Integral $\int_0^1\frac{\arctan^2x}{\sqrt{1-x^2}}\mathrm dx$

Question

Is it possible to evaluate this integral in a closed form? $$I=\int_0^1\frac{\arctan^2x}{\sqrt{1-x^2}}\mathrm dx$$ It also can be represented as $$I=\int_0^{\pi/4}\frac{\phi^2}{\cos \phi\,\sqrt{\cos 2\phi}}\mathrm d\phi$$

Answer 1

Okay, finally I was able to prove it.

Step 0. Observations. In view of the following identity

$$ \int_{0}^{\frac{\pi}{2}} \arctan (r \sin\theta) \, d\theta = 2 \chi_{2} \left( \frac{\sqrt{1+r^{2}} - 1}{r} \right), $$

Vladimir's result suggests that there may exists a general formula connecting

$$ I(r, s) = \int_{0}^{\frac{\pi}{2}} \arctan (r \sin\theta) \arctan (s \sin\theta) \, d\theta $$

and the Legendre chi function $\chi_{2}$. Indeed, inspired by Vladimir's result, I conjectured that

$$ I(r, s) = \pi \chi_{2} \left( \frac{\sqrt{1+r^{2}} - 1}{r} \cdot \frac{\sqrt{1+s^{2}} - 1}{s} \right). \tag{1} $$

I succeeded in proving this identity, so I post a solution here.

Step 1. Proof of the identity $\text{(1)}$. It is easy to check that the following identity holds:

$$ \arctan(ab) = \int_{1/b}^{\infty} \frac{a \, dx}{a^{2} + x^{2}}. $$

So it follows that

\begin{align*} I(r, s) &= \int_{1/r}^{\infty} \int_{1/s}^{\infty} \int_{0}^{\frac{\pi}{2}} \frac{\sin^{2}\theta}{(x^{2} + \sin^{2}\theta)(y^{2} + \sin^{2}\theta)} \, d\theta dy dx \\ &= \int_{1/r}^{\infty} \int_{1/s}^{\infty} \int_{0}^{\frac{\pi}{2}} \frac{1}{x^{2} - y^{2}} \left( \frac{x^{2}}{x^{2} + \sin^{2}\theta} - \frac{y^{2}}{y^{2} + \sin^{2}\theta} \right) \, d\theta dy dx \\ &= \frac{\pi}{2} \int_{1/r}^{\infty} \int_{1/s}^{\infty} \frac{1}{x^{2} - y^{2}} \left( \frac{x}{\sqrt{x^{2} + 1}} - \frac{y}{\sqrt{y^{2} + 1}} \right) \, dy dx. \end{align*}

For the convenience of notation, we put

$$ \alpha = \frac{\sqrt{r^{2} + 1} - 1}{r} \quad \text{and} \quad \beta = \frac{\sqrt{s^{2} + 1} - 1}{s}. $$

Then it is easy to check that $\mathrm{arsinh}(1/r) = - \log \alpha$ and likewise for $s$ and $\beta$. Thus with the substitution $x \mapsto \sinh x$ and $y \mapsto \sinh y$, we have

\begin{align*} I(r, s) &= \frac{\pi}{2} \int_{-\log\alpha}^{\infty} \int_{-\log\beta}^{\infty} \frac{\sinh x \cosh y - \sinh y \cosh x}{\sinh^{2}x - \sinh^{2}y} \, dy dx. \end{align*}

Applying the substitution $e^{-x} \mapsto x$ and $e^{-y} \mapsto y$, it follows that

\begin{align*} I(r, s) &= \pi \int_{0}^{\alpha} \int_{0}^{\beta} \frac{dydx}{1 - x^{2}y^{2}} \\ &= \pi \sum_{n=0}^{\infty} \left( \int_{0}^{\alpha} x^{2n} \, dx \right) \left( \int_{0}^{\beta} y^{2n} \, dx \right) = \pi \sum_{0}^{\infty} \frac{(\alpha \beta)^{2n+1}}{(2n+1)^{2}} \\ &= \pi \chi_{2}(\alpha \beta) \end{align*}

as desired, proving the identity $\text{(1)}$.

