The inverse tangent integral at two minus root three

The inverse tangent integral \(\operatorname{Ti}_2(z) = \int_0^z \frac{\arctan x}{x}\,dx\) equals Catalan's constant \(G = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^2}\) at \(z = 1 = \tan\left(\frac{\pi}{4}\right)\). At the half-angle argument \(z = 2-\sqrt{3} = \tan\left(\frac{\pi}{12}\right)\), it satisfies a remarkable closed-form relation discovered by Ramanujan:

\[ \int_0^{2-\sqrt{3}}\frac{\arctan(x)}{x}\,dx = \frac{\pi}{12}\ln\left(2-\sqrt{3}\right) + \frac{2}{3}G. \]

Integration by parts and the Fourier series of \(\ln(\tan\theta)\)

Let \(I\) denote the integral. Integrating by parts with \(u = \arctan x\) and \(dv = \frac{dx}{x}\), and then substituting \(x = \tan\theta\) (using \(\tan\frac{\pi}{12} = 2-\sqrt{3}\)), we obtain

\[ \begin{aligned} I &= \Bigl[ \ln(x)\arctan(x) \Bigr]_0^{2-\sqrt{3}} - \int_0^{2-\sqrt{3}}\frac{\ln (x)}{1+x^2}\,dx \\ &= \frac{\pi}{12}\ln\left(2-\sqrt{3}\right) - \int_0^{\frac{\pi}{12}} \ln(\tan\theta)\,d\theta. \end{aligned} \tag{1} \]

For \(\theta \in \left(0, \frac{\pi}{2}\right)\), recall the standard Fourier cosine expansions

\[ \begin{aligned} \ln(2\sin \theta) &= -\sum_{k=1}^\infty \frac{\cos(2k \theta)}{k}, \\ \ln(2\cos \theta) &= \sum_{k=1}^\infty (-1)^{k+1}\frac{\cos(2k\theta)}{k}. \end{aligned} \]

Subtracting the second series from the first cancels all even-indexed terms and doubles the odd-indexed terms (\(k = 2m+1\)):

\[ \ln(\tan\theta) = \ln(2\sin\theta) - \ln(2\cos\theta) = -2\sum_{k=0}^\infty \frac{\cos\bigl(2(2k+1)\theta\bigr)}{2k+1}. \]

Term-by-term integration and multisection of the series

Substituting the Fourier series of \(\ln(\tan\theta)\) into equation (1) and integrating term by term gives

\[ \begin{aligned} I &= \frac{\pi}{12}\ln\left(2-\sqrt{3}\right) + 2\int_0^{\frac{\pi}{12}}\left(\sum_{k=0}^\infty \frac{\cos\bigl(2(2k+1)\theta\bigr)}{2k+1} \right) d\theta \\ &= \frac{\pi}{12}\ln\left(2-\sqrt{3}\right) + 2\sum_{k=0}^\infty\frac{1}{2k+1}\int_0^{\frac{\pi}{12}} \cos\bigl(2(2k+1)\theta\bigr)\,d\theta \\ &= \frac{\pi}{12}\ln\left(2-\sqrt{3}\right) + \sum_{k=0}^\infty\frac{\sin\left( (2k+1)\frac{\pi}{6}\right)}{(2k+1)^2}. \end{aligned} \]

To evaluate the remaining Dirichlet-type series, examine the periodic numerator \(s_k = \sin\left((2k+1)\frac{\pi}{6}\right)\) as \(k\) varies modulo \(3\):

When \(k \equiv 0\) or \(2 \pmod 3\) (so \(2k+1\) is not divisible by \(3\)), \(s_k = \frac{1}{2}(-1)^k\).
When \(k = 3m+1\) (so \(2k+1 = 6m+3 = 3(2m+1)\) is an odd multiple of \(3\)), \(s_{3m+1} = \sin\left((2m+1)\frac{\pi}{2}\right) = (-1)^m = -(-1)^k\).

On the full alternating sum \(\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^2} = G\), separating out the subsequence \(k = 3m+1\) where \((-1)^{3m+1} = -(-1)^m\) gives

\[ \sum_{k \not\equiv 1 \pmod 3} \frac{(-1)^k}{(2k+1)^2} = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^2} + \sum_{m=0}^\infty \frac{(-1)^m}{(6m+3)^2} = G + \frac{G}{9}. \]

Multiplying those terms by \(\frac{1}{2}\) and adding the \(k = 3m+1\) terms (which have coefficient \((-1)^m\)) yields

\[ \begin{aligned} I &= \frac{\pi}{12}\ln\left(2-\sqrt{3}\right) + \frac{1}{2}\left( \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^2} + \sum_{m=0}^\infty \frac{(-1)^m}{(6m+3)^2}\right) + \sum_{m=0}^\infty \frac{(-1)^m}{(6m+3)^2} \\ &= \frac{\pi}{12}\ln\left(2-\sqrt{3}\right) + \frac{1}{2}\left(G + \frac{G}{9}\right) + \frac{G}{9} \\ &= \frac{\pi}{12}\ln\left(2-\sqrt{3}\right) + \frac{2}{3}G. \end{aligned} \]