, we set the denominator equal to $0$ and solve for $z$. After doing some algebra to manipulate the equation into a quadratic and using a calculator, the Quadratic Formula yields these simple poles:

$$ \begin{align} z_1 &:= \frac{1}{18}\sqrt{-324-162624\left(t+\frac{233i}{88\sqrt{21}}\right)^{2}}+\frac{44i}{3}\sqrt{\frac{7}{3}}t-\frac{233}{28}; \\ z_2 &:=-\frac{1}{18}\sqrt{-324-162624\left(t+\frac{233i}{88\sqrt{21}}\right)^{2}}+\frac{44i}{3}\sqrt{\frac{7}{3}}t-\frac{233}{28}; \\ z_3 &:= \frac{1}{18}\sqrt{-324-162624\left(t-\frac{233i}{88\sqrt{21}}\right)^{2}}-\frac{44i}{3}\sqrt{\frac{7}{3}}t-\frac{233}{28}; \\ z_4 &:= -\frac{1}{18}\sqrt{-324-162624\left(t-\frac{233i}{88\sqrt{21}}\right)^{2}}-\frac{44i}{3}\sqrt{\frac{7}{3}}t-\frac{233}{28}. \end{align} $$

Notice for $t \in [1,\infty)$ that only $z_1$ and $z_3$ are contained in the interior of the disk bounded by $C$. Let $\displaystyle f(z)=\frac{9z^{2}+233z+9}{\left(z-z_{1}\right)\left(z-z_{2}\right)\left(z-z_{3}\right)\left(z-z_{4}\right)}$. Using Cauchy's Integral Formula, we get $J$ to be

$$ \begin{align} J &= \frac{88\sqrt{21}}{i}\int_{1}^{\infty}\oint_{C}\frac{9z^{2}+233z+9}{\left(z-z_{1}\right)\left(z-z_{2}\right)\left(z-z_{3}\right)\left(z-z_{4}\right)}dzdt \\ &= \frac{88\sqrt{21}}{i}\int_{1}^{\infty}2\pi i\left(\operatorname{Res}\left(f(z),z=z_{1}\right)+\operatorname{Res}\left(f(z),z=z_{3}\right)\right)dt \\ &= 176\sqrt{21}\pi\int_{1}^{\infty}\left(\lim_{z \to z_1}\left(z-z_{1}\right)f(z)+\lim_{z \to z_3}\left(z-z_{3}\right)f(z)\right)dt \\ &= \frac{44\sqrt{21}\pi}{9i}\operatorname{\Large\int_{1}^{\infty}}\left(\frac{1}{\sqrt{1+\left(\frac{\sqrt{162624}}{18}t+\frac{233}{18}i\right)^{2}}}-\frac{1}{\sqrt{1+\left(\frac{\sqrt{162624}}{18}t-\frac{233}{18}i\right)^{2}}}\right)dt. \\ \end{align} $$

The following indefinite integrals should be trivial:

$$\operatorname{\Large \int}\frac{dt}{\sqrt{1+\left(\frac{\sqrt{162624}}{18}t+\frac{233}{18}i\right)^{2}}}=\frac{3}{44}\sqrt{\frac{3}{7}}\operatorname{arcsinh}\left(\frac{44\sqrt{21}}{9}t+\frac{233}{18}i\right)+C$$ $$\operatorname{\Large \int}\frac{dt}{\sqrt{1+\left(\frac{\sqrt{162624}}{18}t-\frac{233}{18}i\right)^{2}}}=\frac{3}{44}\sqrt{\frac{3}{7}}\operatorname{arcsinh}\left(\frac{44\sqrt{21}}{9}t-\frac{233}{18}i\right)+C.$$

By the Fundamental Theorem for Contour Integrals, we get

$$ \begin{align} J =& -\frac{44\sqrt{21}\pi}{9i}\cdot\frac{3}{44}\sqrt{\frac{3}{7}}\left[\operatorname{arcsinh}\left(\frac{44\sqrt{21}}{9}t+\frac{233}{18}i\right)-\operatorname{arcsinh}\left(\frac{44\sqrt{21}}{9}t-\frac{233}{18}i\right)\right]_{1}^{\infty} \\ =& -\frac{44\sqrt{21}\pi}{9i}\cdot\frac{3}{44}\sqrt{\frac{3}{7}}\left[2i\Im\operatorname{arcsinh}\left(\frac{44\sqrt{21}}{9}t+\frac{233}{18}i\right)\right]_{1}^{\infty} \\ =& -2\pi\left[\operatorname{arg}\left(\frac{i}{18}\left(233-88i\sqrt{21}t\right)+\sqrt{1+\frac{1}{324}\left(88\sqrt{21}t+233i\right)^{2}}\right)\right]_{1}^{\infty} \\ =& -2\pi \lim_{t \to \infty}\operatorname{arg}\left(\frac{i}{18}\left(233-88i\sqrt{21}t\right)+\sqrt{1+\frac{1}{324}\left(88\sqrt{21}t+233i\right)^{2}}\right) \\ &+2\pi \operatorname{arg}\left(\frac{i}{18}\left(233-88i\sqrt{21}\right)+\sqrt{1+\frac{1}{324}\left(88\sqrt{21}+233i\right)^{2}}\right) \\ &= -2\pi\cdot0+2\pi\arctan\left(\frac{\Re\left(\frac{i}{18}\left(233-88i\sqrt{21}t\right)+\sqrt{1+\frac{1}{324}\left(88\sqrt{21}t+233i\right)^{2}}\right)}{\Im\left(\frac{i}{18}\left(233-88i\sqrt{21}t\right)+\sqrt{1+\frac{1}{324}\left(88\sqrt{21}t+233i\right)^{2}}\right)}\right) \\ &= 2\pi\arctan\left(\frac{\frac{1}{18}\left(233+88\sqrt{7}\right)}{\sqrt{\frac{108497}{108}+\frac{10252\sqrt{7}}{27}}}\right) \\ &= \frac{\pi^{2}}{3} \\ \end{align} $$ where in taking the limit, we used the fact that the vector represented by the expression inside $\operatorname{arg}$ gets closer to the real axis as $t \to \infty$.

Putting everything together, we conclude with



By symmetry,

$$\begin{align*} & \int_0^1\frac1{\sqrt{1-x^2}}\arctan\frac{88\sqrt{21}}{215+36x^2}\,dx \\ &= \frac12 \int_{-1}^1\frac1{\sqrt{1-x^2}}\left(\arctan\frac{11+6x}{4\sqrt{21}} + \arctan\frac{11-6x}{4\sqrt{21}}\right)\,dx \\ &= \int_{-1}^1\frac1{\sqrt{1-x^2}}\arctan\frac{11\pm6x}{4\sqrt{21}}\,dx = \boxed{\frac{\pi^2}6} \end{align*}$$

where the (negative case of the) last integral is evaluated here.


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