An identity involving integrals of the form $\int_{0}^{\alpha} \arctan (\sqrt{5}\tan\theta) \, d\theta$.

Question

I want to establish the following identity

$$ \int_0^{\pi/3} \arctan (\sqrt{5} \tan \theta) \, d\theta - 2 \int_0^{\pi/6} \arctan (\sqrt{5} \tan \theta) \, d\theta = \frac{\pi^{2}}{30}. \tag{1} $$

by a direct calculation, that is, not relying on some geometry.

Background. This integral is a by-product of the Schläfli's formula for some orthoscheme (special kind of tetrahedron) in the 3-sphere, according to the problem posed by Prof. H. S. M. Coxeter in AMM 95 (1988). He originally asked to find the value of

$$ \int_{1}^{6} \frac{\operatorname{arcsec} t}{(t+2)\sqrt{t+1}} \left[ \frac{1}{\sqrt{t+3}} + 2 \right] \, dt = \frac{2\pi^{2}}{15} $$

without appealing to geometry or the computer. Then another literature shows that it is also written as

$$ \int_{0}^{\frac{\pi}{3}} \arccos\left( \frac{\cos\theta}{1 + 2\cos\theta} \right) \, d\theta = \frac{2\pi^{2}}{15}. $$

(I guess they are related by the volume of the same geometric object, but I'm not sure since the literature is written in French which I can hardly read.) Applying some manipulations to the last integral, we obtain the four times of the left-hand side of $\text{(1)}$.

My trial. Numerical tests show that the constant $\sqrt{5}$ in $\text{(1)}$ is very special; it seems that no other square root of integer (other than 1) gives a rational multiple of $\pi^{2}$. I tried some basic techniques, all of which turned out to be futile.

So I want to post this problem so that everyone can share it.

Answer 1

This doesn't answer the question. But I believe someone can hopefully take it from here and proceed.

Let $$I(a) = \int_{\pi/6}^{\pi/3} \arctan(a\tan(t))dt - \int_0^{\pi/6} \arctan(a\tan(t))dt$$ Note that $$I(1) = \int_{\pi/6}^{\pi/3} tdt - \int_0^{\pi/6} tdt = \left.\dfrac{t^2}2 \right \vert_{\pi/6}^{\pi/3} - \left.\dfrac{t^2}2 \right \vert_{0}^{\pi/6} = \dfrac{\pi^2/9}2 - \dfrac{\pi^2}{36} = \dfrac{\pi^2}{36}$$ We then have $$I'(a) = \int_{\pi/6}^{\pi/3} \dfrac{\tan(t)}{a^2 \tan^2(t)+1}dt - \int_0^{\pi/6} \dfrac{\tan(t)}{a^2 \tan^2(t)+1}dt$$ $$2I'(a) = \int_{\pi/6}^{\pi/3} \dfrac{\sin(2t)}{a^2 \sin^2(t)+\cos^2(t)}dt - \int_{0}^{\pi/6} \dfrac{\sin(2t)}{a^2 \sin^2(t)+\cos^2(t)}dt$$ Setting $\sin^2(t) = x$, we get $\sin(2t)dt = dx$. Hence, $$2I'(a) = \int_{1/4}^{3/4} \dfrac{dx}{(a^2-1)x+1} - \int_{0}^{1/4} \dfrac{dx}{(a^2-1)x+1}$$ Hence, $$2I'(a) = \dfrac{\log(3a^2+1)}{a^2-1} - 2\dfrac{\log(a^2+3)}{a^2-1} + \dfrac{\log(4)}{a^2-1}$$ Integrating the right hand side gives the expressions in terms of $\log$ and polylogarithm function $\text{Li}_2$. Hopefully, this can be simplified further to get to the answer.

Answer 2

I managed to solve the other integral that Kirill posted. A similar approach might work on your integral though I am not sure.