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EDIT. I found a much simpler and intuitive proof of $\text{(1)}$. We first observe that $\text{(1)}$ is equivalent to the following identity

$$ \int_{0}^{\frac{\pi}{2}} \arctan\left( \frac{2r\sin\theta}{1-r^{2}} \right) \arctan\left( \frac{2s\sin\theta}{1-s^{2}} \right) \, d\theta = \pi \chi_{2}(rs). $$

Now we first observe that from the addition formula for the hyperbolic tangent, we obtain the following formula

$$ \operatorname{artanh}x - \operatorname{artanh} y = \operatorname{artanh} \left( \frac{x - y}{1 - xy} \right) $$

which holds for sufficiently small $x, y$. Thus

\begin{align*} \arctan\left( \frac{2r\sin\theta}{1-r^{2}} \right) &= \frac{1}{i} \operatorname{artanh}\left( \frac{2ir\sin\theta}{1-r^{2}} \right) = \frac{\operatorname{artanh}(re^{i\theta}) - \operatorname{artanh}(re^{-i\theta})}{i} \\ &= 2 \Im \operatorname{artanh}(re^{i\theta}) = 2 \sum_{n=0}^{\infty} \frac{\sin(2n+1)\theta}{2n+1} r^{2n+1}. \end{align*}

We readily check this holds for any $|r| < 1$. Therefore

\begin{align*} &\int_{0}^{\frac{\pi}{2}} \arctan\left( \frac{2r\sin\theta}{1-r^{2}} \right) \arctan\left( \frac{2s\sin\theta}{1-s^{2}} \right) \, d\theta \\ &\quad = 4 \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{r^{2m+1}s^{2n+1}}{(2m+1)(2n+1)} \int_{0}^{\frac{\pi}{2}} \sin(2m+1)\theta \sin(2n+1)\theta \, d\theta\\ &\quad = 2 \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{r^{2m+1}s^{2n+1}}{(2m+1)(2n+1)} \int_{0}^{\frac{\pi}{2}} \{ \cos(2m-2n)\theta - \cos(2m+2n+2)\theta \} \, d\theta\\ &\quad = \pi \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{r^{2m+1}s^{2n+1}}{(2m+1)(2n+1)} \delta_{m,n} \\ &\quad = \pi \chi_{2}(rs). \end{align*}

Answer 2

This is a beautiful integral with a beautiful result: $$\int_0^1\frac{\arctan^2x}{\sqrt{1-x^2}}dx=\frac\pi2\Big(\operatorname{Li}_2\left(3-\sqrt8\right)-\operatorname{Li}_2\left(\sqrt8-3\right)\Big).$$ One could expect it to have an elegant solution too, giving an insight into why the result is of this form.

Alas, I was not able to find such a solution. Mine was a long, semi-manual with a lot of assistance from Mathematica, lookups into Gradshteyn-Ryzhik Table of Integrals and huge intermediate expressions not fitting on a single screen. I would rather not post it here now. Instead, I hope somebody comes up with an elegant one.

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Update: An alternative and perhaps simpler form of the result is: $$\frac{7\pi^3}{48}-\frac\pi8\,\ln^22-\frac\pi2\,\ln(1+\sqrt2)\cdot\ln2-\pi \operatorname{Li}_2\left(\tfrac1{\sqrt{2}}\right)$$

