Prove that:
$$\int_0^1 (\arctan x)^2 \, {\rm d}x = \frac{\pi^2}{16}+ \frac{\pi \ln 2}{4} - \mathcal{G}$$
where $\mathcal{G}$ denotes the Catalan constant.
Solution
Integrating by parts twice yields:
$$ \begin{aligned} \int_{0}^{1} (\arctan x)^{2} \, {\rm d}x &= x (\arctan x)^{2} \Big|^{1}_{0} - 2 \int_{0}^{1} \frac{x \arctan x}{1+x^{2}} \,{\rm d}x \\ &= \frac{\pi^{2}}{16} - 2 \int_{0}^{1} \frac{x \arctan x}{1+x^{2}} \,{\rm d}x \\ &= \frac{\pi^{2}}{16} - \arctan(x) \ln(1+x^{2}) \Big|^{1}_{0} + \int_{0}^{1} \frac{\ln (1+x^{2})}{1+x^{2}} \, {\rm d}x \\ &= \frac{\pi^{2}}{16} - \frac{\pi}{4} \ln 2 + \int_{0}^{1} \frac{\ln(1+x^{2})}{1+x^{2}} \, {\rm d}x \end{aligned}$$
Making the changeof variables $x=\tan t$ at the last integral we have that:
$$\begin{aligned}\int_{0}^{1} (\arctan x)^{2} \, {\rm d}x &=\frac{\pi^{2}}{16} - \frac{\pi}{4} \ln 2 - 2 \int_{0}^{\pi /4} \ln (\cos t) \, {\rm d}t \\ &= \frac{\pi^{2}}{16} - \frac{\pi}{4} \ln 2 -2 \int_{0}^{\pi /4} \left( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \cos (2nt) - \ln 2 \right) \,{\rm d}t \\ &= \frac{\pi^{2}}{16} - \frac{\pi}{4} \ln 2 - 2 \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\int_{0}^{\pi /4} \cos (2nt) \ dt + \frac{\pi}{2} \ln 2 \\ &= \frac{\pi^{2}}{16} + \frac{\pi}{4} \ln 2 - \sum_{n=1}^{\infty} \frac{\sin \left(\frac{\pi n}{2} \right)}{n^{2}} \\ &= \frac{\pi^{2}}{16} + \frac{\pi}{4} \ln 2 - \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2}} \\ &= \frac{\pi^{2}}{16} + \frac{\pi}{4} \ln 2 - \mathcal{G} \end{aligned}$$
$$\int_0^1 (\arctan x)^2 \, {\rm d}x = \frac{\pi^2}{16}+ \frac{\pi \ln 2}{4} - \mathcal{G}$$
where $\mathcal{G}$ denotes the Catalan constant.
Solution
Integrating by parts twice yields:
$$ \begin{aligned} \int_{0}^{1} (\arctan x)^{2} \, {\rm d}x &= x (\arctan x)^{2} \Big|^{1}_{0} - 2 \int_{0}^{1} \frac{x \arctan x}{1+x^{2}} \,{\rm d}x \\ &= \frac{\pi^{2}}{16} - 2 \int_{0}^{1} \frac{x \arctan x}{1+x^{2}} \,{\rm d}x \\ &= \frac{\pi^{2}}{16} - \arctan(x) \ln(1+x^{2}) \Big|^{1}_{0} + \int_{0}^{1} \frac{\ln (1+x^{2})}{1+x^{2}} \, {\rm d}x \\ &= \frac{\pi^{2}}{16} - \frac{\pi}{4} \ln 2 + \int_{0}^{1} \frac{\ln(1+x^{2})}{1+x^{2}} \, {\rm d}x \end{aligned}$$
Making the changeof variables $x=\tan t$ at the last integral we have that:
$$\begin{aligned}\int_{0}^{1} (\arctan x)^{2} \, {\rm d}x &=\frac{\pi^{2}}{16} - \frac{\pi}{4} \ln 2 - 2 \int_{0}^{\pi /4} \ln (\cos t) \, {\rm d}t \\ &= \frac{\pi^{2}}{16} - \frac{\pi}{4} \ln 2 -2 \int_{0}^{\pi /4} \left( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \cos (2nt) - \ln 2 \right) \,{\rm d}t \\ &= \frac{\pi^{2}}{16} - \frac{\pi}{4} \ln 2 - 2 \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\int_{0}^{\pi /4} \cos (2nt) \ dt + \frac{\pi}{2} \ln 2 \\ &= \frac{\pi^{2}}{16} + \frac{\pi}{4} \ln 2 - \sum_{n=1}^{\infty} \frac{\sin \left(\frac{\pi n}{2} \right)}{n^{2}} \\ &= \frac{\pi^{2}}{16} + \frac{\pi}{4} \ln 2 - \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2}} \\ &= \frac{\pi^{2}}{16} + \frac{\pi}{4} \ln 2 - \mathcal{G} \end{aligned}$$
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