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\begin{align*} \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \arctan\left(\sqrt{3}\tan\theta\right) \; d\theta -\int_{0}^{\frac{\pi}{6}} \arctan\left(\sqrt{3}\tan\theta\right) \; d\theta &= \frac{\pi^2}{12} \end{align*}

Proof

Let $J$ denote the expression on the left hand side of the equation. The substutution $\sqrt{3}\tan \theta = \tan \varphi$ transforms $J$ into \begin{align*} J &= \frac{\sqrt{3}}{2}\left( \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\frac{\varphi}{1+\frac{1}{2}\cos(2\varphi)}d\varphi -\int_{0}^{\frac{\pi}{4}}\frac{\varphi}{1+\frac{1}{2}\cos(2\varphi)}d\varphi \right) \end{align*} For $0\leq \phi \leq \frac{\pi}{2}$, consider the following generalized integral: \begin{align*} I(\phi) &= \frac{\sqrt{3}}{2}\int_{0}^{\phi}\frac{\varphi}{1+\frac{1}{2}\cos(2\varphi)}d\varphi \end{align*} Using the series identity, \begin{align*} 1+2\sum_{k=1}^\infty x^k \cos(k\theta) &= \frac{1-x^2}{1-2x\cos(\theta) + x^2} \quad |x|<1 \end{align*} we easily deduce that \begin{align*} \frac{1}{1+\frac{1}{2}\cos(2\varphi)} &= \frac{2}{\sqrt{3}}\left(1+2\sum_{k=1}^\infty (-1)^k \rho^k \cos(2k\varphi) \right) \end{align*} where $\rho = 2-\sqrt{3}$. Therefore, \begin{align*} I(\phi) &= \int_0^\phi \varphi \left( 1+2\sum_{k=1}^\infty (-1)^k \rho^k \cos(2k\varphi) \right)d\varphi \\ &= \frac{\phi^2}{2} + 2\sum_{k=1}^\infty (-1)^k \rho^k \int_0^\phi \varphi \cos(2k\varphi)\; d\varphi \\ &= \frac{\phi^2}{2} + 2\sum_{k=1}^\infty (-1)^k \rho^k \left[\frac{\phi \sin(2k\phi)}{2k}-\frac{1-\cos(2k\phi)}{4k^2} \right] \\ &= \frac{\phi^2}{2} -\phi \sum_{k=1}^\infty (-1)^{k+1}\frac{\rho^k\sin(2k\phi)}{k}+\frac{1}{2}\sum_{k=1}^\infty (-1)^{k+1} \frac{1-\cos(2k\phi)}{k^2}\rho^k \\ &= \frac{\phi^2}{2} -\phi \arctan\left(\frac{\rho \sin(2\phi)}{1+\rho\cos(2\phi)} \right) +\frac{1}{2}\sum_{k=1}^\infty (-1)^{k+1} \frac{1-\cos(2k\phi)}{k^2}\rho^k \tag{1} \end{align*} We used the following identity in the last step: \begin{align*} \sum_{k=1}^\infty (-1)^{k+1}\frac{x^k}{k}\sin(k\theta) &= \arctan\left(\frac{x \sin(\theta)}{1+x\cos(\theta)}\right) \quad |x| \leq 1 \end{align*} Using equation (1), we get \begin{align*} I\left(\frac{\pi}{2}\right) &= \frac{\pi^2}{8}+\chi_2\left(\rho\right) \tag{2}\\ I\left(\frac{\pi}{3}\right) &= \frac{\pi^2}{36}+\frac{1}{12}\text{Li}_2\left( -\rho^3\right) -\frac{3}{4}\text{Li}_2\left(-\rho \right) \tag{3}\\ I\left(\frac{\pi}{4}\right) &= \frac{\pi^2}{96}+\frac{1}{2}\chi_2\left(\rho\right)-\frac{1}{4}\chi_2\left(\rho^2\right) \tag{4} \end{align*} Now, we have \begin{align*} J &= I\left(\frac{\pi}{2}\right)-I\left(\frac{\pi}{3}\right)-I\left(\frac{\pi}{4}\right) \\ &= \frac{25\pi^2}{288} + \left[-\frac{1}{4}\text{Li}_2(\rho)+\frac{1}{2}\text{Li}_2(\rho^2)+\frac{1}{12}\text{Li}_2(\rho^3)-\frac{1}{16}\text{Li}_2(\rho^4)-\frac{1}{24}\text{Li}_2(\rho^6) \right]\tag{5} \end{align*} To complete the proof, we must prove that \begin{align*} -\frac{1}{4}\text{Li}_2(\rho)+\frac{1}{2}\text{Li}_2(\rho^2)+\frac{1}{12}\text{Li}_2(\rho^3)-\frac{1}{16}\text{Li}_2(\rho^4)-\frac{1}{24}\text{Li}_2(\rho^6) &= -\frac{\pi^2}{288} \tag{6} \end{align*} This can be done by using the following dilogarithm identites (refer Chapter 5 of Polylogarithms and Associated Functions by Leonard Lewin): \begin{align*} \text{Li}_2 \left(\tan a, \frac{\pi}{2}-2a \right) &= a^2 +\frac{3}{4}\text{Li}_2(\tan^2 a) -\frac{1}{8}\text{Li}_2(\tan^4 a) \\ \text{Li}_2\left(x,\frac{\pi}{3}\right) &= \frac{1}{6} \text{Li}_2(-x^3) -\frac{1}{2}\text{Li}_2(-x) \end{align*} where the notation $\text{Li}_2(x,\theta)$ is used for the real part of $\text{Li}_2(xe^{i\theta})$. Substituting $a=\frac{\pi}{12}$ and $x=\tan\left(\frac{\pi}{12}\right)=2-\sqrt{3}$ gives: \begin{align*} \frac{1}{6}\text{Li}_2(-\rho^3)-\frac{1}{2}\text{Li}_2(-\rho) &= \frac{\pi^2}{144} + \frac{3}{4}\text{Li}_2(\rho^2)-\frac{1}{8}\text{Li}_2(\rho^4) \\ \implies \frac{1}{12}\text{Li}_2(\rho^6)-\frac{1}{6}\text{Li}_2(\rho^3) -\frac{1}{4}\text{Li}_2(\rho^2) +\frac{1}{2} \text{Li}_2(\rho)&= \frac{\pi^2}{144} + \frac{3}{4}\text{Li}_2(\rho^2)-\frac{1}{8}\text{Li}_2(\rho^4) \tag{7} \end{align*} Now, dividing the above equation by 2 and rearranging some terms proves equation (6). $\blacksquare$

