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

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 $$I=\int_0^{\pi/4}\frac{\phi^2}{\cos \phi\,\sqrt{\cos 2\phi}}\mathrm d\phi$$

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 $\text{(1)}$.

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*}

• Very nice result! Nov 11, 2013 at 22:05
• How did you realize it would be incredibly convenient to define the parameters $\alpha$ and $\beta$ in such a manner? I would never have thought of that. Nov 20, 2013 at 21:21
• @RandomVariable As I revealed at the beginning of my answer, the choices of $\alpha$ and $\beta$ came out of the observation from a 1-variable version of this formula as an analogy. When I first derived this, however, I was not sure if this analogy does hold indeed. Nov 20, 2013 at 22:33
• I was assuming that your evaluation of the 1-variable version was similar. A while ago I evaluated $I(1)$ by differentiating inside of $I(r)$. But I didn't actually evaluate $I(r)$ because at the time it didn't seem possible. mymathforum.com/… Nov 20, 2013 at 23:25
• +1. Great answer which represents a lot of work. Thanks. Jun 30, 2014 at 20:05

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.

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

• Note also that the integral is equivalent to the polylogarithmic ladder $$\int_0^1\frac{\arctan^2x}{\sqrt{1-x^2}}dx=\frac{\pi}{2}\left(2\operatorname{Li}_2(\beta^2)-\tfrac{1}{2}\operatorname{Li}_2(\beta^4)\right)$$ where $\beta=1-\sqrt{2}$. Oct 16, 2016 at 15:58