Answer 3

$$ I(r, s) = \int_{0}^{\frac{\pi}{2}} \int_{0}^{r} \int_{0}^{s} \frac{4\sin(u)(v^2+1)}{[v^4+(4\sin(u)^2-2)v^2+1]} \frac{4\sin(u)(y^2+1)}{[y^4+(4\sin(u)^2-2)y^2+1]} , du , dv , dy $$$$ = \int_{0}^{\frac{\pi}{2}} \int_{0}^{r} \int_{0}^{s} \frac{4\sin(u)^2(v^2+1)(y^2+1)}{[v^4+(4\sin(u)^2-2)v^2+1][y^4+(4\sin(u)^2-2)y^2+1]} , du , dv , dy $$$$ = \int_{0}^{\frac{\pi}{2}} 4(v^2+1)(y^2+1) \int_{0}^{\infty} \frac{\sin(u)^2}{[v^4+1+2v^2(2\sin(u)^2-1)][y^4+1+2y^2(2\sin(u)^2-1)]} , du , dv , dy $$$$ = \int_{0}^{\infty} 4(v^2+1)(y^2+1)(y^4-2y^2+1) \frac{1}{[4v^2y^4-(4v^4+4)y^2+4v^2]} \int_{0}^{\infty} \frac{dz}{[(y^4+2y^2+1)z^2+y^4-2y^2+1]} , dv , dy $$$$ K = \int_{0}^{\infty} 4(v^2+1)(y^2+1)\frac{(y^4-2y^2+1)}{[4v^2y^4-(4v^4+4)y^2+4v^2]} \frac{dz}{[(y^4+2y^2+1)z^2+y^4-2y^2+1]} $$$$ = \int_{0}^{\infty} (v^2+1)(y^2+1) \frac{(y^2-1)^2}{[v^2y^4-(v^4+1)y^2+v^2]} \frac{dz}{[(y^2+1)^2z^2+(y^2-1)^2]} $$$$ = \left( \lim_{z \to \infty} (v^2+1)(y^2+1)\frac{(y^2-1)}{[v^2y^4-(v^4+1)y^2+v^2]} \frac{1}{(y^2+1)(y^2-1)}\arctan\left[\frac{(y^2+1)z}{(y^2-1)}\right] - \lim_{z \to 0} (v^2+1)(y^2+1)\frac{(y^2-1)}{[v^2y^4-(v^4+1)y^2+v^2]} \frac{1}{(y^2+1)(y^2-1)}\arctan\left[\frac{(y^2+1)z}{(y^2-1)}\right] \right) $$$$ = \left( \lim_{z \to \infty} \frac{(v^2+1)(y^2-1)}{[v^2y^4-(v^4+1)y^2+v^2]}\arctan\left[\frac{(y^2+1)z}{(y^2-1)}\right] - \lim_{z \to 0} \frac{(v^2+1)(y^2-1)}{[v^2y^4-(v^4+1)y^2+v^2]}\arctan\left[\frac{(y^2+1)z}{(y^2-1)}\right] \right) $$$$ = \frac{\pi}{2}\frac{(v^2+1)(y^2-1)}{[v^2y^4-(v^4+1)y^2+v^2]} $$$$ L = \int_{0}^{\infty} (v^2+1)(y^2+1)(v^4-2v^2+1)\frac{[4v^2y^4-(4v^4+4)y^2+4v^2]}{[(v^4+2v^2+1)z^2+v^4-2v^2+1]} , dz $$$$ = \int_{0}^{\infty} (v^2+1)(y^2+1)(v^4-2v^2+1)\frac{[v^2y^4-(v^4+1)y^2+v^2]}{[(v^4+2v^2+1)z^2+v^4-2v^2+1]} , dz $$$$ = \int_{0}^{\infty} (v^2+1)(y^2+1)(v^2-1)^2\frac{[v^2y^4-(v^4+1)y^2+v^2]}{[(v^2+1)^2z^2+(v^2-1)^2]} , dz $$$$ = \left( \lim_{z \to \infty} (v^2+1)(y^2+1)(v^2-1)^2\frac{1}{[(v^2+1)(v^2-1)]}\arctan\left[\frac{(v^2+1)z}{(v^2-1)}\right] - \lim_{z \to 0} (y^2+1)(v^2-1)^2\frac{1}{[(v^2+1)(v^2-1)]}\arctan\left[\frac{(v^2+1)z}{(v^2-1)}\right] \right) $$$$ = \left( \lim_{z \to \infty} \frac{(y^2+1)(v^2-1)}{[v^2y^4-(v^4+1)y^2+v^2]}\arctan\left[\frac{(v^2+1)z}{(v^2-1)}\right] - \lim_{z \to 0} \frac{(y^2+1)(v^2-1)}{[v^2y^4-(v^4+1)y^2+v^2]}\arctan\left[\frac{(v^2+1)z}{(v^2-1)}\right] \right) $$$$ = \frac{\pi}{2}\frac{(y^2+1)(v^2-1)}{[v^2y^4-(v^4+1)y^2+v^2]} $$$$ \text{So, } I(r, s) = \int_{0}^{r} \int_{0}^{s} \frac{\pi}{2} \frac{(v^2+1)(y^2-1)}{[v^2y^4-(v^4+1)y^2+v^2]} , dv , dy - \int_{0}^{r} \int_{0}^{s} \frac{\pi}{2} \frac{(y^2+1)(v^2-1)}{[v^2y^4-(v^4+1)y^2+v^2]} , dv , dy $$