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Another interesting identity, though not related to the problem above, is obtained using Hill's two-variable relation for the dilogarithm: \begin{align*} \text{Li}_2(xy) &= \text{Li}_2(x) + \text{Li}_2(y) + \text{Li}_2\left(-x\left(\frac{1-y}{1-x}\right) \right)+\text{Li}_2\left(-y\left(\frac{1-x}{1-y}\right)\right)+\frac{1}{2}\log^2\left(\frac{1-x}{1-y}\right) \end{align*} Substitute $x=-y = e^{-i\frac{\pi}{4}}\frac{\sqrt{3}-1}{\sqrt{2}}$. Then $\frac{1-x}{1+x}=ix = e^{i\frac{\pi}{4}}\frac{\sqrt{3}-1}{\sqrt{2}}$ and $ix^2 =\rho=2-\sqrt{3}$. \begin{align*} \text{Li}_2(-x^2) &= \text{Li}_2(x) + \text{Li}_2(-x) + \text{Li}_2(i)+\text{Li}_2(ix^2)+\frac{1}{2}\log^2\left(ix\right) \\ \implies \text{Li}_2(i\rho)&= \frac{1}{2}\text{Li}_2(-i\rho)+ \text{Li}_2(i) + \text{Li}_2(\rho) +\frac{1}{2}\log^2\left(e^{i\frac{\pi}{4}}\frac{\sqrt{3}-1}{\sqrt{2}} \right) \end{align*} Extracting the real part from the above equation gives: \begin{align*} \frac{\text{Li}_2(\rho^4)}{16}-\frac{\text{Li}_2(\rho^2)}{8} -\text{Li}_2(\rho) &= -\frac{5\pi^2}{96}+\frac{1}{8}\log^2(\rho) \end{align*} Note that $\frac{5\pi^2}{96}$ is equal to the Ahmed Integral.

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Note

I feel that Coxeter's integrals might be related to dilogarithm identities known as "ladders". These identities are discussed in the book Structural Properties of Polylogarithms (Mathematical Surveys & Monographs) by Leonard Lewin.

You might find the paper, The Dilogarithm in Algebraic Fields (also by Lewin), to be interesting.

