An integral with arctan

Evaluate the following integral:
$$\int_0^{2-\sqrt{3}}\frac{\arctan x}{x}\, {\rm d}x$$

The result is due to Ramanujan.

Solution

We are invoking Fourier Series and integration by parts so we successively have:

$$\begin{aligned} \int_0^{2-\sqrt{3}}\frac{\arctan x}{x}\,{\rm d}x &= \frac{\pi}{12}\log \left( 2-\sqrt{3}\right)- \int_0^{2-\sqrt{3}}\frac{\log x}{1+x^2}\,{\rm d}x \\ &= \frac{\pi}{12}\log \left( 2-\sqrt{3}\right)- \int_0^{\pi/12}\log \left(\tan \theta \right) \,{\rm d}\theta  \\ &= \frac{\pi}{12}\log \left( 2-\sqrt{3}\right) +2 \int_0^{\pi/12} \sum_{n \text{ odd}}^\infty \frac{\cos(2 n\theta)}{n} \ d\theta \\ &= \frac{\pi}{12}\log \left( 2-\sqrt{3}\right) + \sum_{n \text{ odd}}^\infty \frac{\sin \left(\frac{\pi n}{6} \right)}{n^2} \\ &= \frac{\pi}{12}\log \left(2-\sqrt{3} \right) + \frac{1}{2} \left( \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} +\sum_{n=0}^\infty \frac{(-1)^n}{(6n+3)^2}\right)+\frac{G}{9} \\ &= \frac{\pi}{12}\log \left(2-\sqrt{3} \right) +\frac{2}{3}G \end{aligned}$$

Hence:

$$ \text{Ti}_2 \left( 2-\sqrt{3}\right) = \frac{\pi}{12}\log \left(2-\sqrt{3} \right) +\frac{2}{3}G \tag{1}$$

where $ \displaystyle \text{Ti}_2 =\int_0^x \frac{\arctan t}{t}\, {\rm d}t$.

Differentiating we get that:

$$\text{Ti}_2 (y)-\text{Ti}_2 \left(\frac{1}{y} \right)=\frac{\pi}{2}\log y \quad y>0 \tag{2}$$

Plugging $x=2-\sqrt{3}$ and using $(1)$ in $(2)$ we get that:

$$ \text{Ti}_2 \left( 2+\sqrt{3}\right) = \frac{5\pi }{12} \log \left( 2+\sqrt{3}\right)+\frac{2}{3}G$$

where $G$ denotes the Catalan's constant.