Integral ∫10arctan2x√1−x2dx\int_0^1\frac{\arctan^2x}{\sqrt{1-x^2}}\mathrm dx

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

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