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Let \( \displaystyle \text{Ti}_2(x) = \int_0^x \frac{\arctan t}{t}\, {\rm d}t \) . Ramanujan proved that:

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

where \( \mathcal{G} \) is the Catalan's constant.

Can you provide a proof for the result?

We are invoking \(\displaystyle{\rm{Fourier}}\) series and integration by parts, so :

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

Hence :

$$\text{Ti}_2(2-\sqrt{3})=\dfrac{\pi}{12}\,\ln\,\left(2-\sqrt{3}\right)+\dfrac{2}{3}\,\mathcal{G}$$

Thank you Vaggelis. Ramanujan also proved that:
$$\begin{equation} \text{Ti}_2 (y)-\text{Ti}_2 \left(\frac{1}{y} \right)=\frac{\pi}{2}\log y, \quad y>0 \end{equation}$$

Setting \( 2-\sqrt{3} \) at \( (1) \) we get that:

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

Another special value of this function is known, namely \( \text{Ti}_2(1) \). It is equal to \( \mathcal{G} \) where \( \mathcal{G} \) is the Catalan constant.

Any other special values? Is anyone aware of any other values? I am , denifitely, not.