$$ I(r, s) = \int_{0}^{r} \int_{0}^{s} \frac{\pi}{2} \int \left[ \frac{(v^2+1)(y^2-1)-(v^2+1)(y^2-1)}{v^2y^4-(v^4+1)y^2+v^2} \right] , dv , dy $$$$ = \int_{0}^{r} \int_{0}^{s} \frac{\pi}{2} \int \left[ \frac{(v^2+1)(y^2-1)-(v^2+1)(y^2-1)}{v^2y^2(y^2-v^2)-(y^2-v^2)} \right] , dv , dy $$$$ = \int_{0}^{r} \int_{0}^{s} \frac{\pi}{2} \int \left[ \frac{2(y^2-v^2)}{v^2y^2-1} \right] , dv , dy $$$$ = \int_{0}^{r} \int_{0}^{s} \frac{\pi}{2} \int \left[ \frac{2}{v^2y^2-1} \right] , dv , dy $$$$ = \int_{0}^{r} \int_{0}^{s} \frac{\pi}{2} \int \left[ \frac{2}{(vy+1)(vy-1)} \right] , dv , dy $$$$ = \int_{0}^{r} \int_{0}^{s} \frac{\pi}{2} \int \left[ \frac{1}{vy-1} \right] , dv , dy - \int_{0}^{r} \int_{0}^{s} \frac{\pi}{2} \int \left[ \frac{1}{vy+1} \right] , dv , dy $$$$ = \int_{0}^{r} \left[ -\frac{\pi}{2} \int \left[ \frac{\ln(1-sv)}{v} - \frac{\ln(1)}{v} \right] , dv \right] , dy - \int_{0}^{r} \left[ \frac{\pi}{2} \int \left[ \frac{\ln(1+sv)}{v} - \frac{\ln(1)}{v} \right] , dv \right] , dy $$$$ = \int_{0}^{r} \left[ -\frac{\pi}{2} \left( \frac{\ln(1-sv)}{v} - \frac{\ln(1)}{v} \right) \right] , dy - \int_{0}^{r} \left[ \frac{\pi}{2} \left( \frac{\ln(1+sv)}{v} - \frac{\ln(1)}{v} \right) \right] , dy $$$$ = \int_{0}^{r} \left[ -\frac{\pi}{2} \left( \frac{\ln(1-sv)}{v} - \frac{\ln(1)}{v} \right) \right] , dy - \int_{0}^{r} \left[ \frac{\pi}{2} \left( \frac{\ln(1+sv)}{v} - \frac{\ln(1)}{v} \right) \right] , dy $$$$ = -\frac{\pi}{2} \left( \operatorname{Li}_2(rs) - 0 \right) - \frac{\pi}{2} \left( \operatorname{Li}_2(-rs) - 0 \right) $$$$ = -\frac{\pi}{2} \left( -\operatorname{Li}_2(rs) \right) - \frac{\pi}{2} \left( -\operatorname{Li}_2(-rs) \right) $$$$ = \frac{\pi}{2} \left( \operatorname{Li}_2(rs) + \operatorname{Li}_2(-rs) \right) $$$$ = \frac{\pi}{2} \cdot \frac{1}{2} \operatorname{Li}_2 \left( (rs)^2 \right) $$$$ = \frac{\pi}{4} \operatorname{Li}_2 \left( r^2 s^2 \right) $$