Inverse tangent integral

Remarks

For $\abs{x} \le 1$, integrating the series of $\arctan t / t$ term by term gives $$\operatorname{Ti}_2(x) = \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k+1}}{(2k+1)^2}.$$ The function is odd, and $\operatorname{Ti}_2(x) = \operatorname{Im} \operatorname{Li}_2(ix)$ for real $x$, where $\operatorname{Li}_2$ is the dilogarithm.

Example: $\operatorname{Ti}_2(1) = 1 - \frac19 + \frac1{25} - \cdots = G$, Catalan's constant.

Inversion check: for $x > 0$, the derivative of $\operatorname{Ti}_2(x) - \operatorname{Ti}_2(1/x)$ is $\frac{\arctan x + \arctan(1/x)}{x} = \frac{\pi}{2x}$, and the difference is $0$ at $x = 1$, so $$\operatorname{Ti}_2(x) - \operatorname{Ti}_2\left(\tfrac1x\right) = \frac{\pi}{2} \log x.$$

Non-example: $\int_0^x \frac{\arctan t}{t^2} \, dt$ is not a variant of this function; it diverges at $t = 0$, because $\arctan t / t^2 \sim 1/t$ there.