What is up with the antiderivative of $\arctan^2(x)$?

Question

Integrating $\arctan^2(x)$ has been a mystery to me since I first learned to compute indefinite integrals. When I plug it into integral calculator or WolframAlpha, I get a primitive written in terms of complex numbers and polylogarythms; but $\arctan^2(x)$ is continuous and defined $\forall\space x\in\mathbb{R}$, the area under the curve $(x,\arctan^2(x))$ is well-defined and obvioulsy real. In fact, I can integrate that function numerically and get the area without using Barrow's Rule at all.

I thought maybe the complex part of that expression is constant, therefore applying Barrow's Rule always yields a real number anyway, but then I could subtract a constant equal to its imaginary part times $i$ and get a real and still valid antiderivative. Anyway, I have no idea of how to compute that antiderivative in the first place, so I can't check for myself.

Can a real, continuous function have a complex primitive but not a real primitive? And, can anyone give me a hint to help me compute its antiderivative by myself?

Answer 1

You are totally correct.

Using the result, compute it for $x=0$ and you will get $\frac{i \pi ^2}{12}$; do it for any arbitrary value of $x$ and you will get the decimal representation of this number.

What you could also do is to perform a Taylor expansion of the result around $0$ and get $$\frac{i \pi ^2}{12}+\frac{x^3}{3}-\frac{2 x^5}{15}+\frac{23 x^7}{315}-\frac{44 x^9}{945}+O\left(x^{11}\right)$$

Compute its derivative to get $$x^2-\frac{2 x^4}{3}+\frac{23 x^6}{45}-\frac{44 x^8}{105}+O\left(x^{10}\right)$$ while $$\tan^{-1}(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}+O\left(x^{10}\right)$$ from which $$\left[\tan ^{-1}(x)\right]^2=x^2-\frac{2 x^4}{3}+\frac{23 x^6}{45}-\frac{44 x^8}{105}+O\left(x^{10}\right)$$

Answer 2

WA assumes complex functions by default and provides answers more general than you expect.

But when plugging a real variable, the given antiderivative should simplify to a real function (plus a possibly complex constant).

Answer 3

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You can see an integration of $\ds{\arctan^{2}\pars{x}}$ over $\ds{\pars{0,1}}$ in this link which I guess it's simpler than the present one.

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\begin{align} \int\arctan^{2}\pars{x}\,\dd x & = x\arctan^{2}\pars{x} - \int x\bracks{2\arctan\pars{x}\,{1 \over x^{2} + 1}} \dd x \\[5mm] & = x\arctan^{2}\pars{x} - \ln\pars{x^{2} + 1}\arctan\pars{x} + \int{\ln\pars{x^{2} + 1} \over x^{2} + 1}\,\dd x \end{align}

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\begin{align} \int{\ln\pars{x^{2} + 1} \over x^{2} + 1}\,\dd x & = 2\,\Re\int\ln\pars{x + \ic}\pars{{1 \over x - \ic} - {1 \over x + \ic}}{1 \over 2\ic}\,\dd x \\[5mm] & = \Im\ \underbrace{\int{\ln\pars{x + \ic} \over x - \ic}\,\dd x} _{\ds{\mbox{Set}\,\,\, t = x + \ic}}\ -\ \Im\ \underbrace{\int{\ln\pars{x + \ic} \over x + \ic}\,\dd x} _{\ds{=\ {1 \over 2}\,\ln^{2}\pars{x + \ic}}} \\ & = -\,\Im\int{\ln\pars{t} \over 2\ic - t}\,\dd t - {1 \over 2}\,\Im\ln^{2}\pars{x + \ic} \\[5mm] & = -\,\Im\ \underbrace{\int{\ln\pars{2\ic\braces{t/\bracks{2\ic}}} \over 1 - t/\pars{2\ic}}\,{\dd t \over 2\ic}} _{\ds{\mbox{Set}\,\,\, z = {t \over 2\ic}}}\ -\ {1 \over 2}\,\Im\ln^{2}\pars{x + \ic} \\[5mm] & = -\,\Im\int{\ln\pars{2\ic z} \over 1 - z}\,\dd z - {1 \over 2}\,\Im\ln^{2}\pars{x + \ic} \\[5mm] & = \Im\braces{\ln\pars{1 - z}\ln\pars{2\ic z} - \int{\ln\pars{1 - z} \over z}} - {1 \over 2}\,\Im\ln^{2}\pars{x + \ic} \\[5mm] & = \Im\bracks{\ln\pars{1 - z}\ln\pars{2\ic z} + \,\mrm{Li}_{2}\pars{z}} - {1 \over 2}\,\Im\ln^{2}\pars{x + \ic} \\[5mm] & = \Im\bracks{\ln\pars{1 - {t \over 2\ic}}\ln\pars{t} + \,\mrm{Li}_{2}\pars{t \over 2\ic}} - {1 \over 2}\,\Im\ln^{2}\pars{x + \ic} \\[5mm] & = \Im\bracks{\ln\pars{{\ic \over 2}\bracks{x - \ic}}\ln\pars{x + \ic} + \,\mrm{Li}_{2}\pars{1 - x\ic \over 2}} - {1 \over 2}\,\Im\ln^{2}\pars{x + \ic} \\[5mm] & = \Im\bracks{\bracks{{\pi \over 2}\,\ic - \ln\pars{2}}\ln\pars{x + \ic} + \,\mrm{Li}_{2}\pars{1 - x\ic \over 2}} - {1 \over 2}\,\Im\ln^{2}\pars{x + \ic} \\[5mm] & = {\pi \over 2}\,\Re\ln\pars{x + \ic} - \ln\pars{2}\Im\ln\pars{x + \ic} + \Im\mrm{Li}_{2}\pars{1 - x\ic \over 2} - {1 \over 2}\,\Im\ln^{2}\pars{x + \ic} \\[5mm] & = {\pi \over 4}\ln\pars{x^{2} + 1} + \ln\pars{2}\arctan\pars{x} + \Im\mrm{Li}_{2}\pars{1 - x\ic \over 2} - {1 \over 2}\,\Im\ln^{2}\pars{x + \ic} \end{align}