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  "body": "\n## Definition\n\nThe *inverse tangent integral* is the function $\\operatorname{Ti}_2 \\colon \\R \\to \\R$ given by\n$$\\operatorname{Ti}_2(x) = \\int_0^x \\frac{\\arctan t}{t} \\, dt.$$\nHere $\\arctan$ is the [[arctangent|0CE768XNVP]], and the integrand has the value $1$ at $t = 0$.\n\n## Remarks\n\nFor $\\abs{x} \\le 1$, integrating the series of $\\arctan t / t$ term by term gives\n$$\\operatorname{Ti}_2(x) = \\sum_{k=0}^{\\infty} \\frac{(-1)^k x^{2k+1}}{(2k+1)^2}.$$\nThe function is odd, and $\\operatorname{Ti}_2(x) = \\operatorname{Im} \\operatorname{Li}_2(ix)$ for real $x$, where $\\operatorname{Li}_2$ is the [[dilogarithm|G0X0NWSDB7]].\n\nExample: $\\operatorname{Ti}_2(1) = 1 - \\frac19 + \\frac1{25} - \\cdots = G$, Catalan's constant.\n\nInversion check: for $x \u003e 0$, 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\n$$\\operatorname{Ti}_2(x) - \\operatorname{Ti}_2\\left(\\tfrac1x\\right) = \\frac{\\pi}{2} \\log x.$$\n\nNon-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.\n",
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