Answer 3

Here’s an analytic approach:

If we use the integral representation $\displaystyle \frac{\arctan(xy)}{x}=\int_0^y\frac{1}{1+x^2t^2}\, dt$, we can turn the integral into:

$$I=\int_0^{\sqrt 5}\frac{\ln\left(\frac{x^2+3}{2\sqrt{1+3x^2}}\right)}{1-x^2}\, dx$$

I turned the integral into such a form and sent it to one of my friends, he was able to answer the question. However, he doesn't have a MSE account, so I will answer here. His solution is the following: $$I=\int_0^{1}\frac{\ln\left(\frac{x^2+3}{2\sqrt{1+3x^2}}\right)}{1-x^2}\, dx-\int_{1/\sqrt{5}}^1\frac{ \ln\left(\frac{1+3x^2}{2x\sqrt{3+x^2}}\right)}{1-x^2}\, dx$$ $$=\underbrace{\int_0^1\frac{\ln\left(\frac{x(3+x^2)^{3/2}}{(1+3x^2)^{3/2}}\right)}{1-x^2}\, dx}_{J}+{\int_0^{1/\sqrt{5}}\frac{ \ln\left(\frac{1+3x^2}{2x\sqrt{3+x^2}}\right)}{1-x^2}\, dx}$$ $J$ is equal to $\displaystyle \int_0^1 \frac{\ln x}{1-x^2}\, dx+\frac{3}{2}\int_0^1 \frac{\ln\left(\frac{3+x^2}{1+3x^2}\right)}{1-x^2}$. The first integral is equal to $-\frac{\pi^2}{8}$, we focus on the latter integral: substitute $x\mapsto \frac{1-x}{1+x}$,

$$\int_0^1 \frac{\ln\left(\frac{3+x^2}{1+3x^2}\right)}{1-x^2}=\int_0^1\frac{1}{2x}\ln\left(\frac{1+x+x^2}{1-x+x^2}\right)\, dx$$ $$=\frac{1}{2}\int_{-1}^1\int_0^1 \frac{1}{1+tx+x^2}\, dx\, dt=\int_{-1/2}^{1/2}\frac{\cos^{-1} x}{2\sqrt{1-x^2}}\, dx=\frac{\pi^2}{12}$$ That is, $J=0$. $$I=\frac{1}{2}\int_0^{1/\sqrt{5}}\frac{1}{1-y^2}\ln\left(\frac{(1+3y^2)^2}{(2y\sqrt{3+y^2})^2}\right)\, dy=\frac{1}{2}\int_0^{1/\sqrt{5}}\int_0^{\frac{1+y^2}{2y^2}}\frac{dx}{(1+x)(1+y^2x)}\, dy$$ $$=\frac{1}{2}\int_3^{\infty}\frac{1}{1+x}\int_0^{\frac{1}{\sqrt{2x-1}}}\frac{dy}{1+xy^2}\, dx=\frac{1}{2}\int_3^{\infty}\frac{\arctan\left(\sqrt{\frac{x}{2x-1}}\right)}{\sqrt{x}(1+x)}\, dx$$ set $x\mapsto 1/x^2$, $$I=\int_0^{1/\sqrt{3}}\frac{1}{1+x^2}\arctan\left(\frac{1}{\sqrt{2-x^2}}\right)\, dx$$ set $x=\frac{v}{\sqrt{2+v^2}}$ $$I=\int_0^1 \frac{\arctan\sqrt\frac{2+v^2}{4+v^2}}{(1+v^2)\sqrt{2+v^2}}\, dv=\int_0^1\int_0^1 \frac{1}{(1+x^2)(1+y^2)\sqrt{3+x^2+y^2}}\, dx\, dy \tag{1}$$ $$=\frac{2}{\sqrt{\pi}}\int_0^{\infty}\int_0^1\int_0^1 \frac{e^{-z^2(3+x^2+y^2)}}{(1+x^2)(1+y^2)}\, dx\, dy\, dz$$ Where $$\int_0^1 \frac{e^{-x^2 z^2}}{1+x^2}\, dx=\int_0^{\infty}e^{-t}\int_0^1 e^{-x^2(z^2+t)}\, dx\, dt$$ $$=\int_0^{\infty}\frac{\sqrt \pi}{2}\frac{e^{-t}}{\sqrt{z^2+t}}\mathrm{erf}(\sqrt{z^2+t})\, dt=\frac{\pi}{4}e^{z^2}(1-\mathrm{erf}^2(z))$$ So $$I=\frac{2}{\sqrt{\pi}}\int_0^{\infty}e^{-3z^2}\frac{\pi^2}{16}e^{2z^2}(1-\mathrm{erf}^2(z))^2\, dz$$ $$=\frac{\pi^2}{16}\int_0^1 (1-t^2)^2\, dt=\frac{\pi^2}{30}$$

I believe that we can use an alternative approach when we arrive at $(1)$ but I haven’t been able to come up with yet.